Question

A small block oscillates back and forth on a smooth concave surface of radius R. Find the time period of small oscillation.

Answer 1

For the oscillation of small block on a smooth concave surface, following figure can be drawn, where R is the radius of the concave surface.

`therefore` Restoring torque `tau=-mg R "sin"theta`
As `I=mR^(2)implies tau=-mgRtheta `(since sin `theta~= theta`)
`therefore Ialpha=-mgRthetaimpliesalpha=-(mgR)/(I)theta=-omega^(2)theta`
Time period of such oscillation is
`T=2pisqrt((I)/(mgR))=2pisqrt((mR^(2))/(mgR))impliesT=2pisqrt((R)/(g))`