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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
We could rewrite this equation as
deltaKs = (1/2)*I*(w0 - wf)^2
where
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
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
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.
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.
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