0.30 Phy1300: angular momentum -- rotational kinetic energy and  (Page 10/11)

Let r1 = 0.9*r2 and let the mass be unchanged.

Find the velocity.

Solution:

Given that the inner radius is 0.9 times the outer radius and the mass is unchanged, we can approximate the rotational inertia for this pulley with

I = (1/2)*M*((0.9*R)^2 + R^2)

The rotational inertia

Let's compute the rotational inertia for this pulley.

I = (1/2)*1kg*((0.9*1m)^2 + (1m)^2), or

I = 0.905 (m^2)*kg

As you can see, the rotational inertia for this configuration is almost double the rotational inertia for the uniform disk configuration.

Start with the original equation

Go back and get the original equation that contains numeric values but still has the rotational inertia as a variable.

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

Substitute the rotational inertia

Replace the rotational inertia, I, with the approximate rotational inertia for our new pulley.

v = ((2*(2kg - 1kg)*(9.8m/s^2)*0.5m)/((2kg+1kg) + (0.905 (m^2)*kg)*(1/(1m)^2)))^(1/2)

Once again, if I managed to get through all of that without making an error, the velocity is

v = 1.58 m/s

Analysis

Comparing this value with the velocity from part 2 , , we see that the velocity was reduced from 1.67 m/s to 1.58 m/s due to the increase in the rotationalinertia of the pulley.

This is what we should expect. Increasing the rotational inertia of the pulley causes it to take longer to accelerate given the sameexternal force that is causing it to change its angular velocity and acquire rotational kinetic energy.

A pulley and two objects, part 5

Let's make one more adjustment to the geometry of the pulley and observe the effect that it has on the system.

Keep all of the parameters the same as part 4 except let the radius of the pulley, R, be 10m.

Find the velocity.

Solution:

Begin by computing the rotational inertia of the new pulley.

I = (1/2)*M*((0.9*R)^2 + R^2), or

I = (1/2)*1kg*((0.9*10m)^2 + (10m)^2), or

I = 90.5 (m^2)*kg

Note that the rotational inertia is proportional to the square of the radius. Therefore, increasing the radius by a factor of 10 caused the rotational inertiato be increased by a factor of 100.

Generally speaking, moving the concentration of mass further from the axis of rotation will increase the rotational inertia.

Start with the original equation as before

Going back and getting the original equation that contains numeric values but still has the rotational inertia as a variable,we have.

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

Substitute the new rotational inertia

Replacing the rotational inertia, I, with the rotational inertia for our new pulley yields.

v = ((2*(2kg - 1kg)*(9.8m/s^2)*0.5m)/((2kg+1kg) + (90.5 (m^2)*kg)*(1/(1m)^2)))^(1/2)

Solving this equation with the Google calculator tells us that the new velocity value is

v = 0.323 m/s

which is much lower than the velocity for either part 2 or part 4 .

An Atwood machine

In case you want to learn more about this topic, the arrangement of the pulley and the objects that we have been discussing is commonly called an Atwoodmachine. The device was invented in 1784 by Rev. George Atwood as a laboratory experiment to verify the mechanical laws of motion.