Wheel A has mass m A = 0.815 kg and radius r A = 7.92 cm, and wheel B has mass m

Question

Wheel A has mass mA = 0.815 kg and radius rA = 7.92 cm, and wheel B has mass mB = 0.862 kg and radius rB = 8.85 cm. Each wheel is free to spin about a frictionless axles that passes through its center of mass.

Wheel B is initially at rest, while wheel A is spinning with angular speed 0 = 3.10 rad/s about its axle. Wheel B is then coupled to wheel A by means of a light-weight belt that does not stretch, as shown below. The belt does not slip along the edges of the wheels after the wheels have reached their final states of motion.

(a) What are the final angular speeds of the two wheels?


(b) Assuming that the wheels are uniform solid discs, by how much does the internal energy of the wheels-belt system increase as the belt couples the two wheels together?
Eint =  J

Explanation / Answer

Here,

moment of inertia of disk A , IA = 0.5 * mA * rA^2

IA = 0.5 * 0.815 * 0.0792^2

IA = 2.56 *10^-3 Kg.m^2

moment of inertia of disk B , IB = 0.5 * 0.862 * 0.0885^2

IB = 3.376 *10^-3 Kg.m^2

angular speed , w0 = 3.1 rad/s

let the final angular speed of disk A is wA

final angular speed of disk B is wB

as the belt don't slip

rA * wA = rB * wB

7.92 * wA = 8.85 * wB

wA = wB * 1.117 ----(1)

using conservation of angular momentum

2.56 *10^-3 * 3.10 = 2.56 *10^-3 * wA + 3.376 *10^-3 * wB ---(2)

solving from 1 and 2

wA = 1.42 rad/s

wB = 1.27 rad/s

the final angular speed of the disk A is 1.42 rad/s

the final angular speed of disk B is 1.27 rad/s

part B)

incerase in internal energy = initial kinetic energy - final kinetic energy

incerase in internal energy = 0.5 * 2.56 *10^-3 * 3.10^2 - 0.5 * 2.56 *10^-3 * 1.42^2 - 0.5 * 3.376 *10^-3 * 1.27^2

incerase in internal energy = 6.99 *10^-3 J

the increase in internal energy is 6.99 *10^-3 J

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