Question

A disc, starting from rest, undergoes a constant angular acceleration for a time t_1 and reaches an angular velocity of omega_1. It then undergoes a constant negative acceleration and comes to rest in a time t_2. Find an expression in terms of the quantities t_1, t_2, and omega_1 for the total number of revolutions of the disc in the time t_1 + t_2.

Answer 1

Step 1: Find revolutions made by disk when speeding up

Using 1st rotational kinematic equation:

w1 = w0 + *t1

01

Given that disk starts from rest, So w0 = 0 rad/sec

w1 = 0 + *t1

01

= w1/t1 = angular acceleration of disk when speeding up

01

Now Using 2nd rotational kinematic equation:

$\theta_1$ = w0*t1 + (1/2)**t1^2

01

$\theta_1$ = 0*t1 + (1/2)*(w1/t1)*t1^2

$\theta_1$ = (w1*t1)/2

Step 2: Now find number of revolutions when it undergoes negative constant angular acceleration

Using 1st rotational kinematic equation:

w2 = w1 - *t2

02

Given that disk finally stops, So w2 = 0 rad/sec

0 = w1 - *t2

02

= w1/t2 = angular de-acceleration of disk when speeding down

02

Now Using 2nd rotational kinematic equation:

$\theta_2$ = w1*t2 - (1/2)**t2^2

02

$\theta_2$ = w1*t2 - (1/2)*(w1/t2)*t2^2

$\theta_2$ = (w1*t2)/2

So total angular displacement of disks will be:

$\theta$ = $\theta_1$ + $\theta_2$

$\theta$ = (w1*t1)/2 + (w1*t2)/2

$\theta$ = w1*(t1 + t2)/2

Let me know if you've any query.