A solid uniform sphere of mass m and radius r is placed on...

A solid uniform sphere of mass, M, and radius, R, is placed on an inclined plane at a distance, h, from the base of the incline. The inclined plane makes an angle, θ, with the horizontal. The sphere is released from rest and rolls down the incline without slipping. The moment of inertia of the sphere is I = (2/5)MR^2. Determine the translational kinetic energy of the sphere when it reaches the bottom of the inclined plane. Determine the rotational kinetic energy of the sphere when it reaches the bottom of the inclined plane. Determine the linear acceleration of the sphere when it is on the inclined plane. Determine the friction force acting on the sphere when it is rolling down the inclined plane. If the solid sphere is replaced with a hollow sphere of identical mass and radius, how would that change the answer to question (b)? Explain.

Transcript

00:01 So here in this problem we have a sphere of mass big m and radius big r with moment of inertia which is equal to 2 over 5 m r squared.

00:12 This sphere is going downwards along and inclined the plane of height h and angle of inclination which is equal to theta.

00:23 So for part a we need to get the translational kinetic energy of this sphere.

00:30 At the bottom of the inclined plane.

00:33 So here in part a we use the energy conservation.

00:37 So here mghh which is the potential energy of the sphere at altitude h and this sphere and this sphere was at risk so it's both kinetic translational and the kinetic rotational and the kinetic rotational energy are zeros.

00:57 This is at the beginning of the motion at the end of the motion or at the bottom of the inclined plane we have kinetic energy translational plus the rotational the rotational kinetic energy which is one half i omega squared and the potential energy at the bottom is zero so here this is equal to one half mb squared which is the translational kinetic energy plus one half times in the moment of inertia which is two over five mr squared times omega squared which is here omega is the angular velocity and this is equal to v over r and here v is the linear velocity of the sphere so we cancel the masses from both side of the equation so here gh is equal to v squared over 2 plus v squared over 5 and this is equal to 7 over 10 v squared so here v squared is equal to 10 gh over 7 so the translational kinetic energy is equal to one half m times v squared which is equal to 10 over 7 gh so here the translational kinetic energy is 5 over 7 mgh now for part b we need to get the rotational kinetic energy at the bottom of the inclined plane so the rotational kinetic energy is equal to one half i which is the moment of inertia times times omega squared which is the angular velocity squared and this is equal to one half times two over five m r squared times v over r squared and this is equal to one half mv squared and this is equal to one one over five times m times ten over seven gh so here the rotational kinetic energy is equal to two over seven m gh for part c we need to get the linear acceleration of the sphere during its motion on the inclined plane.

03:01 So here this is a free body diagram of the sphere over the inclined plane.

03:07 So here the force due to gravity is m .g.

03:11 Sine theta...

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