Question

A solid cylinder is released from the top of an inclined plane of height 0.682 m. From what height on the incline should a solid sphere of the same mass and radius be released to have the same speed as the cylinder at the bottom of the hill? Assume that both objects roll down the incline without slipping. m

Answer 1

= 1 Apply the conservation of energy for solid cylinder. Potential energy Translational K.E + Rotational K.E 1 Mgh==Mv? + My? +-10 2 2 1 MR Mv + I = 2 2 R 1 MR2 2 2 R Mgh = 25= 12 = 2gh= v. + 2 4gh 3 14gh Vc 3 Similarly, Apply the conservation of energy for solid cylinder. Potential energy = Translational K.E + Rotational K.E Mgh = 5MP +-102 2 1 MR? Mgh = - M ? + 2 2 R 1 2 [ 1-2970,o=", 2 2 2gh V sp + 2v. sp 5

2_10ghi 1 sp 7 Vsp SP 110gh 7 Now, for the speed of the cylinder and sphere is same. V=V 4gh 10 gh 3 7 28h 3 28h h 30 h = 0.93h = 0.93 x 0.682 = 0.634m Thus, at the height 0.634 m both have same speed. 10h =

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