Circular Motion - AP Physics C Mechanics

1. An object undergoes uniform circular motion with radius $r$ and period $T$. What is the magnitude of its centripetal acceleration?
   Answer: $4\pi^2 r / T^2$ The speed of the object is the circumference divided by the period: $v = 2\pi r / T$. The centripetal acceleration is $a_c = v^2/r$. Substituting the expression for $v$ gives $a_c = (2\pi r / T)^2 / r = (4\pi^2 r^2 / T^2) / r = 4\pi^2 r / T^2$.
2. A car travels around a banked curve with banking angle $\theta = 25°$. The coefficient of static friction between the tires and road is $\mu_s = 0.40$. For a curve of radius 200 m, what is the maximum speed the car can travel without slipping?
   Answer: $38 \text{ m/s}$ For maximum speed, friction acts down the incline. The centripetal force equation becomes: $mg\sin\theta + \mu_s mg\cos\theta = \frac{mv^2}{r}$. Solving: $v = \sqrt{gr(\sin\theta + \mu_s\cos\theta)} = \sqrt{9.8 \times 200 \times (\sin25° + 0.40\cos25°)} = 38 \text{ m/s}$. Choice A ignores the banking angle contribution. Choice B uses only the banking term without friction. Choice D incorrectly adds the terms under separate square roots.
3. An astronaut is in a spacecraft orbiting the Earth at an altitude where the gravitational acceleration is $g/4$. The astronaut's spacecraft is in a stable circular orbit. What is the astronaut's apparent weight?
   Answer: Zero, because the astronaut is in a state of free fall. Apparent weight is the force an object exerts on its support, which is equal in magnitude to the normal force acting on it. Both the astronaut and the spacecraft are in a state of free fall around the Earth, accelerating towards it at the same rate ($g/4$). Since they are accelerating together, the astronaut does not press against the walls or floor of the spacecraft, and the normal force is zero. This state of weightlessness is a result of being in continuous free fall.
4. A car of mass $m$ navigates a circular turn of radius $R$ that is banked at an angle $θ$ to the horizontal. At what speed can the car navigate the turn without relying on friction?
   Answer: $\sqrt{gR \tan\theta}$ The forces on the car are gravity ($mg$, down) and the normal force ($N$, perpendicular to the bank). The vertical component of the normal force must balance gravity: $N \cos\theta = mg$. The horizontal component of the normal force provides the centripetal force: $N \sin\theta = mv^2/R$. Dividing the second equation by the first gives $\tan\theta = v^2/(gR)$. Solving for $v$ yields $v = \sqrt{gR \tan\theta}$.
5. A car of mass $m$ travels at a constant speed $v$ around a flat, unbanked circular track of radius $R$. What is the magnitude of the net force acting on the car?
   Answer: $mv^2/R$ The car is undergoing uniform circular motion, which means it has a centripetal acceleration of magnitude $a_c = v^2/R$ directed toward the center of the circle. According to Newton's second law, the net force is $F_{net} = ma_c = mv^2/R$. This force is provided by the static friction between the tires and the road.
6. A particle moves in a horizontal circle of radius 2.0 m. The particle's angular velocity increases linearly from 0 to 6.0 rad/s in 3.0 s. What is the magnitude of the particle's total acceleration when $t = 2.0 \text{ s}$?
   Answer: $32.2 \text{ m/s}^2$ When a particle undergoes circular motion with changing angular velocity, you need to consider both centripetal and tangential acceleration components. The total acceleration is the vector sum of these perpendicular components. First, find the angular acceleration. Since angular velocity increases linearly from 0 to 6.0 rad/s in 3.0 s: $\alpha = \frac{\Delta\omega}{\Delta t} = \frac{6.0 - 0}{3.0} = 2.0 \text{ rad/s}^2$ At $t = 2.0$ s, the angular velocity is: $\omega = \omega_0 + \alpha t = 0 + (2.0)(2.0) = 4.0 \text{ rad/s}$ The tangential acceleration is: $a_t = r\alpha = (2.0)(2.0) = 4.0 \text{ m/s}^2$ The centripetal acceleration is: $a_c = r\omega^2 = (2.0)(4.0)^2 = 32 \text{ m/s}^2$ Since these accelerations are perpendicular, the total acceleration magnitude is: $a_{total} = \sqrt{a_t^2 + a_c^2} = \sqrt{(4.0)^2 + (32)^2} = \sqrt{16 + 1024} = \sqrt{1040} = 32.2 \text{ m/s}^2$ Choice A ($4.0 \text{ m/s}^2$) gives only the tangential acceleration component. Choice B ($32 \text{ m/s}^2$) gives only the centripetal acceleration component. Choice C ($36 \text{ m/s}^2$) incorrectly adds the components arithmetically rather than using vector addition. Study tip: In circular motion problems with changing speed, always check whether both tangential and centripetal accelerations are present. Use the Pythagorean theorem to find the total acceleration magnitude since these components are always perpendicular.
7. A small object of mass $m$ is attached to a light string of length $L$ to form a conical pendulum. The object revolves in a horizontal circle of radius $r$ with constant speed $v$, and the string makes an angle $θ$ with the vertical. Which of the following is an expression for the speed $v$?
   Answer: $\sqrt{gL \sin\theta \tan\theta}$ Let $T$ be the tension. The vertical component of tension balances gravity: $T\cos\theta = mg$. The horizontal component provides the centripetal force: $T\sin\theta = mv^2/r$. The radius of the circle is $r = L\sin\theta$. From the first equation, $T = mg/\cos\theta$. Substituting this into the second equation: $(mg/\cos\theta)\sin\theta = mv^2/(L\sin\theta)$. This simplifies to $g\tan\theta = v^2/(L\sin\theta)$. Solving for $v$ gives $v = \sqrt{gL\sin\theta\tan\theta}$.
8. A space station rotates to create artificial gravity. The station has a radius of 100 m and rotates such that the acceleration at the rim is $0.4g$, where $g = 9.8 \text{ m/s}^2$. An astronaut walks along the rim in the direction opposite to the station's rotation at 2.0 m/s relative to the station. What acceleration does the astronaut experience?
   Answer: $3.9 \text{ m/s}^2$ Station rim speed: $v_{rim} = \sqrt{0.4gr} = \sqrt{0.4 \times 9.8 \times 100} = 19.8 \text{ m/s}$. Astronaut's speed relative to space: $v_{ast} = 19.8 - 2.0 = 17.8 \text{ m/s}$. Astronaut's acceleration: $a = \frac{v_{ast}^2}{r} = \frac{(17.8)^2}{100} = 3.9 \text{ m/s}^2$. Choice A incorrectly subtracts walking speed from acceleration. Choice C uses the original station acceleration. Choice D incorrectly adds the walking contribution.
9. A small ball is whirled in a horizontal circle at the end of a string. If the speed of the ball is doubled while the radius of the circle remains constant, what happens to the tension in the string?
   Answer: It is quadrupled. The tension in the string provides the centripetal force required for the circular motion, so $T = mv^2/r$. Tension $T$ is directly proportional to the square of the speed $v$. If the speed is doubled ($v' = 2v$), the new tension $T'$ will be $m(2v)^2/r = 4(mv^2/r) = 4T$. The tension is quadrupled.
10. A roller coaster car goes through a vertical loop-the-loop of radius $R$. At the bottom of the loop, the normal force on a passenger of mass $m$ is three times their weight. What is the speed of the car at the bottom of the loop?
    Answer: $\sqrt{2gR}$ At the bottom of the loop, the net force is directed upward and provides the centripetal force. The net force is the difference between the upward normal force $N$ and the downward gravitational force $mg$. So, $F_{net} = N - mg = mv^2/R$. We are given that $N = 3mg$. Substituting this in gives $3mg - mg = mv^2/R$, which simplifies to $2mg = mv^2/R$. Solving for $v$ gives $v = \sqrt{2gR}$.
11. Two masses, $m_1 = 3.0 \text{ kg}$ and $m_2 = 2.0 \text{ kg}$, are connected by a light string and move in horizontal circles of different radii on the same turntable. If $m_1$ is at radius $r_1 = 0.8 \text{ m}$ and $m_2$ is at radius $r_2 = 1.2 \text{ m}$, what is the ratio of the tension forces $T_1/T_2$ in their respective strings?
    Answer: $1.0$ Both masses have the same angular velocity $\omega$. For each mass: $T = m\omega^2 r$. Therefore: $\frac{T_1}{T_2} = \frac{m_1\omega^2 r_1}{m_2\omega^2 r_2} = \frac{m_1 r_1}{m_2 r_2} = \frac{3.0 \times 0.8}{2.0 \times 1.2} = \frac{2.4}{2.4} = 1.0$. Choice B uses only the mass ratio. Choice C incorrectly uses the reciprocal of the radius ratio. Choice D incorrectly squares one of the ratios.
12. A small block is placed on a horizontal turntable that rotates with a constant angular speed $ω$. The block remains at rest relative to the turntable at a distance $r$ from the center. If the coefficient of static friction between the block and the turntable is $μ_s$, which of the following expressions must be true?
    Answer: $μ_s g ≥ ω^2 r$ The centripetal force required to keep the block in circular motion is $F_c = m a_c = mω^2 r$. This force is provided by static friction, $f_s$. The maximum possible static friction is $f_{s,max} = μ_s N = μ_s mg$. For the block not to slip, the required centripetal force must be less than or equal to the maximum static friction: $mω^2 r ≤ μ_s mg$. Dividing by $m$ gives $ω^2 r ≤ μ_s g$.
13. A pilot flies an airplane in a vertical loop of radius $R$ at a constant speed $v$. The pilot's apparent weight is equal to the normal force exerted by the seat. At the bottom of the loop, the pilot's apparent weight is
    Answer: greater than their true weight, $mg$ At the bottom of the loop, the net force must be directed upward toward the center of the circle to provide the centripetal force. The forces on the pilot are the upward normal force $N$ from the seat and the downward force of gravity $mg$. So, $F_{net} = N - mg = mv^2/R$. This means the normal force (apparent weight) is $N = mg + mv^2/R$, which is greater than the true weight $mg$.
14. A satellite orbits Earth in a circular path. If the satellite's orbital radius is increased by a factor of 4, by what factor does its orbital period change?
    Answer: Increases by a factor of 8 From Kepler's third law, $T^2 \propto r^3$, so $T \propto r^{3/2}$. If radius increases by a factor of 4, then $T_{new} = T_{old} \times 4^{3/2} = T_{old} \times 8$. Choice A incorrectly uses $T \propto r^{1/2}$. Choice B incorrectly uses $T \propto r$. Choice D incorrectly uses $T \propto r^2$.
15. An object is in uniform circular motion in a horizontal plane. Which of the following statements correctly describes the object's velocity and acceleration vectors?
    Answer: Both velocity and acceleration vectors are constant in magnitude but continuously change direction. In uniform circular motion, the speed (magnitude of velocity) is constant. However, the direction of the velocity vector is always changing as it is tangent to the circular path. The acceleration (centripetal acceleration) is directed towards the center of the circle and also continuously changes direction as the object moves. Its magnitude, $v^2/r$, is constant.
16. A 2.0 kg mass is attached to a string and moves in a horizontal circle of radius 1.5 m on a frictionless table. If the string makes an angle of 30° below the horizontal with the attachment point above the table, what is the speed of the mass?
    Answer: $3.6 \text{ m/s}$ The vertical component of tension balances weight: $T\sin(30°) = mg$, so $T = \frac{mg}{\sin(30°)} = \frac{2.0 \times 9.8}{0.5} = 39.2 \text{ N}$. The horizontal component provides centripetal force: $T\cos(30°) = \frac{mv^2}{r}$. Substituting: $39.2 \times \cos(30°) = \frac{2.0v^2}{1.5}$, which gives $v = 3.6 \text{ m/s}$. Choice A uses the wrong angle relationship. Choice C incorrectly uses the full tension for centripetal force. Choice D squares the tension incorrectly in the calculation.
17. A car is driving over a hill that has a circular cross-section of radius $R$. What is the maximum speed the car can have at the top of the hill and remain in contact with the road?
    Answer: $\sqrt{gR}$ At the top of the hill, the net force must be directed downward toward the center of the circular path. The forces are gravity ($mg$, down) and the normal force ($N$, up). So, $F_{net} = mg - N = mv^2/R$. The car remains in contact with the road as long as $N ≥ 0$. The maximum speed occurs at the limit where $N = 0$. In this case, $mg = mv_{max}^2/R$, which gives $v_{max} = \sqrt{gR}$.
18. A car of mass $M$ travels at speed $v$ around a flat circular track of radius $R$. If the car were to travel at the same speed $v$ around a track of radius $2R$, what would be the required centripetal force?
    Answer: The force would be halved. The centripetal force is given by the formula $F_c = Mv^2/R$. The force is inversely proportional to the radius $R$. If the radius is doubled from $R$ to $2R$ while mass $M$ and speed $v$ remain constant, the new force $F_c'$ will be $Mv^2/(2R) = (1/2)(Mv^2/R) = F_c/2$. The force would be halved.
19. A satellite is in a stable circular orbit around Earth. If the radius of its orbit is doubled, what is the ratio of its new orbital speed to its original orbital speed?
    Answer: $1/\sqrt{2}$ For a satellite in a circular orbit, the gravitational force provides the centripetal force: $GmM/R^2 = mv^2/R$. Solving for speed gives $v = \sqrt{GM/R}$. This shows that $v$ is proportional to $1/\sqrt{R}$. If the radius $R$ is doubled, the new speed $v'$ will be $v/\sqrt{2}$. The ratio of the new speed to the original speed is $1/\sqrt{2}$.
20. Two planets, A and B, are in circular orbits around the same star. Planet A has an orbital radius $R$ and an orbital period $T$. Planet B has an orbital radius of $4R$. What is the orbital period of Planet B?
    Answer: $8T$ According to Kepler's third law for circular orbits, the square of the period is proportional to the cube of the orbital radius ($T^2 \propto R^3$). So, $(T_B/T_A)^2 = (R_B/R_A)^3$. Given $R_B = 4R_A$, we have $(T_B/T)^2 = (4R/R)^3 = 4^3 = 64$. Taking the square root, $T_B/T = \sqrt{64} = 8$. Therefore, $T_B = 8T$.

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