Small object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches to a maximum height of $\frac{3v^2}{4g}$ with respect to the initial position. The object is: (2013)
Solution:
Using energy conservation, initial kinetic energy equals potential energy at max height: $\frac{1}{2}mv^2 \left(1 + \frac{I}{mR^2}\right) = mgH$. Substituting $H = \frac{3v^2}{4g}$, we get $1 + \frac{I}{mR^2} = 2$, which gives $\frac{I}{mR^2} = 1$. This corresponds to a ring.
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