A 2.2-kg cylinder (radius = 0.10 m, length = 0.50 m) is released from rest at th

Question

A 2.2-kg cylinder (radius = 0.10 m, length = 0.50 m) is released from rest at thetop of a ramp and allowed to roll without slipping. The ramp is0.67 m high and 5.0 m long. (a) When the cylinder reaches the bottom of theramp, what is its total kinetic energy?
                      J

(b) What is its rotational kinetic energy?
                     J

(c) What is its translational kinetic energy?
                       J (a) When the cylinder reaches the bottom of theramp, what is its total kinetic energy?
                      J

(b) What is its rotational kinetic energy?
                     J

(c) What is its translational kinetic energy?
                       J

Explanation / Answer

mass m = 2.2 kg radius r = 0.1 m length L = 0.5 m height h = 0.67 m Length of the ramp L ' = 5 m sin = h / L '         = 0.134 angle of inclination with horizontal = 7.7degrees (a). total kinetic energy E = mgh   Since from law of conservation of energy,potential energy at top = kinetic energy at bottom So, E = 14.4452 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 ] =14.4452 / 1.5                                                                    = 9.63 J we know Translational K.E + Rotational K.E = E So, Rotational K.E = 14.4452 - 9.63                               = 4.815 J

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