An Atwood’s machine consists of mass m1 = 1.5 kg connected to mass m2 = 2.5 kg b

Question

An Atwood’s machine consists of mass m1 = 1.5 kg connected to mass m2 = 2.5 kg by a light string that passes over a disk–shaped pulley of radius R = 0.2 m and mass M = 2.0 kg. The string does not slip on the rim of the pulley. The pulley rotates, without friction, about a horizontal axis passing through the center of the pulley. For parts [a] through [d] below, consider the instant at which m2 has fallen through a distance of 2.0 m, assuming it started from rest. (Note: Use g = 10 m/s^2)

a. Using conservation of energy, find the speed of m2.
b. What is the magnitude of the radial acceleration of a point such as P on the rim of the pulley?
c. What is the magnitude of the tangential acceleration of point P on the rim of the pulley?
d. What is the magnitude of the net torque on the pulley?

Explanation / Answer

let v are the speeds of hanging masses and w is the angular speed of the pulley.

a) using conservation of energy

loss in gravitaional potential energy = gain in kinetic energy

(m2 - m1)*g*h = (1/2)*(m1 + m2)*v^2 + (1/2)*I*w^2

(m2 - m1)*g*h = (1/2)*(m1 + m2)*v^2 + (1/2)*(0.5*M*R^2)*w^2

(m2 - m1)*g*h = (1/2)*(m1 + m2)*v^2 + (1/4)*M*v^2 (since v = R*w)

(2.5 - 1.5)*9.8*2 = (1/2)*(1.5 + 2.5)*v^2 + (1/4)*2*v^2

==> v = 2.8 m/s

b) radial acceleration of a point P, a_rad = v^2/R

= 2.8^2/0.2

= 39.2 m/s^2

c) let a is the tangential acceleration of point P.

use, v^2 - u^2 = 2*a*h

a = (v^2 - u^2)/(2*h)

= (2.8^2 - 0^2)/(2*2)

= 1.96 m/s^2

d) Net torque on the pulley, Tnet = I*alfa

= I*(a/R)

= (0.5*M*R^2)*(a/R)

= 0.5*M*R*a

= 0.5*2*0.2*1.96

= 0.392 N.m

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