Kevin S. Huang

2010: Problem 13

Topic: Rigid Bodies
Concepts: Conservation of energy, General kinetic energy, Rolling motion

Solution:

Since the ball rolls without dissipation, we know it is rolling without slipping:

v=R\omega

Conserving energy between the top and bottom of the ramp,

Mgh=K_{\text{trans}}+K_{\text{rot}}=\frac{1}{2}Mv^2+\frac{1}{2}I\omega^2

Using $I=\frac{2}{5}MR^2$ and $v=R\omega$ ,

Mgh=\frac{1}{2}Mv^2+\frac{1}{2}\left(\frac{2}{5}MR^2\right)\frac{v^2}{R^2}=\frac{7}{10}Mv^2

v^2=\frac{10}{7}gh

When the ball is launched vertically, only the translational kinetic energy gets converted to gravitational potential energy since the ball keeps rotating:

K_{\text{trans}}=\frac{1}{2}Mv^2=Mgy

Using our earlier expression,

y=\frac{v^2}{2g}=\frac{5}{7}h

so the answer is D.

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Kevin S. Huang

2010: Problem 13

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2010: Problem 13

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