| dW |
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Kinetic energy is energy associated with motion. Rotational kinetic energy is defined as:
K = ½ Iω2
With straight-line motion, usually the only way to vary the kinetic energy is to vary the speed, because the mass is generally fixed. For rotational motion the angular speed can change and, in some circumstances, the moment of inertia can change.
Work links the concepts of energy and torque. Because we'll only deal with one-dimensional rotational situations, any time there is a torque and an angular displacement, work will be done by the torque.
In a rotational situation:
W = ∫ τ dθ
For a constant torque, W = τ Δθ
Work can be positive or negative, positive when the torque is in the same direction as the displacement and negative when these are in opposite directions.
This new type of kinetic energy can easily be incorporated into our conservation of energy equation:
Ui + Ki + Wnc = Uf + Kf
K = ½ mv2 if the object is moving but not rotating.
K = ½ Iω2 when an object is only spinning.
Both kinds of kinetic energy need to be included when an object rotates as it moves.
Power is the rate at which work is done.
| P | = |
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In a rotational situation: P = τ • ω
A grinding wheel in the form of a uniform solid disk is spun up to an angular velocity of ωi. The wheel has a mass m and a radius r. A tool to be sharpened is pressed against the wheel with a normal force of N.
(a) If, after t seconds, the wheel's angular velocity is ωf, what is the coefficient of friction between the tool and the grinding wheel? Assume no other torques act on the wheel.
(b) What is the average rate of kinetic energy loss over this time period?
Step 1. Draw a diagram, and take a couple of minutes to set up the problem. The free-body diagram of the wheel shows the normal force and the force of kinetic friction. The wheel is rotating clockwise - we'll take that to be the positive direction.
Step 2. Apply Newton's Second Law for rotation. The force of friction is the only force giving a torque, and that torque is counter-clockwise:
Στ = Iα
-rfk = Iα
Step 3. Substitute in for the force of kinetic friction:
fk = μkN
Therefore -rμkN = Iα
| μk | = |
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Step 4. Use a constant acceleration equation to find α in terms of the angular velocities:
ωf = ωi + αt
| This gives | α | = |
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| So, μk | = |
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Step 5. For a solid disk, I = ½ mr2
| μk | = |
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= |
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Let's take an example where the disk has a mass of 1.5 kg, a radius of 25 cm, and an initial angular speed of 360 rad/s. The tool to be sharpened is pushed against the wheel with a normal force of 20 N, and after t = 10 seconds the wheel's angular speed is 120 rad/s.
| μk | = |
|
= |
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= | 0.225 |
(b) What is the average rate of kinetic energy loss over the 10 second interval?
The rate of energy loss is the power loss. One way to get the average power is to use:
Pavg = -τ ωavg
Another method is to calculate the initial and final kinetic energies, subtract to find the energy loss, and divide by the time interval.
| Pavg | = |
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= |
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For our numerical example:
| Pavg | = |
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= |
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= | -270 W |