0.30 Phy1300: angular momentum -- rotational kinetic energy and  (Page 9/11)

This tangential speed at the edge of the pulley is related to the angular velocity of the pulley as follows

v = w*R, or

w = v/R

Where

• v is the tangential speed of a point on the edge of the pulley, which is also the speed of the cord, which is also the speed of each object.
• w is the angular velocity of the pulley.
• R is the radius of the pulley.

Substitute, combine terms, and simplify

Substitution of v/R for w yields

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

Combining potential and kinetic energy yields

deltaU + deltaK = -m1*g*h + m2*g*h

+ (1/2)*(m1+m2)*v^2 + (1/2)*I*(v/R)^2 = 0

Simplification yields

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

Solving for v^2 yields

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

The general equation for velocity is

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

A pulley and two objects, part 2

Make the following assumptions:

The pulley is a uniform disk with a mass, M, of 1 kg and a radius, R, of 1 meter.

m1 = 2 kg

m2 = 1 kg

h = 0.5 m

Find the velocity v.

Solution:

Referring back to Examples of rotational inertia , and changing the notation to match that being used in this scenario, we find that the rotational inertia for the pulley is

I = (1/2)*M*(R)^2

The rotational inertia

Let's begin by computing the rotational inertia of the pulley.

I = (1/2)*1kg*(1m)^2, or

I = 0.5*(m^2)*kg

Substitute values other than rotational inertia

Substituting values into the general equation yields

v = ((2*(2kg - 1kg)*(9.8m/s^2)*0.5m)/((2kg+1kg) + I*(1/(1m)^2)))^(1/2)

where we still have the rotational inertia as a variable.

We will use this equation again later in this module.

Substitute the value for rotational inertia

Substituting the value for rotational inertia computed above gives us

v = ((2*(2kg - 1kg)*(9.8m/s^2)*0.5m)/((2kg+1kg) + (0.5*(m^2)*kg)*(1/(1m)^2)))^(1/2)

We will also use this equation again later in this module.

Assuming that I managed to get all of that done without making an error, the velocity is

v = 1.67 m/s

A pulley and two objects, part 3

Assume the same conditions as in part 2 , except that m2 = m1 = 2 kg, what is the velocity?

Solution:

Starting with the equation from above and replacing each occurrence of 1kg with 2kg yields

v = ((2*(2kg - 2kg)*(9.8m/s^2)*0.5m)/((2kg+2kg) + (0.5*(m^2)*kg)*(1/(1m)^2)))^(1/2)

Using the Google calculator to solve this equation tells us that

v = 0 m/s.

This is what we should expect for the two objects having the same mass. Each object has an equal desire to fall toward the earth, so they balance one another, causing thesystem to be in equilibrium.

A pulley and two objects, part 4

Go back to the conditions for part 2 except instead of assuming that the pulley is a uniform disk, assume that the pulley approximates a "thick-walled cylindrical tube with open ends, of inner radius r1, and outer radius r2" .

For this configuration, we can approximate the rotational inertia as

I = (1/2)*m*(r1^2 + r2^2) (see Examples of rotational inertia )

For example, think of a pulley that looks something like a bicycle wheel with very lightweight spokes connecting the outer rim to the axle.