Chapter 10, Problem 1Q

(a)

To determine

The comparison of the magnitude of the tangential acceleration for the instances $a,b,c,$ and $d$ .

Answer to Problem 1Q

The magnitude of the tangential acceleration is maximum at an instant $c$ and then at an instant $a$ . And the tangential acceleration at instants $b$ and $d$ is zero.

Explanation of Solution

Given:

A disk rotating like a marry-go table.

And the angular velocity vs the time graph is

  

Formula used:

The tangential acceleration is directly proportional to the rate of the change in angular velocity and it is written as,

  $a_t=r\frac{d\omega }{dt}$

Here $d\omega$ is the change in angular velocity with respect to the time $t$ and $r$ is the radius of the circular path.

Calculation:

From the given figure,

The angular velocity is constant for instants $b$ and $d$ . So the tangential acceleration for instant $b$ and $d$ will be zero.

The magnitude of the change in angular velocity with respect to the time is more at instant $c$ as compared to instant $a$ . So, the tangential acceleration at instant $c$ is more as compared to instant $a$ .

Hence, the magnitude of the tangential acceleration is maximum at instant $c$ and then at instant $a$ . And the tangential acceleration at instants $b$ and $d$ is zero.

Conclusion:

The magnitude of the tangential acceleration is maximum at instant $c$ and then at instant $a$ . And the tangential acceleration at instants $b$ and $d$ is zero.

(b)

To determine

The comparison of the magnitude of the radial acceleration for the instances $a,b,c,$ and $d$ .

Answer to Problem 1Q

The radial acceleration is maximum at instant $b$ , then the radial acceleration is same at instants $a$ and $c$ . The radial acceleration is minimum or zero at instant $d$ .

Explanation of Solution

Given:

A disk rotating like a marry-go table.

And the angular velocity vs the time graph is

  

Formula used:

The radial acceleration is the product of the square of the angular velocity and the radius. So it is written as

  $a_r=r{\omega }^2$

Here $\omega$ is the angular velocity and $r$ is the radius of the circular path.

Calculation:

Since the radius is constant at every time. Only the angular velocity changes with time.

And the radial acceleration is

  $a_r=r{\omega }^2$

Here, $\omega$ is the angular velocity and $r$ is the radius of the circular path.

So, more angular velocity means more radial acceleration.

Now the angular velocity at instant $b$ is highest, so the radial acceleration will be maximum at instant $b$ .

The angular velocity at instant $d$ is minimum or it is $0$ , so the radial acceleration will be minimum or zero at instant $d$ .

The angular velocity at instants $a$ and $c$ is the same and it is non-zero, so the radial acceleration is the same at instants $a$ and $c$ . And it is more than the radial acceleration at instant $d$ and less than the radial acceleration at instant $b$ .

Conclusion:

The radial acceleration is maximum at instant $b$ , then the radial acceleration is same at instants $a$ and $c$ . The radial acceleration is minimum or zero at instant $d$ .