Question

A group of students perform the same "Conservation of Mechanical Energy" experiment that you performed in lab by allowing a solid sphere and then a solid cylinder to roll down the ramp. The solid sphere was released from a height of 12.9 cm. From what height h_cylinder should the solid cylinder be released so that it has the same speed as the solid sphere when it reaches the bottom of the ramp?

Answer 1

Solid Sphere

KE1 + PE1 = KE2 + PE2

0 + m g h = 0.5 m v^2 + 0.5 I w^2 + 0

I = 0.4 m r^2

m g h = 0.5 m v^2 + 0.5 * 0.4 m r^2 (v/r)^2

m g h = 0.5 m v^2 + 0.2 m v^2

g h = 0.7 v^2

v = sqrt(gh/0.7) = sqrt( 9.81 x 0.129 / 0.7) = 1.345 m/s

Solid Cylinder

KE1 + PE1 = KE2 + PE2

0 + m g h = 0.5 m v^2 + 0.5 I w^2 + 0

I = 0.5 m r^2

m g h = 0.5 m v^2 + 0.5 * 0.5 m r^2 (v/r)^2

m g h = 0.5 m v^2 + 0.25 m v^2

g h = 0.75 v^2

h = 0.75 v^2 / g = 0.75 x (1.345)^2 / 9.81 = 0.1385 m = 13.85 cm

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Height = 13.85 cm

If there is any doubts feel free to ask

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