A uniform wheel of mass 6.0 kg is mounted rigidly on a massless axle through its

Question

A uniform wheel of mass 6.0 kg is mounted rigidly on a massless axle through its center, as shown in the figure below. 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 = 30.0° 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. The wheel-axle combination moves down the surface by 3.00 m.

(a) What is its rotational kinetic energy at this point?
J

(b) What is its translational kinetic energy at this point?
J

Explanation / Answer

there will be a decrease in the potential energy of mgd Cos ?. nd this will be equal to the total kinetic energy gained so we can write mgd Sin?= 1/2( mv 2+ I? 2 ) The rotational kinetic energy at this pointwill be mgd Sin ?= 1/2( I? 2) [(mr2 / I )+1] mgdSin ? = K rot [(mr2 / I ) +1] mgd Sin ? =6 kg * 9.8 m/s2 * 3.00 m* Sin 30.0° =88.2 Krot = mgdSin ? / [(mr^2 / I ) +1] 88.2/(6*0.2^2/0.6+1) =63 The translational kinetic energy at thispoint Ktra = 98J - 63=35.

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