Circular Motion
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AP Physics C: Mechanics › Circular Motion
An object moves at a constant speed in a circular path. Which of the following statements about the work done on the object is correct?
The net force does zero work, and the kinetic energy remains constant.
The net force does negative work, decreasing the kinetic energy.
The work done depends on the displacement for one revolution.
The net force does positive work, increasing the kinetic energy.
Explanation
The net force on an object in uniform circular motion is the centripetal force, which is always directed towards the center of the circle. The object's instantaneous displacement is always tangent to the circle. Therefore, the centripetal force is always perpendicular to the displacement. The work done by a force is given by $$W = Fd\cos\theta$$. Since the angle $$θ$$ between the force and displacement is 90°, the work done is zero. By the work-energy theorem, zero net work means the kinetic energy is constant, which is consistent with constant speed.
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?
$$\sqrt{gR}$$
$$\sqrt{2gR}$$
$$\sqrt{3gR}$$
$$2\sqrt{gR}$$
Explanation
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 $$v = \sqrt{2gR}$$.
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?
$$2$$
$$1/\sqrt{2}$$
$$1/2$$
$$\sqrt{2}$$
Explanation
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 $$1/\sqrt{2}$$.
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 $$v$$?
$$\sqrt{gL \cos\theta}$$
$$\sqrt{g / (L \cos\theta)}$$
$$\sqrt{gL \sin\theta \tan\theta}$$
$$\sqrt{gL \sin\theta}$$
Explanation
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 $$v = \sqrt{gL\sin\theta\tan\theta}$$.
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?
Both velocity and acceleration vectors are constant in magnitude and direction.
The acceleration vector is constant, but the velocity vector continuously changes direction.
Both velocity and acceleration vectors are constant in magnitude but continuously change direction.
The velocity vector is constant, but the acceleration vector continuously changes direction.
Explanation
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.
A centrifuge spins a sample at an angular velocity $$ω$$ in a circle of radius $$r$$. The sample experiences a centripetal acceleration of $$a_c$$. If the angular velocity is tripled to $$3ω$$, what is the new centripetal acceleration in terms of $$a_c$$?
$$9a_c$$
$$6a_c$$
$$3a_c$$
$$a_c / 3$$
Explanation
The centripetal acceleration is given by $$a_c = v^2/r$$. The linear speed $$v$$ is related to the angular velocity $$ω$$ by $$v = ωr$$. Substituting this into the acceleration equation gives $$a_c = (ωr)^2/r = ω^2r$$. This shows that $$a_c$$ is proportional to $$ω^2$$. If the angular velocity is tripled ($$ω' = 3ω$$), the new acceleration $$a_c'$$ will be $$(3ω)^2r = 9ω^2r = 9a_c$$.
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?
$$mv^2/R$$
Zero
$$mv/R$$
$$mg$$
Explanation
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.
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?
$$μ_s g ≥ ω^2 r$$
$$μ_s ω^2 r ≥ g$$
$$μ_s g ≥ ω r$$
$$μ_s g ≤ ω^2 r$$
Explanation
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 $$ω^2 r ≤ μ_s g$$.
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?
$$\sqrt{gR \tan\theta}$$
$$gR \cos\theta$$
$$\sqrt{gR / \tan\theta}$$
$$\sqrt{gR \sin\theta}$$
Explanation
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 $$v = \sqrt{gR \tan\theta}$$.
An object undergoes uniform circular motion with radius $$r$$ and period $$T$$. What is the magnitude of its centripetal acceleration?
$$4\pi^2 r / T^2$$
$$4\pi^2 r^2 / T^2$$
$$2\pi r / T^2$$
$$2\pi r / T$$
Explanation
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 $$a_c = (2\pi r / T)^2 / r = (4\pi^2 r^2 / T^2) / r = 4\pi^2 r / T^2$$.