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A disc of mass 4.8 kg and radius 1 m is ...

A disc of mass 4.8 kg and radius 1 m is rolling on a horizontal surface without sliding with angular velocity of 600 rotations/min. What is the total kinetic energy of the disc ?

A

`1440 pi^(2)J`

B

`360J`

C

`600pi^(2)J`

D

`4000pi^(2)J`

Text Solution

Verified by Experts

The correct Answer is:
A
Total K.E. `(K)=K_("translational")+K_("rotational")`
`=(1)/(2)mv^(2)+(1)/(2)Iomega^(2)" …(1)"`
For the disc, `I=(1)/(2)mr^(2)`
and `omega=600" rotations/min "=(600)/(60)xx2pi" rad/s"`
`=20pi" rad/s"`
and `v=romega`
`therefore" From (1),"`
`therefore K=(1)/(2)mr^(2)omega^(2)+(1)/(2)((1)/(2)mr^(2).omega^(2))`
`=(3)/(4)mr^(2)omega^(2)=(3)/(4)xx4.8xx1^(2)xx400pi^(2)`
`therefore K=4.8xx3xx100=1440pi^(2)J`
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