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A spherical ball of mass m and radius r ...

A spherical ball of mass m and radius r rolls without slipping on a rough concave surface of large radius R. It makes small oscillations about the lowest point. Find the time period.

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Verified by Experts

The correct Answer is:
B

Let the angular velocity of the system about the point is suspension at any time be `omega`
`Sve=(R-r)omega`
`Again ve=romega`
`[where omega_1=`rotatioN/Al velocity of the sphere]
`1-v_c/r=((R-r)/r)omega` ………..1
By energy method total energy in SHM is constant.
`So, `mg(R-r)(1-costgheta)+/2mv_c^2+1/2Iomega^2`
=constant
`:. mg(R-r))1-costheta)+1/2m(R-r)^2omega^2`=constant
`:. mg(R-r)(1-costheta)+1/2m(R-r)^2omega^2+1/2mr^2((R-r)/r)omega^2=constant`:.mg(R-r)(1-costheta)+1/2m(R-r)^2omega^2+1/2mr^2((R-r)/r)omega^2`
=constant
`rarr g(R-r)(1-costheta)+(R-r)omega^I2[1/2+1/2]`
=costant
Taking derivative
`g(R-or)sintheta(dtheta)/(dt)=7/10 (R-r)^22omega (domega)/(dt)`
`rarr gsintheta=2xx(7/10)(R-r)alpha`
`implies gsin theta=(7/5)(R-r)alpha`
`alpha=(5gsintheta)/(7(R-r))`
`:. alpha/theta=omega^2`
`=(5g)/(7(R-r))=constant
So the motion is SHM again
`omega=sqrt((5g)/(7(R-r)))`
`rarr T=2pi sqrt((7(R-r))/(5g))`
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