23052\pi \sqrt{\frac{7(R-r)}{5g}}
Akash Anand Excellent work
A spherical ball of mass m and radius r rolls without slipping on a rough concave surface or large radius R. It makes small oscillations about the lowest point. Find the time period. (H.C. Verma Part-1 S.H.M.-Q. No:39)
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2 Answers
1133Total energy at an angle θ with the vertical
U = \frac{1}{2}mv^{2} + \frac{1}{2}Iw^{2} + mg(R - (R-r)cos\Theta )
Since total energy it conserved at any time, therefore
dUdt = 0
On differentiating and simplifying we get,
\frac{7}{5}va = -g(R-r)\sin \Theta \frac{d\Theta }{dt}
sinθ ≈ θ,
θ = xR-r
dθdt = vR-r
on substituting we get,
a = -\frac{5gx}{7(R-r)}
Therefore time period is 2Ï€√7(R-r)5g
- Akash Anand Excellent work Upvote·0· Reply ·2013-03-14 04:56:23