Question

A solid ball rolls down a parabolic path ABC from a height h as shown in figure. portion AB of the path is rough while BC is smooth .How high will the ball climb in BC.

Answer 1

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At B , total kinetic energy = mgh

Here,

m = mass of ball

The ratio of rotational to kinetic energy would be ,

Kr/Kt = 2/5

where, Kr = 2/7mgh

and Kt = 5/7 mgh

In portion BC, friction is absent . Therefore, rotational K.E will remain constant and Translational K.E will convert into potential energy.

Hence, if H be the height to which ball climbs in BC, then

mgH = Kt

mgH = 5/7mgh

H = 5/7h

happy to help !

:)

Answer 2

Explanation:

At B , total kinetic energy = mgh

Here,

m = mass of ball

The ratio of rotational to kinetic energy would be ,

Kr/Kt = 2/5

where, Kr = 2/7mgh

and Kt = 5/7 mgh

In portion BC, friction is absent . Therefore, rotational K.E will remain constant and Translational K.E will convert into potential energy.

Hence, if H be the height to which ball climbs in BC, then

mgH = Kt

mgH = 5/7mgh

H = 5/7h

happy to help !