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A grindstone increases in angular speed ...

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 he angular acceleration is constant ?

A

8.00 rad

B

12.0 rad

C

16.0 rad

D

32.0 rad

Text Solution

Verified by Experts

The correct Answer is:
D
(d) The angular displacement will be
`Delta theta = omega_(avg) Delta t = ((omega_f + omega_i)/(2)) Delta t`
=`((12.00 rad//s + 4.00 rad//s)/(2))(4.00 s) = 32.0 rad`.
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