Rotational Motion and Torque - AP Physics C: Electricity and Magnetism

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Question

A 0.18 m long wrench is used to turn the nut on the end of a bolt. A force of 85 N is applied downward to the end of the wrench, as shown in the figure. The angle between the force and the handle of the wrench is 65 degrees.

What is the magnitude and direction of the torque (around the center of the bolt) due to this force?

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Answer

To calculate the magnitude of the torque,

$\tau =rF\sin\theta$

where the radius is the distance between the center of rotation and the location of the force and the angle $\theta$ is between the radius and the force .

The magnitude is thus,

$\tau =(0.18: m) \cdot(85: N)\cdot\sin(65^{\circ})=14: N\cdot m$

The direction of torque is perpendicular to the plane of the radius and the force , and is given by Right Hand Rule by crossing the radius vector into the force vector. For this situation, the radius vector is left and downward and the force is downward, resulting in the direction of the torque out of the page.

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Question

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?

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Answer

Remember the torque equation:

$\tau = ||r|| * ||F|| * \sin \theta$

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

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

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Question

Doing which of the following would allow you to find the center of mass of an object?

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Answer

Center of mass can be found by spinning an object. It will naturally spin around its center of mass, due to the concept of even distribution of mass in relation to the center of mass. Shape and mass are important factors in this property, but the most improtant factor is the mass distribution.

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Question

If two masses, $m_{1}=50kg$ and $m_{2}=200kg$ are placed on a seesaw of length , where must the fulcrum be placed such that the seesaw remains level?

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Answer

This question asks us to find the center of mass for this system. We know that the center of mass resides a distance $x_{1}$ from the first mass such that:

$m_{1}x_{1}=m_{2}(L-x_{1})$

In this case:

$m_{1}x_{1}+m_{2}x_{1}=m_{2}L$

$x_{1}(m_{1}+m_{2})=m_{2}L$

Plug in known values and solve.

$x_{1}=\frac{m_{2}L}{m_{1}+m_{2}}=\frac{200}{250}L=\frac{4}{5}L$

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Question

If the fulcrum of a balanced scale is shifted to the left, what type of adjustment must be made to rebalance the scale?

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Answer

Changing the position of the fulcrum by moving it to the left means the center of mass will be to the right of the new position. Therefore, the scale will tip right. Adding more mass to the left end will rebalance the scale. None of the other options make sense. Adding more mass to the new fulcrum position will not change the balance of the scale because that mass is a negligible distance from the new fulcrum position and does nothing to change the masses on either side.

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Question

Three point masses are at the points , and

and a point mass is at the point .

How far from the origin is the center of mass of the system?

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Answer

To find the center of mass, we have to take the weighted average of the x coordinates and the y coordinates.

Measures: Measures

First, we take the weighted measurement of the x-axis:

$x_c_m = \frac{(7.07\cdot40:g)+(-7.07 \cdot 40:g) + (0\cdot 40:g)+(0\cdot90:g)}{(3\cdot40)+90}$

We can see that the result of the x-axis contribution is equal to .

Now, let's look at the y-axis contribution:

$y_c_m = \frac{(0\cdot40:g)+(0 \cdot 40:g) + (-7.07\cdot 40:g)+(7.07\cdot90:g)}{(3\cdot40)+90}$

This equals to

Now that we have the x and y components, we take the root of squares to get the final answer:

$\uptext{center}= \sqrt{x^2+y^2}$

This will give us

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Question

A mouse sits at the edge of a $\small 40cm$ diameter record, rotating on a turntable at fifteen revolutions per minute. If the mouse walks straight inwards to a point $\small 10cm$ from the center, what will be its new angular velocity? Assume the mass of the record is negligible.

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Answer

Relevant equations:

$L = mr^2\omega$

Write an expression for the initial angular momentum.

$L_i = mr_i^2\omega_i=m(20cm)^2(15rpm)$

Write an expression for the final angular momentum.

$L_f = mr_f^2\omega_f=m(10cm)^2\omega_f$

Apply conservation of angular momentum, setting these two expressions equal to one another.

$m(20cm)^2(15rpm)=m(10cm)^2\omega_f$

Solve for the final angular velocity.

$w_f=\frac{(20cm)^2(15rpm)}{(10cm)^2}=60rpm$

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Question

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 .

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Answer

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.

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Question

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

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Answer

By the conservation of angular momentum, the angular momentum , 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:

$L_{initial}=m_{i}v_{i}r$

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

$L_{initial}=L_{final}$

Substitute.

$m_{i}v_{i}r=m_{f}v_{f}r$

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

$m_{i}v_{i}r=\frac{1}{2}m_{i}v_{f}r$

$v_{f}=2v_{i}$

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Question

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, $\small \omega_0$ . The child begins to walk slowly toward the edge of the platform. Which quantity will decrease as the child walks?

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Answer

Since the platform is frictionless, as the child walks, angular momentum is conserved. However, since there is an $\small \omega ^{2}$ term in the kinetic energy expression, as $\small \omega$ 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.

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Question

What is the rotational equivalent of mass?

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Answer

The correct answer is moment of inertia. For linear equations, mass is what resists force and causes lower linear accelerations. Similarly, in rotational equations, moment of inertia resists torque and causes lower angular accelerations.

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Question

In rotational kinematics equations, what quantity is analogous to force in linear kinematics equations?

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Answer

Just as force causes linear acceleration, torque causes angular acceleration. This can be seen most in the linear-rotational comparison of Newton's second law:

$\tau=I\alpha$

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Question

A boot is put in a stick which is attached to a rotor. The rotor turns with an angular velocity of $30\pi \frac{rad}{s}$ . What is the linear velocity of the boot?

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Answer

Linear (tangential) velocity, $v_{T}$ is given by the following equation:

$v_{T}=\omega r$

Here, $\omega$ is the angular velocity in radians per second and is the radius in meters.

Solve.

$\omega = 30\pi \frac{rad}{s}$

$\omega = 30\pi\frac{m}{s}$

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Question

A particle is moving at constant speed in a straight line past a fixed point in space, c. How does the angular momentum of the particle about the fixed point in space change as the particle moves from point a to point b?

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Answer

The angular momentum of a particle about a fixed axis is $\small L=mvr\cos(\theta )$ . As the particle draws nearer the fixed axis, both $\small r$ and $\theta$ change. However, the product $r\cos( \theta)$ remains constant. If you imagine a triangle connecting the three points, the product $r\cos( \theta)$ represents the "of closest approach", labeled "" in the diagram.

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Question

Two solid cylinderical disks have equal radii. The first disk is spinning clockwise at $18\frac{rad}{s}$ and the second disk is spinning counterclockwise at $4\frac{rad}s{}$ . 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.

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Answer

For the first disk, we have the information below.

$\omega_1=18\ \frac{rad}{s}$

$m_1=\text{mass of 1st disk}$

$R_1=\text{radius of 1st disk}$

$I_1=\frac{1}{2}m_1R_1^2$

For the second disk, we have the information below.

$\omega_2=-4\ \frac{rad}{s}$

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

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

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.

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

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.

$L_f=I_1\omega+I_2\omega=\frac{1}{2}m_1R_1^2\omega+\frac{3}{2}m_1R_1^2\omega=2m_1R_1^2\omega$

Set this equal to the initial angular momentum.

$\frac{1}{2}m_1R_1^2\omega_1+\frac{3}{2}m_1R_1^2\omega_2=2m_1R_1^2\omega$

Simplify and solve for $\omega$ .

$\omega=\frac{1}{4}\omega_1+\frac{3}{4}\omega_2$

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

$\omega=1.5\ \frac{rad}{s}$

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Question

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 $5000\frac{m}{s}$ . When the debris is from the planet, it travels at $5000\frac{km}{hr}$ . How close does the debris get to the planet in its orbit?

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Answer

The equation for conservation of angular momentum is:

$mv_{1}r_{1}=mv_{2}r_{2}$

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:

$v_{1}r_{1}=v_{2}r_{2}$

Plug in known values.

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

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

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

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

Solve for $r_{1}$ :

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

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Question

An object starts from rest and accelerates to an angular velocity of $12\frac{rad}{s}$ in three seconds under a constant torque of $50\ N\cdot m$ . How many revolutions has the object made in this time?

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Answer

Since it is experiencing a constant torque and constant angular acceleration, the angular displacement can be calculated using:

$\Delta \theta =\omega_ot+\frac{1}{2}\alpha t^2$

The angular acceleration is easily calculated using the angular velocity and the time:

$\alpha=\frac{\Delta\omega}{\Delta t}=\frac{12\frac{rad}{s}-0\frac{rad}{s}}{3s}=4\frac{rad}{s^2}$

Using this value, we can find the angular displacement:

$\Delta \theta =(0\frac{rad}{s})(3s)+\frac{1}{2}(4\frac{rad}{s^2})(3s)^2$

$\Delta \theta=18rad$

Convert the angular displacement to revolutions by diving by $2\pi$ :

$\frac{18rad}{2\pi\frac{rad}{rev}}\approx2.9rev$

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Question

A circular disk of radius 0.5m and mass 3kg has a force of 25N exerted perpendicular to its edge, causing it to spin. What is the angular acceleration of the disk?

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Answer

We can find the angular acceleration using the rotaional motion equivalent of Newton's second law. In rotational motion, torque is the product of moment of inertia and angular acceleration:

$\tau = I\alpha$

The moment of inertia for a circular disk is:

$I=\frac{1}{2}mr^2$

The tourque is the product of force and distance (in this case, the radius):

$\tau=F\cdot r$

We can plug these into our first equation:

$(F)(r)=(\frac{1}{2}mr^2)(\alpha)$

Simplify and rearrange to derive an equation for angular acceleration:

$\alpha=\frac{Fr}{\frac{1}{2}mr^2}=\frac{2F}{mr}$

Use our given values to solve:

$\alpha=\frac{2(25N)}{(3kg)(0.5m)}=33.3\frac{rad}{s^2}$

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Question

A meter stick is nailed to a table at one end and is free to rotate in a horizontal plane parallel to the top of the table. Four forces of equal magnitude are applied to the meter stick at different locations. The figure below shows the view of the meter stick from above.

You may assume the forces and are applied at the center of the meter stick, and the forces and are applied at the end opposite the nail.

What is the relationship among the magnitudes of the torques on the meter stick caused by the four different forces?

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Answer

Torque is given by,

$\tau =rFsin\theta$

Since all of the forces are equal in magnitude, the magnitude of the torque is then influenced by the radius r and the angle theta between the radius and the force.

For ,

$\tau_{1} =\left (\frac{1}{2}L \right )F\sin, 45^{\circ}=0.354\cdot L\cdot F$

For ,

$\tau_{2} =\left (\frac{1}{2}L \right )F\sin, 90^{\circ}=0.5\cdot L \cdot F$

For ,

$\tau_{3} =LF\sin, 30^{\circ}=0.5 \cdot L \cdot F$

For ,

$\tau_{4} =LF\sin, 90^{\circ}=L\cdot F$

Combining this information yields the relationship,

$\tau _{1} < \tau _{2} = \tau _{3} < \tau _{4}$

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Question

A man sits on the end of a long uniform metal beam of length . The man has a mass of and the beam has a mass of .

What is the magnitude of the net torque on the plank about the secured end of the beam? Use gravity .

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Answer

The net torque on the beam is given by addition of the torques caused by the weight of the man and the weight of the beam itself, each at its respective distance from the end of the beam:

$\tau _{net}=\tau _{man}+\tau _{beam}$

Let's assign the direction of positive torque in the direction of the torques of the man's and the beam's weights, noting that they will add together since they both point in the same direction.

$\tau _{net}=L\cdot W_{man}\cdot\sin 90^{\circ}+\left ( \frac{1}{2} L\right )\cdot W_{beam}\cdot\sin, 90^{\circ}$

$\tau _{net}=L\cdot m_{man}\cdot g+\left ( \frac{1}{2} L\right )\cdot m_{beam}\cdot g$

We can further simplify by combining like terms:

$\tau _{net}=L\cdot g \cdot (m_{man}+\frac{1}{2} m_{beam})$

Using the given numerical values,

$\tau _{net}=(3.05: m)\cdot(9.8: m/s^{2})\cdot(85: kg+\frac{1}{2} 40: kg)=3140: N\cdot m$

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