An object that rolls without slipping will have both translational and rotational kinetic energy. If it rolls down an incline, the gravitational potential, mgh (where h is the height of the incline at the top) is converted to both forms of kinetic energy: ?=1/2 ??2+1/2??2 where m is the mass of the object, v is its instantaneous velocity, I is the moment of inertia, and omega is the angular velocity. The relationship between the angular velocity and the tangential velocity is ?=?/? where R is the radius of the rolling object.  Using Energy Conservation and the equations above, solve for an algebraic expression for the final velocity squared, in terms of gravitational acceleration g, height of the incline h, and moment of inertia I.

Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter10: Rotational Motion
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Problem 18P: Rigid rods of negligible mass lying along the y axis connect three particles (Fig. P10.18). The...
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An object that rolls without slipping will have both translational and rotational kinetic energy. If it rolls down an incline, the gravitational potential, mgh (where h is the height of the incline at the top) is converted to both forms of kinetic energy: ?=1/2 ??2+1/2??2 where m is the mass of the object, v is its instantaneous velocity, I is the moment of inertia, and omega is the angular velocity. The relationship between the angular velocity and the tangential velocity is ?=?/?
where R is the radius of the rolling object. 

Using Energy Conservation and the equations above, solve for an algebraic expression for the final velocity squared, in terms of gravitational acceleration g, height of the incline h, and moment of inertia I. 

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