Question

10 Example 7: A ball of mass, m, and radius, r, is released from the very edge of a hemispherical bowl of radius, R. What is the speed of the center of the ball when it gets to the bottom of the bowl.

Answer #1

Initially only Potential energy (PE) is there in ball of mass m , which is at top of hemisphere at height R from base level

Initial Total mechanical Energy Ki= PE = mgR

When the ball reaches bottom of hemisphere it has Kinetic energy (KE)of its center of mass and rotational KE about center of mass ( assuming that it is in " pure rolling " motion )

Final Total mechanical Energy = Kf = 0.5 mVcm^2 + 0.5 I

= 0.5 m Vcm^2 + 0.5 (2/5) mr^2( )

= 0.5 m Vcm^2 + 0.5 (2/5) mr^2( Vcm^2/r^2)

= 0.5 m Vcm^2 + 0.5 (2/5) m Vcm^2

= 0.7 mVcm^2

So by conservation of energy

mgR = 0.7 m Vcm^2

**Vcm = Sqrt ( gR/0.7)**

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