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A wheel is at rest in the horizontal pos...

A wheel is at rest in the horizontal position. Its moment of inertia about a vertical axis passing through its centre is `"15 kgm"^(2)`. A constant torque of 300 N-m is now applied to it for 5 second. What is the change in its kinetic energy?

A

`3xx10^(4)J`

B

`6.5xx10^(4)J`

C

`7xx10^(3)J`

D

`7.5xx10^(4)J`

Text Solution

Verified by Experts

The correct Answer is:
D
`"Initial K.E"=(1)/(2)Iomega_(0)^(2)=0" "because omega_(0)=0`
The angular acceleration produced by the torque is
`alpha=(tau)/(I)=(300)/(15)="20 rad/s"^(2)" "(because tau=Ialpha)`
The angular velocity at the end of 5 s is given by
`omega=omega_(0)+alphat =0+20xx5="100 rad/s"`
`therefore" Its K.E. "=(1)/(2)Iomega^(2)=(1)/(2)xx15xx100xx100=7.5xx10^(4)J`
`therefore" Change in K.E. "=7.5 xx10^(4)J`
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