A uniform wheel of mass 12.0 kg is mounted rigidly on a massless axle through it
ID: 1297153 • Letter: A
Question
A uniform wheel of mass 12.0 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
A uniform wheel of mass 12.0 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 theta = 30.0 A degree 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 2.00 m. (a) what is its rotational kinetic energy at this point? _____j (b) what is its translational kinetic energy at this point? _____jExplanation / Answer
Conservation of energy is usually a good starting point. The PE lost must equal the total KE gained. I'm assuming the mass of wheel + axle is 12kg although the question is ambiguous.
PE lost, Ep = m.g.h = 12 x 9.8 x 2sin(30) = 117.6J
The rotational KE, Er = 0.5 Iz w^2
(where 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
117.6= 0.5 w^2
w^2 = 235.5
a) rotational KE = 0.5 Iz w^2 = 0.5 x 0.6 x 235.2= 70.56J
b) translational KE = 117.6- 70.56= 47.04J
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