A solid cylinder of mass M and radius R rolls without slipping on a horizontal surface with a linear velocity v of its center of mass. It then rolls up an inclined plane. What is the maximum vertical

Question

A solid cylinder of mass M and radius R rolls without slipping on a horizontal surface with a linear velocity v of its center of mass. It then rolls up an inclined plane. What is the maximum vertical height h it can reach on the inclined plane?

TestHub · Accepted Answer

When the solid cylinder rolls without slipping, its total kinetic energy is the sum of its translational and rotational kinetic energies.

Total initial kinetic energy $K_{t o t a l} = K_{t r an s l a t i o na l} + K_{ro t a t i o na l} = \frac{1}{2} M v^{2} + \frac{1}{2} I ω^{2}$ . For a solid cylinder, $I = \frac{1}{2} M R^{2}$ . Since it rolls without slipping, $v = R ω$ , so $ω = v / R$ . Substituting these values: $K_{t o t a l} = \frac{1}{2} M v^{2} + \frac{1}{2} (\frac{1}{2} M R^{2}) (\frac{v}{R})^{2} = \frac{1}{2} M v^{2} + \frac{1}{4} M v^{2} = \frac{3}{4} M v^{2}$ .

As the cylinder rolls up the inclined plane, this kinetic energy is converted into gravitational potential energy. At the maximum height $h$ , all kinetic energy is converted to potential energy, $K_{t o t a l} = M g h$ . $\frac{3}{4} M v^{2} = M g h ⟹ h = \frac{3 v ^{2}}{4 g}$

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