Using Torque Equations - AP Physics C: Electricity and Magnetism

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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. A force of is applied perpendicularly to its length at a distance from the nailed end.

Calculate the resulting angular acceleration experienced by the meter stick.

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Answer

For a net torque applied to an object free to rotate,

$\tau _{net}=I\alpha$

The net torque for this problem is supplied by the force applied at a distance from the pivot. The torque of a force is calculated by

$\tau =Fr\sin, \theta$

The moment of inertia for a long uniform thin rod about its end is

$I_{rod,end}=\frac{1}{3}mL^{2}$

Combining above,

$Fr\sin90^{\circ}=(\frac{1}{3}mL^{2})\alpha$

Solving for the angular acceleration,

$\alpha =\frac{Fr\sin, 90^{\circ}}{\frac{1}{3}mL^{2}}=\frac{16N\cdot0.67: m\cdot1}{\frac{1}{3}\cdot2, kg\cdot(1:m)^{2}}=16: rad/s^{2}$

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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

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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Question

Two children sit on the opposite sides of a seasaw at a playground, doing so in a way that causes the seasaw to balance perfectly horizontal. The child on the left is from the pivot.

What is the mass of the second child if she sits from the pivot?

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Answer

A torque analysis is appropriate in this situation due to the inclusion of distances from a given pivot point. Generally,

$\tau_{net}=I\alpha$

This is a static situation. There are two torques about the pivot caused by the weights of two children. We will note that these weights cause torques in opposite directions about the pivot, such that

$\tau _{net}=0=\tau _{1}-\tau _{2}$

Consequently,

$\tau _{1}=\tau _{2}$

$r_{1}\cdot m_{1} \cdot g \cdot \sin: 90^{\circ}=r_{2}\cdot m_{2}\cdot g \cdot \sin: 90^{\circ}$

Or more simply,

$r_{1}\cdot m_{1}=r_{2} \cdot m_{2}$

Solving for ,

$m_{2}=\frac{m_{1}r_{1}}{r_{2}}=\frac{36: kg\cdot1.8: m}{2.4: m}}=27: kg$

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Question

A gymnast is practicing her balances on a long narrow plank supported at both ends. Her mass is . The long plank has a mass of .

Calculate the force the right support provides upward if she stands from the right end. Use gravity .

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Answer

A torque analysis is appropriate in this situation due to the inclusion of distances from a given pivot point. Generally,

$\tau_{net}=I\alpha$

This is a static situation. As such, any pivot point can be chosen about which to do a torque analysis. The quickest way to the unknown force asked for in the question is to do the torque analysis about the left end of the plank. There are three torques about this pivot: two clockwise caused by the weights of the gymnast and the plank itself, and one counterclockwise caused by the force from the right support. Designating clockwise as positive,

$\tau _{net}=0=\tau _{gymnast}+\tau _{plank}-\tau _{right}$

Consequently,

$\tau _{gymnast}+\tau _{plank}=\tau _{right}$

$r _{gymnast}\cdot W_{gymnast}\cdot\sin: 90^{\circ}+r _{plank}\cdot W_{plank}\cdot \sin: 90^{\circ}=r _{right}\cdot F_{right}\cdot \sin: 90^{\circ}$

This simplifies to

$r _{gymnast}\cdot m_{gymnast}\cdot g+r _{plank}\cdot m_{plank}\cdot g=r _{right}\cdot F_{right}$

Solving for $F_{right}$ ,

$F_{right} = \frac{(r _{gymnast}\cdot m_{gymnast}+r _{plank}\cdot m_{plank})\cdot g}{r _{right}}$

Solving with numerical values,

$F_{right} = \frac{(2: m\cdot 55: kg+3: m\cdot 23: kg)\cdot 9.8: m/s^{2}}{6: m}=290: N$

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Question

A square of side lengths and mass is shown with possible axes of rotation.

Which statement of relationships among the moments of inertia is correct?

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Answer

The moments of inertia for both axes and are equal because both of these axes are equivalently passing through the center of the mass.

By the parallel axis theorem for moments of inertia ( $I = I_{cm}+ MD^2$ ), the moment of inertia for axis is larger than or

because it is located a distance

$\frac{L}{2}$ away from the center of mass.

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Question

A wind catcher is created by attaching four plastic bowls of mass each to the ends of four lightweight rods, which are then secured to a central rod that is free to rotate in the wind. The four lightweight rods are of lengths , , , and .

Calculate the moment of inertia of the four bowls about the central rod. You may assume to bowls to be point masses.

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Answer

The moment of inertia for a point mass is .

To calculate the total moment of inertia, we add the moment of inertia for each part of the object, such that

$I_{total}=I_{1}+I_{2}+I_{3}+I_{4}$

$I_{total}=m_{1}R_{1}^{2}+m_{2}R_{2}^{2}+m_{3}R_{3}^{2}+m_{4}R_{4}^{2}$

The masses of the bowls are all equal in this problem, so this simplifies to

$I_{total}=m (R_{1}^{2}+R_{2}^{2}+R_{3}^{2}+R_{4}^{2})$

Plugging in and solving with numerical values,

$I_{total}=(0.6: kg) [(0.8: m)^{2}+(0.6: m)^{2}+(0.4: m)^{2}+(0.2: m)^{2}]=0.72: kg\cdot m^{2}}$

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Question

A long uniform thin rod of length has a mass of .

Calculate the moment of rotational inertia about an axis perpendicular to its length passing through a point from one of its ends.

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Answer

For a long thin rod about its center of mass,

$I_{cm}=\frac{1}{12}mL^{2}$

According to the parallel axis theorem,

$I_{//}=I_{cm}+md^{2}$

where is defined to be the distance between the center of mass of the object and the location of the axis parallel to one through the center of mass. For this problem, is the difference between the given distance and half the length of the rod.

Combining the above,

$I=\frac{1}{12}mL^{2}+m(\frac{1}{2}L-l)^{2}$

Inserting numerical values,

$I=\frac{1}{12}(12)(1.82)^{2}+(12)(\frac{1.82}{2}-.036)^{2}=6.9: kg\cdot m^{2}$

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Question

The moment of inertia of a long thin rod about its end is determined to be .

What is the new value if the mass and length of the rod are both reduced by a factor of ?

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Answer

The moment of inertia for a long uniform thin rod about its end is given by

$I_{rod,end}=\frac{1}{3}mL^{2}$

Reducing the mass and the length by a factor of four introduces the following factors into the equation,

$I_{new}=\frac{1}{3}(\frac{1}{4}m)(\frac{1}{4}L)^{2}=\frac{1}{3}(\frac{1}{4})m(\frac{1}{64})L^{2}$

Simplifying,

$I_{new}=(\frac{1}{64})\frac{1}{3}mL^{2}=\frac{1}{64}I_{0}$

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Question

An art sculpture comprised of a long hollow painted cylinder is designed to pivot about an axis located along its long edge. The cylinder has mass , radius , and length .

Calculate the moment of inertia of the sculpture about this axis.

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Answer

The moment of inertia for a hollow cylinder about its center is given by

$I_{cyl,hollow}=mR^{2}$

The parallel axis is applied here because the pivot axis is located on the outside edge of the cylinder, a distance from the center of mass:

$I=I_{cm}+md^{2}=mR^{2}+mR^{2}=2\cdot mR^{2} = 2\cdot234: kg\cdot(2.2: m)^{2}=2,265: kg\cdot m^{2}$

Incidentally, for this problem, the length does not contribute to the moment of inertia.

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Question

A meter stick is secured at one end so that it is free to rotate in a vertical circle. It is held perfectly horizontally, and released.

Calculate the instantaneous angular acceleration the moment it is released.

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Answer

For a net torque applied to an object free to rotate,

$\tau _{net}=I\alpha$

The net torque for this problem is supplied by the force applied at a distance from the pivot. The torque of a force is calculated by

$\tau =Fr\sin, \theta$

The moment of inertia for a long uniform thin rod about its end is

$I_{rod,end}=\frac{1}{3}mL^{2}$

Combining above,

$Fr\sin90^{\circ}=(\frac{1}{3}mL^{2})\alpha$

where the force causing the torque is the meter stick's weight, which is applied at the center of mass of the meter stick - a distance of half the length away from its end.

Solving for the angular acceleration,

$\alpha =\frac{Fr\sin, 90^{\circ}}{\frac{1}{3}mL^{2}}=\frac{W\cdot\frac{1}{2}L\cdot \sin, 90^{\circ}}{\frac{1}{3}mL^{2}}$

Solving numerically,

$\alpha =\frac{(2: kg\cdot 9.8: m/s^{2})\cdot 0.5: m\cdot 1}{\frac{1}{3}\cdot 2, kg\cdot (1:m)^{2}}=15: rad/s^{2}$

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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?

Tap to reveal answer

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?

Tap to reveal answer

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 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?

Tap to reveal answer

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 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?

Tap to reveal answer

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 .

Tap to reveal answer

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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Question

Two children sit on the opposite sides of a seasaw at a playground, doing so in a way that causes the seasaw to balance perfectly horizontal. The child on the left is from the pivot.

What is the mass of the second child if she sits from the pivot?

Tap to reveal answer

Answer

A torque analysis is appropriate in this situation due to the inclusion of distances from a given pivot point. Generally,

$\tau_{net}=I\alpha$

This is a static situation. There are two torques about the pivot caused by the weights of two children. We will note that these weights cause torques in opposite directions about the pivot, such that

$\tau _{net}=0=\tau _{1}-\tau _{2}$

Consequently,

$\tau _{1}=\tau _{2}$

$r_{1}\cdot m_{1} \cdot g \cdot \sin: 90^{\circ}=r_{2}\cdot m_{2}\cdot g \cdot \sin: 90^{\circ}$

Or more simply,

$r_{1}\cdot m_{1}=r_{2} \cdot m_{2}$

Solving for ,

$m_{2}=\frac{m_{1}r_{1}}{r_{2}}=\frac{36: kg\cdot1.8: m}{2.4: m}}=27: kg$

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