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 ? = 30.9o 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 8.90 m, what are (a) its rotational kinetic energy and (b) its translational kinetic energy? Please show all work. Thanks!

Explanation / Answer

By law of conservation of the energy

PEi = KEf

PEi = KEtrans + KErot

Mgh = 1/2Mv^2 + 1/2I2

Mgdsin = 1/2M2r2 + 1/2I2

10*9.8*8.90*sin30.9 = 1/2*10*2*0.2^2 + ½*0.6*2     => = 29.93 rad/s

KErot = 1/2I2 = 1/2*0.6*29.93^2 = 268.7 J

KEtrans = PEi – KErot = Mgdsin - KErot = 10*9.8*8.90*sin30.9 - 268.7 = 179.2 J

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