A uniform wheel of mass 10.0 kg and radius 0.400 m is mounted rigidly on an axle

Question

A uniform wheel of mass 10.0 kg and radius 0.400 m is mounted rigidly on an axle through its center (see the figure). The radius of the axle is 0.200 m, and the rotational inertia of the wheel-axle combination about its central axis is 0.600 kg·m2. The wheel is initially at rest at the top of a surface that is inclined at angle ? = 16.3o with the horizontal; the axle rests on the surface while the wheel extends into a groove in the surface without touching the surface. Once released, the axle rolls down along the surface smoothly and without slipping. When the wheel-axle combination has moved down the surface by 9.68 m, what are (a) its rotational kinetic energy and (b) its translational kinetic energy?

Explanation / Answer

m =10 kg , g =9.8 m/s^2, h = 9.68 m ,Iz = 0.6kg.m^2

From conservation of energy the PE lost must equal the total KE gained.

PE lost, Ep = m.g.h = 10 * 9.8 * 9.68 sin(16.3) = 266.25 J

The rotational KE, Er = 0.5 Iz w^2

Here Iz is the moment of inertia of the wheel-axle and w is the angular velocity

The translational KE, Et = 0.5 m (w.r)^2

(where m is the mass of wheel + axle and r is the axle radius and w is angular velocity. w.r is the velocity of the axle in the direction of the slope)

The total KE, Es = Er + Et = 0.5 ( Iz + m.r^2 ).w^2

Es = 0.5 ( 0.6 + 10 x 0.2^2 ) w^2

Es = 0.5 w^2

Conservation of energy dictates that Es = Ep

266.25 = 0.5 w^2

w^2 = 532.5

a) rotational KE = 0.5 Iz w^2 = 0.5 x 0.6 x 532.5 = 159.75 J

b) translational KE = 266.25 -159.75 = 106.5 J

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