A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rad/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to:
A) 17 s B) 15 s C) 19 s D) 21 s E) 23 s

Question

A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rad/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to:
A) 17 s B) 15 s C) 19 s D) 21 s E) 23 s

raffle_snaffle · Answer

Use kinematics with angular velocity.

shamim · Answer

use
angular acceleration =(final angular velocity -initial angular velocity) /time

shamim · Answer

anyway welcome to openstudy

rachvanschaick · Answer

But how?? Is it possible to see the problem worked out?

irishboy123 · Answer

you will know the equations of motion for **linear** motion, eg one simple version is 

$v = u + at$
$x = u t + \frac{1}{2} a t^2$
$v^2 = u^2 + 2 a x$

And if this were about a machine moving in a line I am sure you would have no problems


For **rotary** motion, these equations have analogues:

$\omega_2 = \omega_1 + \alpha t$
$\theta  = \omega_1 t + \frac{1}{2} \alpha t^2$
$\omega_2^2 = \omega_1^2 + 2 \alpha \theta$

look at the pattern matching, eg $a \to \alpha$, to see what has happened.  We're just switching letters around :-)  the big idea is still motion with constant [here, angular] acceleration.