# Terms in kinetic energy

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A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rolls
down a smooth surface to the ground, then climbs up another hill of height 30 m and
finally rolls down to a horizontal base at a height of 20 m above the ground. What is the
velocity attained by the ball .

I know to solve this problem we have to conserve energy .
That js potential energy = kinetic energy
But my doubtvis that in kinetic energy do we both of these ($\frac {1}{2}mv^2 + \frac {1}{2}I\omega^2$) or only $\frac {1}{2}mv^2$

asked Mar 2, 2017
Please can you explain the reason for your doubt? Which do you think it should be and why do you doubt it?
As it is rolling so we should consider first case(Transational KE + Rotational KE) . But in the answer it is given second case .
"The velocity attained by the ball" means only its final translational velocity. But to find this you need to consider rotational KE as well as translational KE (and PE of course).

If this does not answer your difficulty please upload the answer.
That solution takes no account of rotational KE, so I think it must be incorrect. None of the other options is correct either.

I do not like the problem anyway. If the surface is smooth the ball cannot roll, it can only slide. The answer treats the ball as though it slides, but the question twice says that the ball "rolls". Such inconsistencies are a sign that the problem has not been checked thoroughly.

## 1 Answer

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You are correct. A rolling sphere gains rotational as well as translational KE as it falls. When it arrives at its destination its translational KE of the CM is not equal to its loss of PE, because some of this energy has been converted into rotational motion about the CM.

The given solution is not correct. It treats the sphere as though it slid the whole way, although the problem statement says twice that it rolls.

answered Mar 4, 2017 by (27,948 points)
selected Mar 4, 2017 by koolman