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Lesechka [4]
4 years ago
13

A torque applied to a flywheel causes it to accelerate uniformly from a speed of 161 rev/min to a speed of 853 rev/min in 5.0 se

conds. determine the number of revolutions n through which the wheel turns during this interval. (suggestion: use revolutions and minutes for units in your calculations.)
Physics
1 answer:
Temka [501]4 years ago
3 0
First of all, let's convert the time interval into minutes. Since
60 s: 1 min = 5 s: x
we find
\Delta t =  \frac{5.0 s}{60 s/min}=0.083 min

Then we can find the angular acceleration of the flywheel:
\alpha =  \frac{\omega _f - \omega_i}{\Delta t}= \frac{853 rpm-161 rpm}{0.083 min}=8337 rev/min^2

At this point, we can use the law of motion of an uniformly accelerated rotational motion. The angular displacement after a time \Delta t is given by
\theta (\Delta t)= \omega_i t +  \frac{1}{2} \alpha t^2 =
=(161 rpm)(0.083 min)+ \frac{1}{2}(8337 rev/min^2)(0.083 min)^2 =42.1 rev
So, the flywheel covers 42.1 revolutions.
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