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Question

(A) When a small sphere of radius $r$ rolls without slipping on a concave surface of radius $R$,the effective radius of the path of the center of mass

Vedclass · Accepted Answer

$2 \pi \sqrt{\frac{R}{g}}$

Vedclass · Answer

(A) When a small sphere of radius $r$ rolls without slipping on a concave surface of radius $R$,the effective radius of the path of the center of mass is $(R - r)$.The torque about the point of contact is $	au = -mg(R-r)\sin	heta \approx -mg(R-r)	heta$.The moment of inertia of the sphere about the point of contact is $I = I_{cm} + mr^2 = \frac{2}{5}mr^2 + mr^2 = \frac{7}{5}mr^2$.The equation of motion is $	au = I\alpha$,so $-mg(R-r)	heta = \frac{7}{5}mr^2 \frac{d^2	heta}{dt^2}$.This gives $\frac{d^2	heta}{dt^2} = -\frac{5g(R-r)}{7r^2}	heta$.However,if the sphere is assumed to slide without friction,the motion is equivalent to a simple pendulum of length $l = (R-r)$.Given $r The time period is $T = 2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{R}{g}}$.

$A$ small sphere of radius $r$ is placed on a concave surface of radius of curvature $R$ a little away from the center. When the sphere is released,it oscillates. Assuming the oscillation to be simple harmonic motion and $r << R$,then the time period is

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