A 2.4-kg cylinder (radius = 0.08 m, length = 0.50 m) is released from rest at th

Question

A 2.4-kg cylinder (radius = 0.08 m, length = 0.50 m) is released from rest at thetop of a ramp and allowed to roll without slipping. The ramp is0.66 m high and 5.0 m long. (a) When the cylinder reaches the bottom of theramp, what is its total kinetic energy? (b) What is its rotational kinetic energy?
(c) What is its translational kinetic energy?
(a) When the cylinder reaches the bottom of theramp, what is its total kinetic energy? (b) What is its rotational kinetic energy?
(c) What is its translational kinetic energy?

Explanation / Answer

mass m = 2.4 kg radius r = 0.08 m length L = 0.5 m height h = 0.66 m Length of the ramp L ' = 5 m sin = h / L '         = 0.132 angle of inclination with horizontal = 7.58degrees (a). total kinetic energy E = mgh   Since from law of conservation of energy,potential energy at top = kinetic energy at bottom So, E = 15.5232 J (b). we know potential energy at top = kineticenergy at bottom                        mgh = ( 1/ 2) mv ^ 2 * [ 1+(k/r) ^ 2] For cylinder ( k / r ) ^ 2 = 1/ 2                          mgh = ( 1/ 2) m v ^ 2 * [ 1+ 1/2]                                 = ( 3/2 ) ( 1/ 2) mv ^ 2              14.4452 = 1.5 * translational kinetic energy translational kinetic energy [ ( 1/ 2) mv ^ 2 ] =15.5232 / 1.5                                                                    = 10.3488 J we know Translational K.E + Rotational K.E = E So, Rotational K.E = 15.5232 -10.3488                               = 5.1744 J

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