Question

The tub of a washer is in a spin-dry cycle, running at a constant angular speed of 77 rad/s for a duration of 10s. After 10s, the lid is opened and the safety switch turns off the washer. The tub slows down to rest in 5s. Find the total angular displacement (for entire 15s) in

(a) radians =

(b) revolutions =

Answer 1

Part (a)

Given that washer rotates at 77 rad/s for first 10 sec, So

Angular displacement during first 10 sec will be:

0 = w0*t0

w0 = Initial angular speed of washer = 77 rad/sec

t0 = time interval = 10 sec

0 = (77 rad/s)*10 s = 770 rad

Now after than when washer is turned off then using 1st rotational kinematic equation:

w = w0 + *t1

t1 = time taken by washer to stop = 5 sec

w = final angular velocity = 0 rad/s

So,

= angular acceleration = (0 - 77)/5 = -15.4 rad/sec^2

Now Using 2nd rotational kinematic equation:

1 = w0*t1 + (1/2)**t1^2

1 = 77*5 + (1/2)*(-15.4)*5^2

1 = angular displacement in last 5 sec = 192.5 rad

So, total angular displacement will be:

= total angular displacement = 0 + 1

= 770 + 192.5

= 962.5 rad

Part B.

Now we know that 1 rev = 2*pi rad, So

= 962.5 rad*(1 rev/(2*pi rad)) = 962.5/(2*pi) rev

= 153.186633 rev

= total angular displacement in revolutions = 153.2 rev

Let me know if you've any query.