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A grindstone of moment of inertia 6 kg m...

A grindstone of moment of inertia `6 kg m^(2)` is found to have a speed of `150` rpm. `10 sec`, after starting from rest, torque applied is

A

`3pi` N m

B

3 N

C

`pi/3` N m

D

`4pi` N m

Text Solution

Verified by Experts

The correct Answer is:
A
Here, I=6kg `m^(2)`, t=10s, `omega_(0)=0`
v=150 rpm `=150/60 rps = 5/2 rps`
`omega=2piv= 2pi xx 5/2 = 5pi rads^(-1)`
`alpha = (omega-omega_(0))/(t) = (5pi-0)/(10) = pi/2 rad s^(-2)`
`therefore` Torque, `tau=I alpha= 6 xx alpha/2=3pi` Nm
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