ap physics 1 homework forces in circular motion

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AP Physics 1: Circular Motion and Gravitation Practice Problems +Answers

Over 20 multiple-choice questions on circular motion and gravitation which appear on the AP Physics 1 exam are provided with detailed explanations.  

AP Physics: Circular Motion Solved Problems

Problem (1): A motorcycle weighing $200\,{\rm kg}$ turning around an unbanked circular track of radius $12.5\,{\rm m}$ at a constant speed of $56\,{\rm km/h}$. What is the magnitude of the net force applying on the motorcycle? 

(a) $3.844\times 10^{3}\,\rm N$        (b) $5.0176\times 10^{4}\,\rm N$ (c) $248\rm N$                                  (d) $3.844\times 10^{4}\,\rm N$

Solution : Because the motorcycle moves around a circular path at a constant speed $v$, so it undergoes a centripetal force whose magnitude is given by the following formula \[F_c=m\frac{v^2}{r}\] Plugging the numerical values into above, we get \[F_c=200\times \frac{(15.5)^2}{12.5}=3844\,{\rm N}\] Keep in mind that usually in the AP physics 1 problems on circular motions, the constant speed is given in terms of $\rm \frac{km}{h}$. To use the centripetal acceleration or force formula above, we must convert it into the SI units of speed which is $\rm \frac ms$ by multiplying it by $\frac{10}{36}$. Thus, the given speed in SI units is \[56\times \frac{10}{36}=15.5\,{\rm m/s}\] Thus, the correct answer is (a). 

Problem (2): A $355-\rm g$ ball is fastened at the end of a $1.1-\rm m$ long rope and is whirled in a vertical circle at a constant speed of $3.75\,\rm m/s$. The tensions in the rope when the ball is at the top and the bottom of its path are, respectively, closest to 

(a) $2\, ,\, 7$                (b) $8\, ,\, 1$ (c) $1\, ,\, 8$                (d) $1\, ,\, 9$

Solution : At the top and the bottom of the path the force are exerting on the ball is shown in the free-body diagram below. In both positions, the weight force is always down but at the top the tension is down and at the bottom is up. On the other hand, we know that the centripetal force is always directed toward the center of circle at each position. Applying Newton's second law, we get \begin{gather*} F_c=T+mg \quad \text{top} \\\\ \frac{mv^2}{r}=T+mg \\\\ \Rightarrow T=\frac{0.355\times (3.75)^2}{1.1}-(0.355)(9.8)\\\\ \Rightarrow \boxed{T=1.06\,\rm N} \\\\ F_c=T-mg \quad \text{Bottom}\\\\ \frac{mv^2}{r}=T-mg \\\\ \Rightarrow T=\frac{0.355\times (3.75)^2}{1.1}+(0.355)(9.8) \\\\ \Rightarrow \boxed{T=8.02\,\rm N}\end{gather*} Hence, the correct answer is (c). 

Problem (3): A small bead of mass $20\,\rm g$ is attached to the end of light string $120\,\rm cm$-long and is deflected at an angle of $18\circ$. Once the bead is released, what is the speed of the bead in $\rm m/s$? 

(a) $1.3$                (b) $3.2$ (c) $0.55$              (d) $3.5$

Solution : When the bead is held at your hand at an arbitrary angle, say $18\deg$, the tension in the string is $T\cos\theta=mg$, as you can see in the free-body diagram below. After releasing the bead, under the string's tension, moves around a semi-circle. As a consequence, it experiences a centripetal force whose magnitude is \[T=F_c=\frac{mv^2}{\ell}\] where $\ell$ is the string's length. Substitute from the first equation the tension as $T=\frac{mg}{\cos\theta}$, into the second one, and solve for the speed $v$ \begin{align*} \frac{mg}{\cos\theta}&=\frac{mv^2}{\ell}\\\\ \Rightarrow v&=\sqrt{\frac{g\ell}{\cos\theta}}\\\\&=\sqrt{\frac{1.2\times 9.8}{\cos 18^\circ}}\\\\&=3.5\quad\rm m/s\end{align*} Thus, the correct answer is (d). 

Problem (4): In a horizontal circle of radius $85\,{\rm cm}$, a constant force of $245\,\rm N$ is applied to a $2.5\,\rm kg$ object as it revolves around the track. The speed of the object (in $\rm m/s$) is closest to

(a) $91.3$            (b) $9.13$ (c) $83.3$             (d) $8330$

Solution : All information required to use the centripetal force formula $F_c=\frac{mv^2}{r}$ is given but the object's speed $v$. Thus, rearranging this for $v$ and substituting the known values into it, we get \begin{align*} v&=\sqrt{\frac{rF_c}{m}}\\\\ &= \sqrt{\frac{0.85\times 245}{2.5}}\\\\&=\boxed{9.13\quad \rm m/s}\end{align*} Thus, the correct choice is (b).

Problem (5): At the end of a very light string $2-\rm m$-long, a $500-\rm g$ toy is attached and whirled around a circle on a frictionless surface. Assume the maximum tension that the string can withstand is $350\,\rm N$ without breaking. In this situation, what is the maximum speed the toy can have in $\rm m/s$? 

(a) $26$            (b) $37$ (c) $19$            (d) $42$

Solution : The toy whirled around a circular frictionless path so that it experiences an inward centripetal force which is provided by the tension in the string. The magnitude of this centripetal force which equals to the tension is obtained by $T=F_c=\frac{mv^2}{r}$. Substituting the numerical values into above formula, and solving for $v$, we get \begin{align*} v^2&=\frac{rT}{m}\\\\&=\frac{2\times 350}{0.5}\\\\&=\sqrt{1400}\\\\ \Rightarrow v&=37.5\quad\rm m/s\end{align*} which is closest to (b). 

Problem (6): A race road is designed with a curve of radius $100\,\rm m$ and banked at an angle of $45^\circ$. What maximum speed a race car can have on this road without friction in $\rm m/s$?

(a) $25$            (b) $36$ (c) $31$            (c) $69$ 

Solution : 

Problem (7): A $1000-\rm kg$ motorcycle is moving at a constant speed of $30\,\rm m/s$ around a horizontal unbanked curve whose radius is $100\,\rm m$. To keep the car from sliding, how much friction force is required in $\rm N$?

(a) $9000$            (b) $8000$ (c) $7500$            (d) $6000$

Solution : The motorcycle wants to turn around a circular path, so a centripetal force toward the center of the circle is applied to it whose magnitude is obtained by $F_c=\frac{mv^2}{r}$. This force is provided by the friction force. Without friction, the motorcycle slides and deviates from the main track which is circular. Substituting the given numerical values into above, we get \begin{align*} F_c&=\frac{mv^2}{r}\\\\ &=\frac{1000\times 30^2}{100}\\\\&=9000\quad \rm N \end{align*} Thus, the correct answer is (a). 

Problem (8): We want to create artificial gravity inside a large rotating hollow cylinder of radius $190\,\rm m$ by rotating it about its longitudinal axis. If we want the magnitude of this artificial gravity to be the same as the normal gravity, the time taken for each rotation of such a system is closest to?

(a) $32\,\rm s$            (b) $28\,\rm s$ (c) $30\,\rm s$            (d) $19\,\rm s$

Solution : Any object in a rotating system experiences a centripetal acceleration whose magnitude is found by $a_c=\frac{v^2}{r}$, where $v$ and $r$ are the tangential constant speed and radius of the circular system (path), respectively. 

In this case, we want the magnitude of the artificial gravity to be the same as the Earth's gravitational acceleration that is $g=9.8\,\rm m/s^2$. Substituting this for $a_c$ and solving for $v$, we get the speed required to complete one full circle \[g=\frac{v^2}{r} \Rightarrow v=\sqrt{gr}\]

On the other hand, we know that each full rotation around a circle of radius $r$ corresponds a traveled distance of circumference of that circle, i.e., $\ell=2\pi r$. Hence, the time required to complete a full revolution, $T$, is obtained using the definition of average speed $v=\frac{\ell}{t}$ where $t$ is the time taken to cover the distance $\ell$. \[T=\frac{\ell}{v}=\frac{2\pi r}{v}\] Substituting the constant velocity obtained above into this time taken formula, we get \begin{align*} T&=\frac{2\pi r}{\sqrt{rg}}\\\\ &=2\pi\sqrt{\frac{r}{g}}\\\\ &=2\pi\sqrt{\frac{190}{9.8}}\\\\ &=27.7\quad \rm s\end{align*} In the second line, to eliminate the radical from denominator, we rationalize the fraction by multiplying the numerator and denominator by $\sqrt{r}$. Thus, the correct choice is (b), since it is closest to the obtained value. 

Problem (9): A car moving along a curved track, as shown in the figure below. At the lowest point of the curve, what is the direction of the normal force exerted by the road on the car?

(a) Upward            (b) Downward (c) Forward            (d) Backward

Solution : When an object is in contact with a surface, a force is applied to it by the surface that is perpendicular. This force is known as normal force. At the lowest point of the track, the normal force is toward the center of the circular track. Thus, it is directed upward. Hence, the correct answer is (a). 

Problem (10): A car is moving on a road with many hills and valleys. Assume the radius of these roughnesses to be $R$. At what point on the track, the driver most likely feel weightlessness?

(a) At the top of the hill when $v=\sqrt{Rg}$. (b) At the bottom of the valley when $v>\sqrt{Rg}$ (c) At the top of the hill when $v<\sqrt{Rg}$ (d) At the bottom of the valley when $v=\sqrt{Rg}$

Solution : In physics, weightlessness is when the object loses contact with the surface. In this situation, the normal force becomes zero. At the hills, the normal force is up, and the weight force is downward. Taking up as positive and applying Newton's second law, the net force, which is equal to the centripetal force, at the top of hills is written as \begin{gather*} mg-F_N=\frac{mv^2}{R} \\\\\Rightarrow F_N=mg-\frac{mv^2}{R}\end{gather*} Similarly, at the bottom of the valley, the net force is \begin{gather*} F_N-mg=\frac{mv^2}{R} \\\\\Rightarrow F_N=mg+\frac{mv^2}{R}\end{gather*} Keep in mind that the centripetal force is always toward the center of the circle, for this reason at the hills is down and at the valleys is up. 

As you can see, the normal force at the valleys can never be removed (it is always the sum of two nonzero forces) but at the hills it is possible. So, setting $F_N=0$ at the hills, we get \[v=\sqrt{Rg}\] If the car's speed at the top of the hill is this amount, then the driver will feel weightlessness. Hence, the correct answer is (a). 

Problem (11): A car is traveling along a curved path of radius $r$ at a constant speed $v$ as shown below. What is the apparent weight of the driver at the bottom of the curved path? 

(a) $mg-mv^2/r$            (b) $mg$ (c) $mg+mv^2/r$            (d) none of these 

Solution : the car moves along a circular path, so it undergoes a centripetal force toward the center of the circle. As shown in the free-body diagram below, two external forces are applied to the driver at this position; a downward weight force, and an upward normal force. Take up as positive and apply Newton's second law at this point \begin{gather*} F_N-mg=\frac{mv^2}{r} \\\\ \Rightarrow F_N=mg+\frac{mv^2}{r}\end{gather*} In all AP Physics 1 circular motion problems the apparent weight means the same as the normal force. Thus, the correct answer (c).

Problem (12): A bead is placed on a turntable that is rotating clockwise as in the figure below. Assume after a while, this turntable starts to decrease its spinning speed due to friction in its axis until it ceased. Which arrow best depicts the net force acting on the bead at the instant shown? 

(a) $\downarrow$                (b) $\uparrow$ (c) $\searrow$                (d) $\nearrow$ (e) $\rightarrow$

Solution : The bead is rotating, so a centripetal force toward the center of the circle applies to it which at point $P$ is rightward. On the other side, the turntable is slowing down which means an opposing force (say, friction in the axis) must be applied to the bead. At point $P$, the bead's velocity is upward, so the opposing force must be downward to create a slowing down motion. 

Consequently, the resultant of these two force vectors, which is obtained by the graphical method, is toward $\searrow$. Hence, the correct answer is (c). 

AP Physics: Gravitation Solved Problems 

Problem (13): A satellite weighing $M$ takes a time of $T$ to complete a revolution around a planet. Now assume that the satellite had three times as much mass. What changes would occur for the time of circling around the planet at the same altitude?

(a) $2T$            (b) $4T$ (c) $T$                (d) $\frac{T}{2}$

Solution : One of the most common problems on circular motion and gravitation in the AP Physics 1 exam is about whirling a satellite around a planet. In such AP physics questions, the inward centripetal force that the satellite experiences is provided by the gravity force between the satellite and the planet. \[\frac{GmM}{r^2}=\frac{mv^2}{r}\] where $m$ and $M$ refer to the masses of satellite and planet, respectively.

Usually, asked for the satellite's speed or period in the orbit. In this question, we consider the case of time orbiting around the planet. 

The satellite travels a perimeter of a circle of radius $r$ at a constant speed $v$ in the orbit. This trip takes $T$ which is found using the definition of average speed as below \[T=\frac{2\pi r}{v}\] Substituting this into the previous equation and rearranging it, we get \[T^2=\frac{4\pi^2}{GM}r^3\] This is known as Kepler's third law. As you can see, the mass of the satellite does not appear in the time period formula. Hence, by changing the satellite's mass, its period around the planet does not change. So, the correct answer is (c).

Problem (14): An artificial satellite revolving around the Earth completes each orbit in $120$ minutes. At what altitude is this satellite located in $\rm km$?

(a) $8033$                (b) $9000$ (c) $6033$                (d) $10000$

Solution : The orbital period $T$ of a satellite (or a planet) at altitude $r$ circling around an object of mass $M$ is found using Kepler's third law as below formula \[r^3=\frac{GM}{4\pi^2}T^2\] where with a good approximation we can set, in the AP Physics tests, $\frac{GM}{4\pi^2}=10^{13}$. In this question, the orbital period is given in minutes, which must be converted in seconds by multiplying it by $60$. \[T=120\times 60=7200\,\rm s\] Plugging this into the previous formula and  taking the cube root of both sides, we get \begin{align*} r^3 &=\frac{GM}{4\pi^2}T^2\\\\&=10^{13}\times (7200)^2\\\\&=5.184\times 10^{20} \\\\ & \text{Take the cube root of both sides} \\\\ r&=\left(5.184\times 10^{20}\right)^{\frac 13} \\\\ &=8.033\times 10^6\quad\rm m\end{align*} Note that this result is in meters. By diving it by $1000$ it is converted to $\rm km$. Thus, the answer is $8033\,\rm km$ which is (a). 

Problem (15): A satellite with a circular speed of $v$ revolves around a planet of mass $M$. Now assume if the planet had half as much as mass, how would the orbital speed of the satellite change at the same altitude (distance)?

(a) $\frac{v}{\sqrt{2}}$            (b) $\frac{v}{2}$ (c) $\sqrt{2}v$                                 (d) $2v$

Solution : The satellite moves around the planet in a circular orbit, so it experiences an inward centripetal force which is provided by the attraction gravity force between the Earth and the satellite. Without this gravity force, the satellite cannot whirl around the orbit. Thus, we have \begin{gather*} \frac{GmM_E}{r^2}=\frac{mv^2}{r}\\\\ \Rightarrow \quad v=\sqrt{\frac{GM}{r}} \end{gather*} where $m$ is the satellite's mass. As you can see, the satellite's speed in orbit is proportional to the square root of the planet's mass, i.e., $v\propto \sqrt{M}$. So, the ratio the of satellite's speed at two different cases mentioned above is \begin{gather*} \frac{v'}{v}=\sqrt{\frac{M'}{M}}=\sqrt{\frac{\frac 12 M}{M}}\\\\ \Rightarrow \quad v'=\frac{v}{\sqrt{2}} \end{gather*} where prime denotes the case when the planet's mass is half. Hence, if the planet had half as much as mass, then its speed in the orbit could be a factor of as much as $\frac{v}{\sqrt{2}}$. Therefore, the correct answer is (a).

Problem (16): Assume two planets, say $A$ and $B$, orbiting an unknown star. The planet $A$ is $10$ times farther from the star than is planet $B$. The ratio of their speed $\frac{v_A}{v_B}$ is closest to?

(a) $0.4$                (b) $0.2$ (c) $0.3$                (d) $0.5$

Solution : In the previous question, we realized that the speed of an object of mass $m$ orbiting at a distance of $r$ around a mass $M$ is given by \[v=\sqrt{\frac{GM}{r}}\] This formula shows us that the speed in orbit inversely proportional to the square root of distance from the mass $M$. Thus, their ratio is found to be as below \begin{align*} \frac{v_A}{v_B}&=\sqrt{\frac{r_B}{r_A}}\\\\&=\sqrt{\frac{r_B}{10\,r_B}}\\\\&=\frac{1}{\sqrt{10}}\\\\ &=\boxed{0.31} \end{align*} Hence, the correct answer is (c).

Problem (17): Two bodies attract each other with a force of $F$ gravitationally. Now suppose the masses doubled and the separation between them gets tripled. After these changes, what will the force be?

(a) $4$            (b) $\frac{1}{36}$ (c) $1$            (d) $2$

Solution : According to Newton's universal gravitation law, $F=\frac{m_1 m_2}{r^2}$, the gravitation force between two masses of $m_1$ and $m_2$ is proportional to their product and inversely proportional to the square of distance between them. Thus, after changes have taken place, the force $F'$ is a factor of as much as the original force $F$ which is found as \begin{align*} \frac{F'}{F}&=\frac{m'_1 m'_2}{m_1 m_2}\times \frac{r^2}{r'^2} \\\\ &=\frac{(\frac 12 m_1)(\frac 12 m_2)}{m_1 m_2}\times \frac{r^2}{(3r)^2}\\\\&=\frac{1}{36}\end{align*} Thus, \[F'=\frac{1}{36}F\] the correct answer is (b).

Problem (18): Consider two small spheres of masses $8\,\rm kg$ and $12\,\rm kg$ that are placed $2\,\rm m$ far away from each other. What is the magnitude of gravitational force between them in $\rm N$? (Take $G=\frac{20}{3}\times 10^{-11}\,\rm \frac{N.m^2}{kg^2}$).

(a) $1.6\times 10^{-8}$            (b) $3.2\times 10^{-8}$ (c) $1.6\times 10^{-9}$            (d) $3.2\times 10^{-9}$

Solution : Two objects of masses $m_1$ and $m_2$ at a center-to-center distance of $r$, attract each other with a gravitational force of magnitude \[F=G\frac{m_1 m_2}{r^2}\] where $\rm G$ is the gravitational constant. Substituting the given values into this formula, we get \begin{align*} F&=G \frac{m_1 m_2}{r^2}\\\\&=\left(\frac{20}{3}\times 10^{-11}\right)\times \frac{8\times 12}{2^2}\\\\&=\boxed{1.6\times 10^{-9}\quad \rm N} \end{align*} Hence, the correct answer is (c). 

Problem (19): Two small balls from $20\,\rm m$ from each other, attract one another with a gravitational force of $10^{-8}\rm N$. If one of the balls is $60\,\rm kg$, then what is the mass of the other ball? (Take $G=\frac{20}{3}\times 10^{-11}\,\rm \frac{N.m^2}{kg^2}$)

(a) $10^2$                (b) $10^3$ (c) $5$                        (d) $50$

Solution : Apply Newton's law of universal gravitation formula and solve for the unknown mass \begin{gather*} F=G \frac{m_1 m_2}{r^2}\\\\ 10^{-8}=\left(\frac{20}{3}\times 10^{-11}\right)\times \frac{60\times m_2}{20^2} \\\\ \Rightarrow \boxed{m_2=1000\quad\rm kg} \end{gather*} Hence, the correct answer is (b).

Problem (20): The masses of Earth and the Moon are about $5.98\times 10^{24}\rm kg$ and $7.36\times 10^{22}\,\rm kg$, respectively. The separation between them is also $3.84\times 10^8\,\rm m$. The order of magnitude of the gravitational force between those is closest to?  

(a) $10^{22}$                (b) $10^{25}$ (c) $10^{28}$                (d) $10^{31}$

Solution : In this problem, the order of magnitude of the force is asked, so it is not necessary to get the full answer. First, determine the order of magnitude of each of the known values, then use the universal gravitation law.

Problem (21): Two small objects of masses $m$ and $2m$ at a distance of $r$ applying a gravitational force of $F$ on each other. How many $F$ times is the magnitude of the gravitational force between two objects of masses $3m$ and $4m$ at a distance of $2r$?

(a) $\frac 13$        (b) $3$ (c) $\frac 23$        (d) $\frac 32$

Solution : According to the universal gravitation law, $F=G\frac{m_1 m_2}{r^2}$, the attraction force between two objects is proportional to the product of their masses and inversely proportional to the square of distance between them. If we denote the varied masses and distance by prime, then we have \begin{align*} \frac{F'}{F}&=\frac{m'_1}{m_1} \times \frac{m'_2}{m_2} \times \left(\frac{r}{r'}\right)^2 \\\\&=\frac{3m}{m}\times \frac{4m}{2m}\times \left(\frac{r}{2r}\right)^2 \\\\&=3\times 2\times \frac{1}{4}\\\\&=\boxed{\frac 32} \end{align*} Hence, the correct answer is (d).

Problem (22): The plot of the gravitational force between two small objects in terms of distance between them is depicted below. What is $r$ in this plot?

(a) $8$                    (b) $12$ (c) $16$                (d) $24$

Solution : From the plot, we can find the force between those two masses at an arbitrary distance. 

Problem (23): Two similar particles of masses $m$ at a distance of $r$ exert a gravitational force of magnitude $F$ on each other. What percent of the mass of one should be removed and added to the other so that at the previous distance, the magnitude of the gravitational force between them reduces $25\%$?

(a) $75$            (b) $25$ (c) $50$            (d) $33$

Solution : Assume $x$ as much is removed from one and is added to the other. After this manipulation, the masses becomes $m'_1=m-x$ and $m'_2=m+x$. On the other hand, the force between them $F'$, as a result, gets reduced $25\%$ as much as the original force $F$. This means \[F'=F-0.25F=0.75F\] Now, form their ratio and apply universal gravitation law to find the unknown $x$ \begin{gather*} \frac{F'}{F}=\frac{m'_1}{m_1}\times \frac{m'_2}{m_2} \\\\ \frac 34=\frac{m-x}{m}\times \frac{m+x}{m}\\\\ \frac{3}{4}=\frac{m^2-x^2}{m^2}\\\\ 3m^2=4m^2-4x^2 \\\\ \Rightarrow 4x^2=m^2 \\\\ \Rightarrow x=\frac{m}{2} \end{gather*} This computation shows us that we must subtract by $50\%$ from one and add to the other. Thus, the correct answer is (c). 

Problem (24): Imagine a point between the Moon and Earth where if an object is placed there, the net force applied on it from the Moon and Earth be zero. How many times is the distance of that point from the Earth's center relative to the Moon's center? (Take Earth's mass to be $81$ times the Moon's mass).  

(a) $9$            (b) $10$ (c) $80$          (d) $81$

Solution : Assume the point lies at a distance of $r_E$ from the Earth and $r_M$ from the Moon. We want at this point the gravitational force on an imaginary mass $m$ is zero. So, the gravitational forces $F_E$ from the Earth and $F_M$ from the Moon on $m$ must be balanced.

\begin{gather*} F_E=F_M \\\\ \frac{GmM_E}{r_E^2}=\frac{GmM_M}{r_M^2} \\\\ \Rightarrow \frac{M_E}{r_E^2}=\frac{M_M}{r_M^2}\\\\ \rm{set \ M_E=81M_M} \\\\ \Rightarrow \frac{81M_M}{r_E^2}=\frac{M_M}{r_M^2} \\\\ \Rightarrow r_E^2=81\times r_M^2 \\\\ \Rightarrow \boxed{r_E=9r_M}\end{gather*} Hence, the correct answer is (a). 

Problem (25): An spacecraft of mass $3\times 10^4\,\rm kg$ is in between the distance of Earth and the Moon. What is the net gravitational force applying on this spacecraft at this location? (Take the Earth's mass, $6\times 10^{24}\,\rm kg$, the Moon's mass $7.5\times 10^{22}\,\rm kg$, the distance between Earth and the Moon $4\times 10^5\,\rm km$, and $G=\frac{20}{3}\times 10^{-11}\,\rm \frac{N.m^2}{kg^2}$)

(a) $303.75$            (b) $296.25$ (c) $607.5$                (d) $592.5$

Solution : The spacecraft is positioned halfway between the Earth and the Moon. Thus, its distance from both is \[d=\frac{4\times 10^5}{2}=2\times 10^5\rm km\] Now, using the universal gravitational law find the force exerted by the Earth and the Moon on the spacecraft.

\begin{align*} F_E&=G\frac{M_E m}{d^2}\\\\ &=\left(\frac{20}{3}\times 10^{-11}\right)\times \frac{(3\times 10^4)(6\times 10^{24})}{4\times 10^{16}}\\\\&=3\times 10^2 =300\,{\rm N} \\\\ F_M&=G\frac{M_M m}{d^2}\\\\ &=\left(\frac{20}{3}\times 10^{-11}\right)\times \frac{(3\times 10^4)(7.5\times 10^{22})}{4\times 10^{16}}\\\\&=3.75\,{\rm N} \end{align*} These two forces on the spacecraft are in the same line but in opposite direction, so the net force applying on the spacecraft is their subtraction. \begin{align*} F_{net}&=F_E-F_M \\\\ &=300-3.75\\\\&=\boxed{296.25\quad {\rm N}}\end{align*} Thus, the correct answer is (b). 

Problem (26): The ESA wants to design a satellite to orbit around Earth $5$ times a day. What would be the radius of its orbit in $\rm m$, neglecting the presence of the Moon? (Assume $G=6.67\times 10^{-11}\,\rm \frac{N.m^2}{kg^2}$, and $M_{earth}=5.97\times 10^{24}\,\rm kg$)

(a) $3.45\times 10^7$            (b) $1.44\times 10^7$ (c) $2.11\times 10^7$            (d) $0.69\times 10^7$

Solution : '' $5$ times a day'' means $5$ full revolution every $24$ hours. Thus, one full revolution takes $\frac{24}{5}$ hours or $\frac{24}{5}\times 3600$ in seconds. This is the time period of this designed satellite. Now, apply Kepler's third law of motion to find the radius of orbit \begin{align*} r^3 &=\frac{GM}{4\pi^2}T^2\\\\&=10^{13}\times \left(\frac{24}{5}\times 3600\right)^2 \\\\ & \text{Take the cube root of both sides} \\\\ r&=1.44\times 10^7\quad \rm m \end{align*} Keep in mind that in all the AP physics gravitation questions, we can set $\rm \frac{GM}{4\pi^2}=10^{13}$ with a good approximation. Thus, the correct answer is (b).

Note: the formula of cube root is \[a=\sqrt[3]{b} \ \text{or} \ a= b^{\frac 13}\]

Author : Dr. Ali Nemati Published : 1/1/2022   

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AP Physics 1 Practice Test 11: Circular Motion and Gravitation

• AP Physics 1 Practice Tests

1. A pendulum bob is attached to a string that is tied to the ceiling, and the bob is pulled back and released from different heights. As the bob moves through the bottom of the swing, how is its centripetal acceleration related to its speed?

Questions 2-3 refer to the following information.

The diagram above shows a top view of a pinball moving on a horizontal plane after it is launched from a spring mechanism.

2. While the ball encounters the curved wall at position A, which of the following forces act on the ball?

I. A force along the line pointed from A toward C

II. A force along the line pointed from A toward D

III. A force pointed up and to the left along a line tangent to the curve

3. The curved wall abruptly ends at point B. Which of the following best represents the future path of the ball after it leaves the wall at point B?

4. Two satellites orbit the Earth at the same speed in identical orbits. Satellite A is twice the mass of satellite B. How does the centripetal acceleration of satellite A compare with that of satellite B?

5. A car moves around a curve in a horizontal circle with a radius of 100 m. The tangential speed of the car is constant at a value of 30 m/s. What is the car's acceleration?

6. If the car in the previous question has a mass of 1,000 kg, then what is the force of friction acting on the car?

7. A satellite orbits the Earth at a distance of 100 km from the surface of the Earth. The mass of the satellite is 100 kg, while the mass of the Earth is approximately 6.0 × 10 24 kg. The radius of the Earth is approximately 6.4 × 10 6 m. What is the approximate force of gravity acting on the satellite?

8. A 70-kg astronaut floats at a distance of 10 m from the center of a 50,000-kg spacecraft. What is the force of attraction between the astronaut and spacecraft?

9. The centripetal acceleration on a 1,000-kg car in a turn is 1 × 10 5 m/s 2 . The radius of the turn is 40 m. What is the car's speed?

An empty 150-kg roller coaster cart is approaching a 6.0-meter-tall circular-shaped loop-the-loop as shown in the figure above. In order to complete the loop, determine the minimum speed of the cart when it is upside-down at the top of the loop.

11. A slingshot is spun in a horizontal circle at a constant speed. The sling is 1.5 m long, and the stone has a mass of 50 g. Determine the stone's speed when the tension in the string is 3.3 N.

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AP Physics 1: Uniform Circular Motion, Newton's Law of Gravitation, and Rotational Motion Multiple-Choice Practice Questions

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1. A 60 cm rope is tied to the handle of a bucket which is then whirled in a vertical circle. The mass of the bucket is 3 kg.

At the lowest point in its path, the tension in the rope is 50 N. What is the speed of the bucket?

2. A 60 cm rope is tied to the handle of a bucket which is then whirled in a vertical circle. The mass of the bucket is 3 kg.

What is the critical speed below which the rope would become slack when the bucket reaches the highest point in the circle?

3. An object moves at a constant speed in a circular path of radius r at a rate of 1 revolution per second. What is its acceleration?

4. In an effort to tighten a bolt, a force F is applied as shown in the figure above. If the distance from the end of the wrench to the center of the bolt is 20 cm and F = 20 N, what is the magnitude of the torque produced by F ?

5. In the figure above, what is the torque about the pendulum's suspension point produced by the weight of the bob, given that the length of the pendulum, L , is 80 cm and m = 0.50 kg ?

6. A uniform meter stick of mass 1 kg is hanging from a thread attached at the stick's midpoint. One block of mass m = 3 kg hangs from the left end of the stick, and another block, of unknown mass M , hangs below the 80 cm mark on the meter stick. If the stick remains at rest in the horizontal position shown above, what is M ?

7. If the distance between two point particles is doubled, then the gravitational force between them

8. At the surface of Earth, an object of mass m has weight w . If this object is transported to an altitude that is twice the radius of Earth, then at the new location,

9. A moon of mass m orbits a planet of mass 100 m . Let the strength of the gravitational force exerted by the planet on the moon be denoted by F 1 , and let the strength of the gravitational force exerted by the moon on the planet be F 2 . Which of the following is true?

10. The dwarf planet Pluto has 1/500 the mass and 1/15 the radius of Earth. What is the value of g (in m/s 2 ) on the surface of Pluto?

11. A satellite is currently orbiting Earth in a circular orbit of radius R ; its kinetic energy is K 1 . If the satellite is moved and enters a new circular orbit of radius 2 R , what will be its kinetic energy?

12. A moon of Jupiter has a nearly circular orbit of radius R and an orbit period of T . Which of the following expressions gives the mass of Jupiter?

13. Two large bodies, Body A of mass m and Body B of mass 4 m , are separated by a distance R. At what distance from Body A, along the line joining the bodies, would the gravitational force on an object be equal to zero? (Ignore the presence of any other bodies.)

14. You are looking at a top view of a planet orbiting the Sun in a clockwise direction. Which of the following would describe the velocity, acceleration, and force acting on the planet due to the Sun's pull at point P ?

15. An object of mass m is traveling at constant speed v in a circular path of radius r . How much work is done by the centripetal force during one-half of a revolution?

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AP Physics 1 : Circular, Rotational, and Harmonic Motion

Study concepts, example questions & explanations for ap physics 1, all ap physics 1 resources, example questions, example question #1 : harmonic motion.

This problem deals with the conservation of energy in the form of a spring. The formula for the conservation of energy is:

We can eliminate initial kinetic energy to get:

We have two forms of potential energy in this problem: gravitational and spring. Therefore, we can rewrite the expression to be:

There is no initial gravitational potential energy and no final spring potential energy. Subsituting our expressions for potential and kinetic energy, we get:

Rearranging for final velocity, we get:

We have all of the values, so we can solve:

Example Question #2 : Harmonic Motion

Before the elevator begins to accelerate, the block has a weight of:

While the elevator is accelerating, the block has a weight of:

Now we can say that the weight of the block has effectively increased by 2N.

We can use this increase in force to find out how much the equilibrium position will shift:

Rearranging for x (displacement), we get:

We know that the maximum acceleration of a spring occurs when it is at its maximum displacement. At this point, we can write an expression for the force that the block is under:

Rearranging for the distance, we get:

This is also the point at which the spring is storing its maximum potential energy:

We can substitute our rearranged distance equation into the equation for potential energy:

We also know that the maximum velocity of a spring occurs when it is at the point of equilibrium. Since the spring is on a frictionless surface, we can say that its kinetic energy at this point is equal to the maximum potential energy:

Plugging in the expressions for these variables:

Rearranging for velocity:

We know all of the values, so we can plug and chug:

Example Question #4 : Harmonic Motion

From the problem statement, we can calculate how much potential energy is initially stored in the spring. Since we are neglecting air resistance, we can say that all of this energy is lost through friction as the system comes to a stop.

Calculating initial potential energy:

Then we can write an expression for the force of friction on the mass:

We can then write an expression of the work done by friction:

Setting this equal to the initial potential energy:

Therefore, the mass travels a total distance of 20 meters before coming to rest. We know that the distance between maximum compression and maximum extension of the spring is 4 meters. Therefore, we can say that the mass travels from one maximum displacement to the other five times, passing through the equilibrium once each time. The mass passes through the equilibrium point 5 times.

Example Question #5 : Harmonic Motion

We can calculate the initial potential energy of the spring from what we are given in the problem statement:

Assuming the only energy loss of the spring is through internal friction, we can write:

Example Question #6 : Harmonic Motion

This problem can be solved by the conservation of energy. The block initially has kinetic energy equal to:

When the velocity of the block is zero, all the kinetic energy has been transferred to potential spring energy.

Example Question #7 : Harmonic Motion

A homogenous mass of 0.25kg is fixed to a 0.5kg Hookean spring. When the mass/spring system is stretched 1cm from the equilibrium, it takes 3N of force to hold the mass in place. If the displacement from equilibrium is doubled, the force necessary to keep the system in place will  __________ .

increase by a factor of 2

increase by a factor of 4

remain the same

Since the spring is Hookean, the relationship between the force and displacement from equilibrium of the mass can be expressed by Hooke's Law:

Example Question #8 : Harmonic Motion

To find the answer to this problem, we'll need to consider the case before the object collides with the spring, and also after it collides. Before the collision, all of the object's energy is in the form of kinetic energy.

Upon collision, the spring will be compressed by the moving object. The maximum displacement, however, will occur the instant the object stops moving. At this point, the object will no longer have any kinetic energy because it has stopped moving. Furthermore, all of the energy has now been transferred into the spring, and is held as spring potential energy.

Also, since we are told in the question stem that no friction is involved, we have a situation in which mechanical energy is conserved. In other words, we can relate the initial and final energies as equal to one another.

Set these two equations equal to each other, and solve for the maximum displacement.

Example Question #9 : Harmonic Motion

Hooke's Law is:

We can ignore the negative sign when we determine units, as the negative sign only indicates the direction of the force. 

Example Question #10 : Harmonic Motion

To determine which type of sinusoid is used, we have to look at its starting point and where it goes. It starts at its minimum and moves towards its maxima. The simplest model we can use is a negative cosine model, which will require no phase shifts. 

The only appropriate model is

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Learn AP Physics

Ap physics 1 & 2 - circular motion & gravitation.

Some students find the transition from linear motion to circular motion a little confusing, but you needn't worry. Rotational kinematics, torque, kinetic energy, and angular momentum are all analogous to their linear versions.

Circular Motion Video Lessons & Problems

• Moving in Circles (Mechanical Universe, Episode 9)
• The Law of Falling Bodies (Mechanical Universe, Episode 2)
• The Apple and the Moon (Mechanical Universe, Episode 8)
• Rotational Kinematics
• Universal Gravitation (Flipping Physics)

Multiple-Choice Practice Problems

Scroll down to see multiple choice practice problems in Rotational Motion, Torque, Rolling Objects, Momemt of Inertia, and Angular Momentum.

Find what you need to study

Unit 3 Overview: Circular Motion and Gravitation

9 min read • january 21, 2023

Kashvi Panjolia

Seemingly simple actions around us, such as car tires spinning or a satellite orbiting a planet, have much more complex processes at hand. In Unit 3, you will continue to build upon your understanding of the physical world using representations to form a more complete and complex model of motion as it relates to gravitational mass and inertial mass. Specific preconceptions will be addressed in this unit, such as the idea of a centrifugal force .

This unit makes up 4-6% of the AP exam, and it tends to span over approximately 7-9 forty-five minute class periods.

Image courtesy of makeameme.org

Applicable Big Ideas

Big Idea #1: Systems - Objects and systems have properties such as mass and charge. Systems may have an internal structure.

Big Idea #2: Fields - Fields existing in space can be used to explain interactions.

Big Idea #3: Force Interactions - The interactions of an object with other objects can be described by forces.

Big Idea #4: Change - Interactions between systems can result in changes in those systems.

Key Concepts

Vector Field

Uniform Circular Motion

Centripetal Force

Centrifugal

Gravitational Force

Newton’s Universal Law of Gravitation

Gravitational Field

Gravitational Mass

Inertial Mass

Frame of Reference

Key Equations

3.1 Vector Fields

Vector fields in physics are mathematical representations of a physical quantity that has both magnitude and direction . In the case of uniform circular motion , the vector field represents the velocity of an object moving in a circular path at a constant speed.

The vector field for uniform circular motion can be represented by a vector pointing in the direction of the object's motion, with its magnitude equal to the object's speed. The direction of the vector field changes as the object moves in its circular path, but its magnitude remains constant.

In addition to velocity, vector fields can also be used to represent other physical quantities such as force, acceleration, and magnetic fields. These vector fields can be visualized using arrows, with the direction and length of the arrow representing the direction and magnitude of the physical quantity.

The vector field between two objects is beyond the scope of AP Physics 1, so we simplify the vector field into a single vector. For example, the gravitational force between the Earth and the moon can be represented with two vectors:

Image courtesy of School for Champions .

This image shows how the Earth is pulling on the moon (the vector on the moon pointing left), but the moon is also pulling on the Earth (the vector on the Earth pointing right). However, the moon's pull on the Earth is much weaker than the Earth's pull on the moon, so the moon orbits around the Earth instead of the other way around.

3.2 Fundamental Forces

The four fundamental forces in physics are the gravitational force , the electromagnetic force , the weak force , and the strong force .

The gravitational force is the force of attraction between two masses. It is one of the weakest forces in nature, but it dominates at large mass and distance scales, such as the orbits of planets and galaxies. The electromagnetic force is the force that mediates the interactions of electrically charged particles. It is one of the strongest forces in nature and is responsible for phenomena such as electricity, magnetism, and light. The weak force is the force responsible for certain types of radioactive decay, such as beta decay. The strong nuclear force is the force that holds the protons and neutrons in the nucleus of an atom together. It is the strongest of the four fundamental forces and is responsible for phenomena such as nuclear fusion and fission. For the purposes of AP Physics 1, only need to understand the gravitational force in depth.

Gravity is a long-range force because its effects can be felt over large distances. Unlike the other fundamental forces (electromagnetism, the weak nuclear force , and the strong nuclear force ), gravity does not have a specific range beyond which its effects are negligible. The strength of the gravitational force between two masses decreases with distance, but it never completely disappears.

Gravity dominates at the largest mass and distance scales because its effects are proportional to the mass of the objects involved. As the masses of the objects that are attracted to each other increase (the masses of planets and stars, for example, are very large), the gravitational force between them increases. This relationship is also the reason you are not attracted to the cup of coffee next to you. The cup does exert a gravitational force on you as you exert on the cup, but the magnitude of the force is so small that it is insignificant.

3.3 Gravitational and Electric Forces

The gravitational force of a planet is proportional to the product of the two masses and inversely proportional to the square of the distance between them. The equation for this force is F=

where F is the force of gravity, m1 and m2 are the masses of the two objects, r is the distance between the two objects and G is the gravitational constant. This equation is known as Newton's Universal Law of Gravitation , and it applies to every object in the universe -- even your physics notebook!

The gravitational force is always attractive and is responsible for the creation of galaxies, the general structure of the entire universe, and is the reason the Earth doesn't go hurtling off into space.

3.4 Gravitational Field/Acceleration Due to Gravity on Different Planets

On Earth, we know the gravitational acceleration to be 9.8 m/s/s. However, the gravitational acceleration on other planets is different, and in this topic, we learn how to find the value of the gravitational acceleration for any planet.

The equation

is a mathematical expression for the gravitational acceleration on a planet other than Earth, where G is the gravitational constant, M is the mass of the planet and R is the radius of the planet. It is also known as the "surface gravity" equation, and it describes the acceleration experienced by an object near the surface of a planet due to the planet's gravitational field .

The equation was derived by using Newton's Universal Law of Gravitation, which states that the force of gravity acting on an object is proportional to the product of the object's mass and the gravitational acceleration (g). The equation tells us that the gravitational acceleration on a planet depends on the mass and the radius of the planet.

A gravitational field is a region around an object in which another object experiences a force due to gravity. The gravitational field strength, also known as "g-force" is the acceleration experienced by an object in the field. The strength of a gravitational field is represented by the gravitational acceleration g.

3.5 Inertial vs. Gravitational Mass

Inertial mass and gravitational mass are two different types of mass that are associated with an object. Inertial mass is a measure of an object's resistance to acceleration, while gravitational mass is a measure of an object's strength of interaction with a gravitational field .

A common example used to illustrate the difference between inertial mass and gravitational mass is that of a bowling ball and a feather falling on the Earth and in a vacuum. On the Earth, both the bowling ball weighs more than the feather, so it falls to the ground faster than the feather. However, if we were to drop them in a vacuum, the bowling ball would fall at the same rate as the feather, due to the lack of air resistance.

The value of an object's inertial mass and its gravitational mass is found to always be the same for both types of mass, since inertial and gravitational mass are two ways of measuring the mass of an object.

3.6 Centripetal Acceleration and Centripetal Force

Centripetal acceleration in uniform circular motion is the acceleration of an object moving in a circular path at a constant speed. It is directed towards the center of the circle, and is what keeps the object moving in a circular path. The equation for centripetal acceleration is:

where a_c is the centripetal acceleration , v is the velocity of the object and r is the radius of the circle.

The velocity of an object in uniform circular motion is tangent to the acceleration. This means that the velocity vector is always at a 90-degree angle to the acceleration vector. This is because the direction of the velocity is constantly changing as the object moves in a circular path (the cause of the acceleration), but its magnitude is not, since the velocity remains constant. The velocity vector is pointing in the direction of motion (along the circle) at any given moment, and the acceleration vector is always pointing toward the center of the circle.

The centripetal force in uniform circular motion is the force that acts on an object moving in a circular path at a constant speed, and is responsible for keeping the object moving in a circular path. The centripetal force is not a new force; just another name for the net force in circular motion. This force could be tension, gravity, normal force, or friction. The equation for centripetal force using Newton's Second Law is: F = m a = m v^2/r where F is the centripetal force , m is the mass of the object, v is the velocity of the object and r is the radius of the circle.

3.7 Free-Body Diagrams for Objects in Uniform Circular Motion

A free body diagram (FBD) is a visual representation of the forces acting on an object. It is a powerful tool that can be used to analyze the motion of an object, such as an object undergoing uniform circular motion . You have already drawn countless free-body diagrams for projectile motion, but now it's time to apply the fundamentals you learned to uniform circular motion . When drawing a free body diagram for an object in uniform circular motion , it is important to select an appropriate coordinate system and to accurately represent the forces acting on the object.

Your positive x or y axis should be in the same direction as the centripetal acceleration . Since the centripetal acceleration vector always points toward the center of the circle, your positive axis should also be toward the center of the circle. If any of the forces acting on your object do not align with this coordinate system, you should break them into their x and y components.

Image courtesy of APlusPhysics.

In this image, the car of a rollercoaster is at the top of a loop-the-loop. At the top, the weight of the car is a force vector pointing downwards. Since the normal force is caused by the force of the track on the car, the normal force is also pointing downwards, and both these forces will add to give us our net force , or centripetal force . Notice how the centripetal force is still pointing towards the center of the circle created by the loop-the-loop.

3.8 Applications of Circular Motion and Gravitation

An inertial frame of reference in physics is a reference frame in which an object is at rest or moving with a constant velocity. In an inertial frame of reference , the laws of physics are the same for all observers and all directions. In other words, if an experiment is performed in an inertial frame of reference , the results will be the same regardless of the velocity of the observer or the direction of the experiment.

For example, an observer who is standing in front of a ferris wheel might say the ferris wheel is rotating clockwise. However, a person standing behind the same ferris wheel will say the wheel is rotating counterclockwise.

The linear quantities used to represent position (x), velocity (v), and acceleration (a) all have rotational equivalents . The rotational analog for position is the angle θ (the Greek lowercase theta) in radians . The rotational analog for velocity is angular velocity , the change in angle divided by the change in time, measured in radians per second and represented by ω (the Greek lowercase omega). The rotational analog for acceleration is angular acceleration α (the Greek lowercase alpha), and it measures the change in angular velocity over time in radians per second squared.

The kinematics equations can also be converted into rotational kinematics equations . Just like the kinematics equations only work in situations where there is a constant linear acceleration, the rotational kinematics equations only work in situations where there is a constant angular acceleration.

Key Terms to Review ( 19 )

Angular Acceleration (α)

Angular Velocity (ω)

Centrifugal Force

Centripetal Acceleration

Electromagnetic Force

Free-Body Diagrams (FBD)

Fundamental Forces

Inertial Frame of Reference

Rotational Kinematics Equations

Strong Nuclear Force

Surface Gravity Equation

Weak Nuclear Force

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AP Physics 1 Course Syllabus Important Physics 1 Dates   Test is still Tues 16 Feb in spite of 2 hour delay! --> Text:   College Physics - Explore and Apply (2nd Edition) by Etkina, Planinsic, and Van Heuvelen Text:   Open Stax, College Physics by Paul Peter Urone & Roger Hinrichs Physics 1 Overview from the College Board Physics 1 Course & Exam Description from the College Board Physics 1 Home multiple resources and information from the College Board AP Exam Info and Equation Sheet for Physics 1 2021 AP Exam Day Information --> AP Physics 1 Released Free Response - by year, includes scoring guidelines * Review Sessions May 18 - 20, 3:40 - 4:45 pm, Mr. M's room O102 * --> Astronomy and Physics Tutorials --> Mr. M's favorite internet Links for physics Lab Information and Homework Requirements

Measurement Review  - Units, Uncertainty, Graphical Analysis     Science Skills Assessment - Analyze your skills and consult this table --> solution guide     The following seven links lead to optional material that can be very useful and worthwhile review and reference:     Objectives & Homework     (see also Homework Requirements )     In-Class Practice     Lab Report - the Pendulum     (see also Lab Information )     Graphing - directions, examples, free graph paper     Notes - uncertainty, significant digits, and units     Functions Overview - summary of important types of equations     Review Problems

Kinematics  - a Mathematical Description of Motion     Objectives & Homework - Kinematics     (see also Homework Requirements )         Answer Key     In-Class Practice - Vector Magnitude & Direction     In-Class Practice - Word Problems     In-Class Practice - Position Graphs     Desmos graphs: Bolt Olympics ,   Bolt WR ,   Car & Stoplight     In-Class Practice - Velocity Graphs     Desmos graphs: Fan Cart ,   Block on Ramp     Interactive Desmos Graph:   Kinematics Model - constant acceleration functions, vectors, & animation     Mini-Lab Graph Matching - walk a graph!     Mini-Lab VW Bus - kinematics of a spring powered toy     Mini-Lab Jug Toss - kinematics of an airborne milk jug     Review Problems     Lab Report - Freefall Acceleration     (see also Lab Information )     Class Notes:   Vectors ,   Position, Distance, & Displacement ,   Speed & Velocity ,   Acceleration ,   Formulas ,   Freefall

Vectors  - Two-Dimensional Kinematics     Objectives & Homework - Vectors         Answer Key     Vector Addition PhET - Exploration and inquiry using PhET interactive     In-Class Practice     Boat in River - analysis of the timeless classic...     Projectile Exploration - inquiry with PhET Projectile Motion     Graphing Projectiles - Parametric equations on TI Calculators     Mini-Lab Projectile - tennis ball through photogate     Lab Report     Review Problems Lab Report - Projectile Motion Mini Launcher -->     Optional Lab - Pick one: Projectile Video or Strobe Photograph or Mini Launcher     Lab Report - Projectile Motion Video Analysis-->             Videos:   Nerf Football   4-Square Ball   Superball     Class Notes:   Addition/Graphical ,   Components ,   Addition/Numerical ,   Relative Motion ,   Projectiles

Forces  - Newton's 3 Laws of Motion (Dynamics)     Objectives & Homework - Forces         Answer Key     In-Class Practice     net Force - virtual mini-lab using PhET simulation     MiniLab - Net Force - using Vernier GoDirect Force and Acceleration sensor     MiniLab - Friction - using Vernier GoDirect Motion Detector     Review Problems     Lab Report - Newton's 2nd Law     Lab tips and pics     Class Notes:   Laws 1 & 2 ,   Net Force ,   Weight & Mass ,   3rd Law & Normal Force ,   Friction ,   Components & Inclines

Periodic Motion & Gravity  - SHM, Orbits, & Force Fields     Objectives & Homework - Periodic Motion         Answer Key     Solar System Data - Sun, planets, moons     In-Class Practice     Rotating Frames Desmos Graphs:   Space Station Ball Toss     SHM Summary - a concise overview (more complete notes at link below)     Review Problems     Mini-Lab Mass on Spring - interactive exploration of Hooke's Law and simple harmonic motion using PhET simulation     Lab Report - the Conical Pendulum     Class Notes:   Circular Motioin ,   Gravitation ,   Simple Harmonic Motion

Conservation of Energy & Momentum  - an Alternate Form of Mechanics     Objectives & Homework - Energy Answer Key -->     Objectives & Homework - Momentum Answer Key -->     In-Class Practice - Energy     In-Class Practice - Momentum     Energy Form Examples     Review Problems     Effect of Speed     Mini-Lab: Rubberband Energy     Mini-Lab: Ping Blaster!         Solution Summary     Mini-Lab Spring Energy - interactive exploration of conservation of energy using PhET simulation     Virtual Lab: Momentum & Center of Mass     Lab Report - Mass on a Spring       Lab Setup and Tips             Logger Pro file download:     SHM Energy.cmbl             Doing the report Using Graphical Analysis App on Chromebook (instead of Logger Pro)-->     Class Notes:   Energy & Conservation ,   Work ,   Springs ,   Power & Machines     Class Notes:   Momentum, Force, & Impulse ,   Conservation of Momentum, Elasticity

Rotation  - Mechanics of Rotating Bodies     Objectives & Homework - Rotation Answer Key -->     In-Class Practice     Mini-Lab Cardboard Tube - rotational kinematics - using GoDirect gyro     Mini-Lab Torque - using GoDirect Force and a meter stick     Mini-Lab Gyro1 - rotational kinematics of a gyroscope     Mini-Lab Gyro2 - rotational dynamics of a gyroscope and falling mass     Mini-Lab 2nd Law - 2nd Law for rotation, using Rotary Motion sensor     Mini-Lab Ramp Rollers - the effect of rotational inertia on a ramp     Review Problems     Class Notes:   Rotational Kinematics ,   Torque & Rotational Inertia ,   Rotational Work & Energy, Angular Momentum

Electricity  - Coulomb's Law & DC Circuits     Objectives & Homework - Electricity Answer Key -->     In-Class Practice     The Digital Multimeter - a Primer     Review Problems     Some Notes     Lab Report - Virtual Circuits     Lab Report - Ohm's Law     Class Notes:   Electric Charge & Force ,   Electric Field & Potential     Class Notes:   Current & Power ,   Resistance ,   Series & Parallel ,   Combinations & Kirchoff's Laws

Waves & Interference  - Mechanical Waves and Sound     Objectives & Homework - Waves Answer Key -->     Wave Graphs - Homework Problem 21     Lab Report - Virtual Waves on a String     In-Class Practice     Class Notes:   Wave Parameters ,   Sound & Wave Graphs ,   Interference & Standing Waves     Review Problems

Final Exam!     About the Final Exam . . .     Early Dismissal Form - Turn in ONLY the ONE form that best applies to you     FHS Attendance and AP Incentives     Review Questions and Problems - Multiple Choice     Practice Problems based on DAWN     Practice Problems using Interactive Simulations     Free Response Practice - Actual AP exam questions from the College Board

AP Physics 1

Circular motion, accelerated rotation, waves & sound, electrostatics and circuits.

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