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A wheel rotating at an angular speed of ...

A wheel rotating at an angular speed of 20 rad/s ils brought to rest by a constant trouque in 4.0 secons. If the moiment of inertia of the wheel about the axis of rotation is 0.20 `kg-m^2` find the work done by the torque in the first two seconds.

A

10J

B

20J

C

30J

D

40J

Text Solution

Verified by Experts

The correct Answer is:
C

`omega_(1)= 20` rad/sec, `omega_(2)= 0, t= 4 sec`. So angular retardation `a = (omega_(1)- omega_(2))/(t)= (20)/(4)= 5 "rad"//sec^(2)`
Now angular speed after 2 sec `omega_(2)= omega_(1)- alpha t= 20-5 xx 2= 10` rad/sec
Work done by torque in 2 sec = loss in kinetic energy `= (1)/(2) I (omega_(1)^(2)- omega_(2)^(2))= (1)/(2) (0.20) ((20)^(2) - (10)^(2))`
`=(1)/(2) xx 0.2 xx 300= 30J`
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