0.33 Phy1330: angular momentum -- torque, work and energy  (Page 4/4)

• A moment of inertia, I, equal to (1/2)*M*R^2
• M = mass = 80 kg
• R = radius = 0.0.5 meters

Part 1

Find the amount of work that must be done to bring the wheel from rest to an angular velocity of 8.38 radians/sec

Solution:

Recall from a previous module that the rotational kinetic energy for a rotating object is given by

Ks = (1/2)*I*w^2

• where Ks represents the kinetic energy for the system
• I represents the rotational inertia for the system
• w represents the angular velocity of the system

We could rewrite this equation as

deltaKs = (1/2)*I*(w0 - wf)^2

where

• deltaKs represents the change in kinetic energy
• w0 represents the initial kinetic energy
• wf represents the final kinetic energy

However, since the initial kinetic energy value is zero, that would simplycomplicate the algebra. Therefore, we will stick with the original equation .

We either have, or can calculate values for all of the terms in this equation. Substituting the values given above gives us

Ks = (1/2)*I*w^2 , or

Ks = (1/2)*((1/2)*M*R^2)*w^2 , or

Ks = (1/2)*((1/2)*80kg*(0.5m)^2)*(8.38 radians/sec)^2

Entering this expression into the Google calculator gives us

Ks = 351 joules

This is the amount of work that must be done to bring the wheel from rest to an angular velocity of 8.38 radians/sec

Part 2

If the motor that drives the wheel delivers a constant torque of 10 N*m during this time, how many revolutions does the wheel turn in coming up to speed.

Solution:

We know how to relate the displacement angle and the work for a constant torque using the equation in Figure 2 .

W = T*A

where

• W represents the work done by a constant torque
• T represents the constant torque
• A represents the angle of displacement measured in radians resulting from the application of the constant torque

In this case, we know the amount of work and the value of the torque and need to find the angle. Therefore,

A = W*joules/T*n*m

However, this gives us the angular displacement in radians. We need to scale to convert it to revolutions.

A = (W*joules/T*n*m)/2*pi, or

A = (351joules/10newton meters)/(2*pi), or

A = 5.59 revolutions

This is the number of revolutions that the wheel turns in coming up to speed.

Do the computations

I encourage you to repeat the computations that I have presented in this lesson to confirm that you get the same results. Experiment withthe scenarios, making changes, and observing the results of your changes. Make certain that you can explain why your changes behave as they do.

Resources

I will publish a module containing consolidated links to resources on my Connexions web page and will update and add to the list as additional modulesin this collection are published.

Miscellaneous

This section contains a variety of miscellaneous information.

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