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Given V("CM")=2m/s, m= 2kg,R = 4m ...

Given `V_("CM")=2m/s, m= 2kg,R = 4m`

Find angular momentum of ring about origin if it is in pure rolling

A

`32kgm^(2)//s`

B

`24kgm^(2)//s`

C

`16kgm^(2)//s`

D

`8kgm^(2)//s`

Text Solution

Verified by Experts

The correct Answer is:
A
`L=I_("CM").omega+mV_("CM")r_(1)d`
`omega=(v)/(R)=(2)/(4)="0.5 rad/sec"`
`vecL=MR^(2).omega+MV_("CM").R`
`=2xx16xx0.5+2xx2xx4=16+16`
`vecL=32(kgxxm^(2))/(s)`
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