AP Physics C: Electricity and Magnetism - Using Torque Equations

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?

Explanation

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 ,

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

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 .

Ps1 gymnast

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

Explanation

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,

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

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?

Explanation

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 $2\pi$ :

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

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 ?

$I_{new}=\frac{1}{64}I_{0}$

$I_{new}=\frac{1}{4}I_{0}$

$I_{new}=\frac{1}{12}I_{0}$

$I_{new}=\frac{1}{16}I_{0}$

$I_{new}=\frac{1}{192}I_{0}$

Explanation

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,

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

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.

$16: rad/s^{2}$

$64: rad/s^{2}$

$7.9: rad/s^{2}$

$24: rad/s^{2}$

Explanation

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,

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

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.

$15: rad/s^{2}$

$60: rad/s^{2}$

$45: rad/s^{2}$

$30: rad/s^{2}$

Explanation

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,

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

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 .

$3,140: N\cdot m$

$1,940: N\cdot m$

$3,740: N\cdot m$

$1,350: N\cdot m$

Explanation

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,

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

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

Ps1 square

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

Explanation

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.

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 .

Ps1 wind

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

Explanation

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,

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

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.

Ps1 wrench

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

$14: N\cdot m;out;of;the;page$

$14: N\cdot m;into;the;page$

$14: N\cdot m;downward$

$6.5: N\cdot m;out;of;the;page$

$6.5: N\cdot m;downward$

Explanation

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.

Using Torque Equations

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