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A uniform ridig rod of mass m & length I...

A uniform ridig rod of mass m & length I is released from vertical position on rough surface with sufficient friction for lower end not to slip as shown in figure. When rod makes angle `60^(@)` with vertical then find correct altrnatice/s

A

`alpha = (2g)/(l)`

B

`omega = sqrt((3g)/(2l))`

C

`N=(mg)/(16)`

D

`a_("radial") "of centre"=(3g)/(4)`

Text Solution

Verified by Experts

The correct Answer is:
B, C, D

`Delta K + Delta U =0`
`(1)/(2)I_(0)omega^(2)= -Delta U`
`(1)/(2)(ml^(2))/(3)omega^(2)= -((-mg)(l)/(4))`
`omega = sqrt((3g)/(2l))`
`rArr a_("radial")=omega^(2)(l)/(2)=(3g)/(2l)(l)/(2)=(3g)/(4)`
`rArr tau = I_(0) alpha`
`alpha = ((mg)(l)/(2)sin60^(@))/(m(l^(2))/(3))=(3sqrt(3)g)/(4l)`
`rArr a_(v)=(alpha (l)/(2))sin 60^(@) + omega^(2)(l)/(2)cos 60^(@)`
`a_(v)= (3sqrt(3)g)/(8)(sqrt(3))/(2)+(3g)/(8)`
`a_(v) = (9g)/(16) + (6g)/(16)`
`mg-N=ma_(v)`
`N = (mg)/(16)`
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