Chapter 4.3, Problem 1aT

A T-shaped board of uniform mass density has two small holes as shown. Initially, the pivot is placedthrough the right hole, which corresponds to the center of mass of the board. The board ¡s then held in place.

1. Predict the motion of the board after it is released from rest, Explain.

2. Check your prediction by observing the demonstration.

a. Describe the angular acceleration of the board. Explain how you can tell.

What does your answer imply about the ne, torque about the pivot? Explain.

b. Describe the acceleration of the center of mass of the board. Explain how you can tell.

What does your answer imply about the net force acting on the board? Explain.

3. Explain how your answers about net torque and net force in question 2 would change, if at all, if there is appreciable friction between the board and the pivot and the board remains at rest.

To determine

To Predict: The motion of the board.

Answer to Problem 1aT

The board will perform a circular repetitive motion.

Explanation of Solution

Given:

Gravitational force is acting downward on the edge of the handle.

  

Formula used:

  $\tau =r\times F=r\times mg$

Calculation:

  $\begin{array}{l}\tau =r\times F\\ \Rightarrow \tau =r\times mg\\ \Rightarrow \tau =rmgsin\theta \left(\theta =90°\right)\\ \Rightarrow \tau =rmg\end{array}$

Conclusion:

The gravitational force at the edge of the handle leads to produce torque using equation in formula. r is the distance between the edge of the handle and the pivot around which the board will rotate. Angle between r and force is 90 degree.

To determine

To Describe: The angular acceleration of the board, acceleration of the center of mass and the net torque about the pivot.

Explanation of Solution

Given:

  $\tau =rmg$

  

Formula used:

  $\tau =I\alpha$

Where, I is the moment of inertia and $\alpha$ is the angular acceleration.

Calculation:

From part 1, the net torque is $\tau =rmg$

From the rotational equation of motion:

  $\tau =I\alpha$

  $\alpha =\frac{d\omega }{dt}$

On comparing both the equations:

  $\tau =rmg=I\alpha$

  $\alpha =\frac{rmg}{I}$

The center of mass of the board lies on the axis of rotation, therefore, the acceleration of the center of mass is zero.

The net torque is $\tau =I\alpha$ and the net force is mg .

Conclusion:

Hence, the angular acceleration is $\alpha =\frac{rmg}{I}$ . The acceleration of the center of mass of the board is zero as there is no translation motion of the board.

To determine

The effect of friction on the torque and force.

Answer to Problem 1aT

The net torque and force will be reduced due to friction in effect.

Explanation of Solution

Given:

Gravitational force is acting downward on the edge of the handle.

Formula used:

  $\tau =r\times F-frictionalforce$

  $F_{net}=mg-frictionalforce$

Calculation:

  $\begin{array}{l}\tau =r\times F=r\times mg\\ \tau =rmgsin\theta -\mu F_f\\ \tau =rmg-\mu F_f\end{array}$

Conclusion:

The frictional force acts opposite to the motion of the object. Hence, net force and torque are reduced which eventually reduces the angular acceleration and linear acceleration.