Question

A counterweight of mass ? = 4.00 kg is attached to a light cord that is wound around a pulley as shown in the figure. The pulley is a thin hoop of radius ? = 8.00 cm and mass ? = 2.00 kg. The spokes have negligible mass. (Determine your answers first in terms of these symbols and then numerically evaluate.)    What is the net external torque acting on the system of wheel-cord- counterweight? Hint: Use the formula of torque in terms of tangential force.    Calculate the angular momentum of the system, about the axle of the pulley, in terms of an arbitrary speed ?. (Symbolically only.) Careful: note that the counterweight has an angular momentum even though it travels in a straight line! Hint: Angular momentum of the system= due to rotation of pulley +                                 due to linear motion of counterweight    Use the relation ? = ?L/?t to calculate the acceleration of the counterweight. Hint: Use expression of torque from question 1 and angular momentum from question 2.     Draw free-body diagram

Answer 1

Part A.

We know that System is balance.

Suppose Tension in counterweight is 'T', then

T = m*g

Now Net torque about the axle of pulley will be:

= RxT = R*m*g

R = radius of pulley = 8.00 cm = 0.08 m

m = mass of counterweight = 4.00 kg

= 0.08*4.00*9.81

= 3.14 N-m

Direction = to the right along the axis of rotation

Part B.

Now total angular momentum of system about the axle of pulley will be:

L = Rxm*v + I*w

Rxm*v = R*m*v*sin 90 deg = R*m*v

I = Moment of inertia of wheel rotating about it's center of mass = M*R^2

w = angular velocity = v/R

So,

L = m*v*R + (M*R^2)*(v/R) = (m + M)*R*v

here, M = mass of pulley = 2.00

L = (4.00 + 2.00)*0.08*v

L = (0.48 kg.m)*v

Part C.

Net torque = dL/dt

m*g*R = d[(m + M)*R*v]/dt

m*g*R = (m + M)*R*dv/dt

m*g*R = (m + M)*R*a

a = m*g/(m + M)

a = 4.00*9.81/(4.00 + 2.00)

a = 6.54 m/sec^2

2.

Free Body diagram: