A uniform wheel of mass 10.0 kg and radius 0.400 m is mounted rigidly on a massl

Question

A uniform wheel of mass 10.0 kg and radius 0.400 m is mounted rigidly on a massless axle through its center (Fig. 11-62). 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 m^2. The wheel is initially at rest at the top of a surface that is inclined at angle theta = 30.0 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. When the wheel-axle combination has moved down the surface by 2.00 m. what are (a) its rotational kinetic energy and (b) its translational kinetic energy?

Explanation / Answer

Here

moment of inertia , I = 0.6 Kg.m^2

m = 10 Kg

r = 0.20 m

theta = 30 degree

distance , d = 2m

let the velocity at the bottom is v

angular velocity is w

as v = w * r

Using conservation of energy

0.5 * I * w^2 + 0.5 * m * v^2 = m * g * d * sin(theta)

0.5 * 0.6 * (v/0.20)^2 + 0.5 * 10 * v^2 = 10 * 9.8 * 2 * sin(30)

solving for v

v = 2.8 m/s

a) Rotational kinetic energy = 0.5 * I * (w^2)

Rotational kinetic energy = 0.5 * 0.6 * (2.8/.2)^2

Rotational kinetic energy = 58.8 J

b)

Translational kinetic energy = 0.5 * m* v^2

Translational kinetic energy = 0.5 * 10 * 2.8^2

Translational kinetic energy = 39.2 J

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