Chapter 9, Problem 1P

(a)

To determine

The point moves a greater distance in a given time, if the disk rotates with increasing angular velocity.

Answer to Problem 1P

The particle on the rim covers a greater linear distance as compared to the particle situated half way between the rim and the axis of rotation.

Explanation of Solution

Given:

A disk rotating with increasing angular velocity about an axis passing through its center and perpendicular to its plane.

Distance of the particle 1 (on the rim) from the axis of rotation

  $r_1=r$

Distance of particle 2 (half way between the axis of rotation and the rim) from the axis of rotation

  $r_2=\frac{r}{2}$

Here, the radius of the disk is $r$

Formula used:

The angular acceleration $\alpha$ of the disk is given by

  $\alpha =\frac{d\omega }{dt}$ ……(1)

Here, $\omega$ is the angular velocity of the disk.

The angular acceleration of a point on the rotating disk is related to the linear acceleration $a$ as,

  $a=r\alpha$……(2)

Here, $r$ is the distance of the point from the axis of rotation.

The linear distance $\Delta x$ travelled by a particle in a given time interval $t$ is given by

  $\Delta x=v_{0}t+\frac{1}{2}a{t}^2$……(3)

Calculations:

The disk of radius $r$ rotates with increasing angular velocity about an axis passing through its center and perpendicular to the plane. Point 1 is located at its rim and point 2 is located at a distance $\frac{r}{2}$

from the axis of rotation. This is shown in the figure 1 below:

  

  Figure 1

If the disk rotates with an angular acceleration $\alpha$ , both points 1 and 2 have the same angular acceleration and they would have the same angular displacement during any time interval considered.

Calculate the linear accelerations of the particles located at point 1 and 2 using the equation (2).

  $\begin{array}{l}{a}_1=r_1\alpha \\ a_2=r_2\alpha \end{array}$……(4)

Substitute $r_1=r$ and $r_2=\frac{r}{2}$ in equation (4).

  $\begin{array}{c}{a}_1=r_1\alpha \\ =r\alpha \end{array}$

  $\begin{array}{c}{a}_2=r_2\alpha \\ =\frac{r}{2}\alpha \end{array}$

Therefore, from the above equations,

  $a_2=\frac{1}{2}{a}_1$……(5)

If the disk is assumed to start from rest, both particles would start with their initial velocities $v_0$ which would be equal to zero.

Use equation (3) to calculate the distance travelled by the two points.

  $\begin{array}{c}\Delta x_1=\frac{1}{2}{a}_1{t}^2\\ \Delta x_2=\frac{1}{2}{a}_2{t}^2\end{array}$

From equation (5),

  $\Delta x_1=2\Delta x_2$

Conclusion:

Therefore, the point on the rim travels a greater distance when compared to the point located halfway between the rim and the axis of rotation.

(b)

To determine

The point that turns through a greater angle.

Answer to Problem 1P

Both the points turn through the same angle.

Explanation of Solution

Formula used:

The angular displacement $\Delta \theta$ of a particle in a time interval $t$ is given by

  $\Delta \theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$…..(6)

Here ${\omega }_0$ is the initial angular velocity of the particle.

Calculation:

The points 1 and 2 located at points A and B on the disc rotating with an angular acceleration $\alpha$ in the clockwise direction move to positions $A\text{'}\text{ and }B\text{'}$ in the time $t$ as shown in Figure 2.

  

  Figure 2

At any instant of time, both particles have the same instantaneous angular velocity and angular acceleration. As it can be seen from Figure 2, both particles describe the same angle at an instant of time.

Conclusion:

Thus, the particle located at the rim and the particle located half way between the rim and the axis of rotation turn through the same angle.

(c)

To determine

The point which travels with greater speed.

Answer to Problem 1P

The point on the rim travels with greater speed.

Explanation of Solution

Formula used:

The instantaneous speed $v$ of a point located at a distance $r$ from the axis of rotation is given by

  $v=r\omega$……(7)

Here, $\omega$ is the instantaneous angular velocity.

Calculation:

The disk moves with increasing angular velocity. But, both points at any instant would have the same instantaneous angular velocity, since they turn through the same angle in a given interval of time.

Hence, it can be inferred from equation (7):

  $v\propto r$

Since the point on the rim has the greater value of $r$ when compared to the point located inside, it would have a greater speed.

Conclusion:

Thus, the point on the rim would have a greater speed when compared to the point located half way between the rim and the axis of rotation.

(d)

To determine

The point which has the greater angular speed.

Answer to Problem 1P

Both the particles have the same angular speed.

Explanation of Solution

Formula used:

The angular velocity of a particle is given by

  $\omega =\frac{d\theta }{dt}$……(8)

Calculation:

From Figure 2, it is seen that at any instant of time, both particles 1 and 2 cover the same angles. Hence, the rate of change of their angular displacement $\frac{d\theta }{dt}$ will be the same. Thus, it follows that both the points will have the same angular velocity at any instant of time, according to equation (8).

Conclusion:

Thus, the particle on the rim and the particle located halfway between the rim and the axis of rotation have the same angular velocity.

(e)

To determine

The point which has the greater tangential acceleration.

Answer to Problem 1P

The point on the rim has a greater tangential acceleration when compared to the point located midway between the rim and the axis of rotation:

Explanation of Solution

Formula used:

The tangential acceleration $a_t$ of a point on a rotating disk located at a distance $r$ from the axis of rotation is given by

  $a_t=r\alpha$……(9) Here, $\alpha$ is the angular acceleration of the disk.

Calculation:

If the disk rotates with a varying angular velocity, it has angular acceleration. Assuming that the angular acceleration of the disk remains constant, from equation (9) it can be inferred that

  $a_t\propto r$

The point on the rim has the greater value of $r$ when compared to the point located halfway between the rim and the axis of rotation.

Therefore, the point on the rim has a greater tangential acceleration when compared to any point located inside the rim. This is also proved by the fact that the point on the rim gains a larger tangential velocity when compared to any inner point.

Conclusion:

Thus, the point on the rim has a greater tangential acceleration when compared to the point located midway between the rim and the axis of rotation.

(f)

To determine

The point which has a greater angular acceleration.

Answer to Problem 1P

Both particles have the same angular acceleration.

Explanation of Solution

Formula used:

The angular acceleration $\alpha$ of the disk is given by

  $\alpha =\frac{d\omega }{dt}$……(1)

Calculation:

It has been proved in (d) that at any instant of time, both the particles have the same angular velocity. Therefore, in an interval of time $dt$ , the change in their angular velocities $d\omega$ will also be equal.

Hence, from equation (1), it can be proved that at a given instant of time, both the points will have the same angular accelerations.

Conclusion:

Thus, both particles are found to have the same angular acceleration.

(g)

To determine

The point which has the greater centripetal acceleration.

Answer to Problem 1P

The point on the rim has a greater centripetal acceleration.

Explanation of Solution

Formula used:

The centripetal acceleration of a point located at a distance $r$ from the axis of rotation is given by

  $a_r=r{\omega }^2$……(10)

Calculation:

It has been established in part (d) that at any instant of time, the point on the rim and the point located halfway between the rim and the axis of rotation have the same angular velocity.

Therefore, from equation (10), it can be inferred that

  $a_r\propto r$

The point on the rim has the greater value of $r$ when compared to the point located in between the rim and the axis of rotation. Therefore, the centripetal acceleration of the point on the rim would be greater than the centripetal acceleration of the point located in between the rim and the axis of rotation.

Conclusion:

Thus, the point on the rim has a greater centripetal acceleration.