0.30 Phy1300: angular momentum -- rotational kinetic energy and  (Page 8/11)

• m represents the mass of the rod
• L represents the distance between the centers of mass of the two spheres
• R represents the radius of each sphere
• M represents the mass of each sphere

I will leave it as an exercise for the student to assign typical numeric values to each of those variables and to compute the rotational inertia of adumbbell.

Another good exercise for the student would be to replace the spheres with disks having the same mass and compare the rotational inertia of the twoconfigurations.

A pulley and two objects, part 1

In this scenario, a friction-free pulley with a mass M and a rotational inertia I is suspended from a beam. A cord is threaded around the pulley and twoobjects with masses of m1 and m2 are fastened to the ends of the cord.

The cord will not stretch and will not slip on the surface of the pulley.

When the two objects are held at the same level with the cord taut and released simultaneously, one may move up and the other may move down, dependingof the relative mass values of the two objects. Because the cord cannot stretch, the magnitude of the velocity of each object must be the same.

Ignoring air resistance, write a general equation for the magnitude of the velocity of each object after each object has moved a distance h.

Solution:

With no friction and no air resistance, all forces acting on the system are conservative. Therefore, the mechanical energy of the system must be preserved.

deltaU + deltaK = 0

where

• deltaU is the change in potential energy
• deltaK is the change in kinetic energy

Since the pulley mechanism is in equilibrium, the only potential energy possessed by the system that can change is the gravitational potential energy ofthe two objects. Therefore, when the objects move, the gravitational potential energy of the two objects is converted into translational kinetic energy of theobjects and rotational kinetic energy of the pulley.

Changes in potential energy of the system

In order to keep our algebraic signs straight, we will assume that m1 is greater than (or possibly equal to) m2. As a result, when the objects move, m1 will move downand m2 will move up.

This will result in the following changes in the potential energy of the system.

deltaU1 = -m1*g*h

deltaU2 = +m2*g*h

where

• deltaU1 and deltaU2 represent the changes in gravitational potential energy for m1 and m2 respectively.
• g represents the acceleration of gravity.
• h represents the magnitude of the change in height of each object.

The mechanical energy of the system

The mechanical energy of the system includes the translational kinetic energy of each of the two objects plus the rotational kinetic energy of the pulley.

deltaK = (1/2)*(m1+m2)*v^2 + (1/2)*I*w^2

where

• m1 and m2 are the mass values for each object
• v is the magnitude of the translational velocity of each object
• I is the rotational inertia of the pulley
• w is the angular velocity of the pulley

Translational speed versus tangential speed

Because the cord cannot slip on the pulley, the tangential speed of a point on the edge of the pulley must be equal to the translational speed of theobjects.