Don't Forget that Energy

A disk of mass m m and radius r r is pushed up on an inclined plane of angle θ \theta with initial speed v v . If g = 10 [ m s ] g=10[\frac{m}{s}] and the coefficient of friction is such that the disk can roll without slipping, the maximum height reached by the disk can be expressed as a b v 2 \frac{a}{b}v^2 . Find a + b a+b .


The answer is 43.

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1 solution

At first we have kinetic energy and rotational kinetic energy of a disc, so we have: E 0 = m v 2 2 + I ω 2 2 = m v 2 2 + m r 2 v 2 2 2 r 2 = 3 4 m v 2 E_{0}=\frac{mv^2}{2}+\frac{I\omega^2}{2}=\frac{mv^2}{2}+\frac{mr^2v^2}{2\cdot 2r^2}=\frac{3}{4}mv^2 At maximum height we have: E f = m g h E_{f}=mgh As energy conservates: E f = E 0 E_{f}=E_{0} m g h = 3 4 m v 2 mgh=\frac{3}{4}mv^2 h = 3 4 g v 2 = 3 40 v 2 h=\frac{3}{4g}v^2=\frac{3}{40}v^2

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