AP Physics C Electricity : Mechanics Exam

Study concepts, example questions & explanations for AP Physics C Electricity

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Example Questions

Example Question #1 : Understanding Conservation Of Angular Momentum

An ice skater begins to spin, starting with his arms spread out as far as possible, parallel to the ice. He pulls his arms into his body, and then raises them completely vertically towards the ceiling. As the skater pulls his arms inward, his angular velocity will __________, and as he raises his arms vertically his angular velocity will __________.

Possible Answers:

increase . . . stay the same 

decrease . . . stay the same

decrease . . . increase 

increase . . . decrease

Correct answer:

increase . . . stay the same 

Explanation:

Through conservation of angular momentum, as moment of inertia decreases, the angular velocity increases. Moment of inertia is dependent on the distribution of the spinning body's mass away from the center of mass—as the skater brings his arms in, he lowers his moment of inertia, thus increasing his angular velocity. If he raises his arms completely vertically from this point, it does not change the radius of mass distributions (from the center of his body), thus maintaining his angular velocity.

Example Question #72 : Ap Physics C

Sarah spins a ball of mass \(\displaystyle m\) attached to a string of length \(\displaystyle r\) around her head with a velocity \(\displaystyle v_i\). If the ball splits in half, losing exactly one-half of its mass instantaneously, what is its new velocity, \(\displaystyle v_f\)?

Possible Answers:

\(\displaystyle \frac{v_i}{2}\)

\(\displaystyle 4v_i\)

\(\displaystyle 2v_i\)

\(\displaystyle \frac{v_i}{4}\)

\(\displaystyle v_i\) (no change in velocity)

Correct answer:

\(\displaystyle 2v_i\)

Explanation:

By the conservation of angular momentum, the angular momentum \(\displaystyle L\), is equal to the product of the mass, angular velocity, and radius (or length of the rope in this case). The equation relating these terms is:

\(\displaystyle L_{initial}=m_{i}v_{i}r\)

Here, \(\displaystyle m_{i}\) is the initial mass, \(\displaystyle v_{i}\) is the initial angular velocity, and \(\displaystyle r\) is the length of the rope, which remains constant. Angular momentum must be conserved, thus: 

\(\displaystyle L_{initial}=L_{final}\)

Substitute.

\(\displaystyle m_{i}v_{i}r=m_{f}v_{f}r\)

We are given that \(\displaystyle m_{f}=\frac{1}{2}m_{i}\) and \(\displaystyle v_{f}\) is the final velocity. Plug in and solve.

\(\displaystyle m_{i}v_{i}r=\frac{1}{2}m_{i}v_{f}r\)

\(\displaystyle v_{f}=2v_{i}\)

Example Question #2 : Understanding Conservation Of Angular Momentum

Rotating platform

A child is standing at the center of a frictionless, rotating platform. Both the child and the platform are rotating with and initial angular velocity, \(\displaystyle \small \omega_0\). The child begins to walk slowly toward the edge of the platform. Which quantity will decrease as the child walks?

Possible Answers:

Total momentum

Rotational inertia of the disk

Total rotational kinetic energy

Total angular inertia

Total angular momentum

Correct answer:

Total rotational kinetic energy

Explanation:

Since the platform is frictionless, as the child walks, angular momentum is conserved. However, since there is an \(\displaystyle \small \omega ^{2}\) term in the kinetic energy expression, as \(\displaystyle \small \omega\) decreased due to the increase of inertia, it affects the energy more, and it decreases. From the reference frame of the disk, the child feels an outward directed force, and thus is doing negative work. This is the same principle that allows ice skaters to increase their spinning speed.

Example Question #72 : Mechanics Exam

Two solid cylinderical disks have equal radii. The first disk is spinning clockwise at \(\displaystyle 18\frac{rad}{s}\) and the second disk is spinning counterclockwise at \(\displaystyle 4\frac{rad}s{}\). The second disk has a mass three times larger than the first. If both spinning disks are combined to form one disk, they end up rotating at the same angular velocity and same direction. Find this angular velocity after combination.

Possible Answers:

\(\displaystyle 2.0\frac{rad}{s}\)

\(\displaystyle 10.5\frac{rad}{s}\)

\(\displaystyle 4.5\frac{rad}{s}\)

\(\displaystyle 1.5\frac{rad}{s}\)

\(\displaystyle 8.3\frac{rad}{s}\)

Correct answer:

\(\displaystyle 1.5\frac{rad}{s}\)

Explanation:

For the first disk, we have the information below.

\(\displaystyle \omega_1=18\ \frac{rad}{s}\)

\(\displaystyle m_1=\text{mass of 1st disk}\)

\(\displaystyle R_1=\text{radius of 1st disk}\)

\(\displaystyle I_1=\frac{1}{2}m_1R_1^2\)

For the second disk, we have the information below.

\(\displaystyle \omega_2=-4\ \frac{rad}{s}\)

(The minus sign indicates counterclockwise while the positive indicates clockwise.)

\(\displaystyle m_2=3m_1\)

\(\displaystyle R_1=R_2\)

\(\displaystyle I_2=\frac{1}{2}m_2R_2^2=\frac{1}{2}(3m_1)R_1^2=\frac{3}{2}m_1R_1^2\)

To do this problem, we use conservation of angular momentum. Before the disks are put in contact, the initial total angular momentum is given by the equation below.

\(\displaystyle L_i=I_1\omega_1+I_2\omega_2=\frac{1}{2}m_1R_1^2\omega_1+\frac{3}{2}m_1R_1^2\omega_2\)

It is just the sum of the angular momentums of each disk. When the disks are combined together, then final angular momentum can be found by the following equation.

\(\displaystyle L_f=I_1\omega+I_2\omega=\frac{1}{2}m_1R_1^2\omega+\frac{3}{2}m_1R_1^2\omega=2m_1R_1^2\omega\)

Set this equal to the initial angular momentum.

\(\displaystyle \frac{1}{2}m_1R_1^2\omega_1+\frac{3}{2}m_1R_1^2\omega_2=2m_1R_1^2\omega\)

Simplify and solve for \(\displaystyle \omega\).

\(\displaystyle \omega=\frac{1}{4}\omega_1+\frac{3}{4}\omega_2\)

\(\displaystyle \omega=\frac{1}{4}(18\ \frac{rad}{s})+\frac{3}{4}(-4\ \frac{rad}{s})\)

\(\displaystyle \omega=1.5\ \frac{rad}{s}\)

Example Question #2 : Using Angular Momentum Equations

A piece of space debris is travelling in an elliptical orbit around a planet. At its closest point to the planet, the debris is travelling at \(\displaystyle 5000\frac{m}{s}\). When the debris is \(\displaystyle 10000km\) from the planet, it travels at \(\displaystyle 5000\frac{km}{hr}\). How close does the debris get to the planet in its orbit?

Possible Answers:

\(\displaystyle 3156km\)

\(\displaystyle 7191km\)

\(\displaystyle 5000km\)

\(\displaystyle 1145km\)

\(\displaystyle 2778km\)

Correct answer:

\(\displaystyle 2778km\)

Explanation:

The equation for conservation of angular momentum is:

\(\displaystyle mv_{1}r_{1}=mv_{2}r_{2}\)

This means that the angular momentum of the object at the two points in its orbit must be the same. Since the mass of the debris does not change, this gives us an equality of the product of the velocity and distance of the debris at any two points of its orbit:

\(\displaystyle v_{1}r_{1}=v_{2}r_{2}\) 

Plug in known values.

\(\displaystyle r_{1}5000\frac{m}{s}=10000km \left(5000\frac{km}{hr}\right)\)

There is disagreement between units; the velocities are given in both \(\displaystyle \frac{m}{s}\) and \(\displaystyle \frac{km}{hr}\). Glance down at the answer choices and note that they are all lengths in kilometers.

\(\displaystyle \left(5000\frac{m}{s}\right)\left(\frac{3600s}{1hr}\right)=18\cdot10^6\frac{m}{hr}\)

\(\displaystyle \left(18\cdot 10^6\frac{m}{hr}\right)\left(\frac{1m}{1000m}\right)=18000\frac{km}{hr}\)

Solve for \(\displaystyle r_{1}\)

\(\displaystyle r_{1}=\frac{(10000km)(5000\frac{km}{hr})}{18000\frac{km}{hr}}=\frac{50000}{18}km\approx2778km\)

Example Question #71 : Mechanics Exam

A mechanic is using a wrench to loosen and tighten screws on an engine block and wants to increase the amount of torque he puts on the screws to adjust them more easily. Which of the following steps will not help him to do so?

Possible Answers:

Pulling or pushing at an angle that is less perpendicular to the wrench

Pulling or pushing at an angle that is more perpendicular to the wrench

Increasing the magnitude of the force he puts on the wrench

Using a longer wrench

Using a wrench of a different material

Correct answer:

Pulling or pushing at an angle that is less perpendicular to the wrench

Explanation:

Remember the torque equation:

\(\displaystyle \tau = ||r|| * ||F|| * \sin \theta\)

Exerting force on the wrench at an angle less perpendicular to the wrench will reduce \(\displaystyle \sin \theta\) and thus reduce the torque.

The other answer options will all increase the torque being applied, making the twisting motion easier.

Example Question #1 : Forces

Weightlessness is experienced when the normal force equals __________.

Possible Answers:

zero

the force due to gravity 

the mass of the object

a negative value 

Correct answer:

zero

Explanation:

"Weightlessness" is analogous to free-fall (neglecting air resistance), during which the only force on an object is the force of gravity. When normal force becomes zero, the object loses physical contact with the surface.

When normal force is equal to the force of gravity, the object is in equilibrium and the net vertical force is zero. This results in zero acceleration, but does not result in "weightlessness." Mass is not a force, and negative force values merely indicate relative direction (assuming correct calculations)

Example Question #2 : Normal Force

A book of mass 1kg is held to a vertical wall by a person's hand applying a 20N force directly toward the wall. The wall has a static friction coefficient of 0.3 and a kinetic friction coefficient of 0.2. With the book held at rest, what the is frictional force keeping the book from sliding down the wall? 

Possible Answers:

\(\displaystyle 4N\)

\(\displaystyle 58.8N\)

\(\displaystyle 6N\)

\(\displaystyle 39.2N\)

Correct answer:

\(\displaystyle 6N\)

Explanation:

Frictional force is calculated using the normal force and the friction coefficient. Since the book is at rest (not in motion), we use the coefficient of static friction.

\(\displaystyle F_f=\mu_sF_N\)

The normal force will be equal and opposite the force pushing directly into the surface; in this case, that means the normal force will be 20N.

Solve for the force of friction:

\(\displaystyle F_f=(0.3)(20N)=6N\)

Example Question #72 : Mechanics Exam

\(\displaystyle \begin{align*}&\text{A block of mass }190kg\text{ rests on an incline of }33^{\circ}\\&\text{It is also partially supported by a rope with a tension of }559.17N\\&\text{What is the normal force acting on the block from the surface?}\\&g=9.81\frac{m}{s^2}\end{align*}\)

Possible Answers:

\(\displaystyle 1015.15N\)

\(\displaystyle 1563.2N\)

\(\displaystyle 1094.24N\)

\(\displaystyle 710.61N\)

Correct answer:

\(\displaystyle 1094.24N\)

Explanation:

Normalwrope

\(\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), the force of the rope supporting it, a normal force,}\\&\text{and since it’s on an incline, a frictional force to balance}\\&\text{it all out. Notice that the tension in the rope mitigates some}\\&\text{of the gravitational force (as illustrated in the upper left}\\&\text{corner) to lessen what is applied to the slope:}\\&F_{mass}=m*g\\&F_{normal}=(F_{mass}-F_{rope})*cos(a)\\&F_{friction}=(F_{mass}-F_{rope})*sin(a)\\&F_{normal}=(190kg(9.81\frac{m}{s^2})-559.17N)(cos(33^{\circ}))\\&F_{normal}=1094.24N\end{align*}\)

Example Question #1 : Friction And Normal Force

\(\displaystyle \begin{align*}&\text{A block of mass }233kg\text{ rests on an incline of }81^{\circ}\\&\text{What is the normal force acting on the block from the surface?}\\&g=9.81\frac{m}{s^2}\end{align*}\)

Possible Answers:

\(\displaystyle 357.57N\)

\(\displaystyle 2257.59N\)

\(\displaystyle 2285.73N\)

\(\displaystyle 230.13N\)

Correct answer:

\(\displaystyle 357.57N\)

Explanation:

Normal

\(\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), a normal force, and since it’s on an incline, a frictional}\\&\text{force to balance it all out:}\\&F_{mass}=m*g\\&F_{normal}=m*g*cos(a)\\&F_{friction}=m*g*sin(a)\\&F_{normal}=233kg(9.81\frac{m}{s^2})(cos(81^{\circ}))\\&F_{normal}=357.57N\end{align*}\)

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