Question

2. A grindstone increases in angular speed from 4.00 rad/s to 12.00 rad/s in 4.00 s. Through what angle does it turn during that time interval if the angular acceleration is con- stant? (a) 8.00 rad (b) 12.0 rad (c) 16.0 rad (d) 32.0 rad (e) 64.0 rad

Answer 1

Initial angular speed, $\omega _{0}=4rad/s$

final angular speed ,$\omega=12rad/s$

time,$t=4s$

First find angular acceleration

We know ,$v=u+at$

$\omega =\omega _{0}+\alpha t$

$12rad/s=4rad/s+\alpha *4s$

$12=4+4\alpha$

${\color{Red} \alpha =2rad/s^{2}}$

================

Now angle ,we know $s=ut+1/2at^{2}$

$\theta =\omega _{0}t+1/2\alpha t^{2}$

$\theta =4rad/s*4s+0.5*2rad/s^{2}*(4s)^{2}$

$\theta =4*4+0.5*2*4^{2}$

${\color{Red} \theta =32rad}$

ANSWER: Option (d) 32.0 rad

==========================