Calculate the rotational kinetic energy in the motorcycle wheel (Figure \(10.38\)) if its angular velocity is \(120 \mathrm{rad} / \mathrm{s}\). Assume \(M=12.0 \mathrm{~kg}, R_{1}=0.280 \mathrm{~m}\), and \(R_{2}=0.330 \mathrm{~m}\). Equation Transcription: Text Transcription: 10.38 120 rad/s M=12.0 kg,R_1=0.280 m, and R_2=0.330 m
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Textbook Solutions for College Physics
Question
During a very quick stop, a car decelerates at \(7.00 \mathrm{m} / \mathrm{s}^{2}\)
(a) What is the angular acceleration of its 0.280-m-radius tires, assuming they do not slip on the pavement?
(b) How many revolutions do the tires make before coming to rest, given their initial angular velocity is 95.0 rad/s ?
(c) How long does the car take to stop completely?
(d) What distance does the car travel in this time?
(e) What was the car’s initial velocity?
(f) Do the values obtained seem reasonable, considering that this stop happens very quickly?
Solution
Step 1 of 6
From the data given, we are required to calculate various quantities considering the equations of rotational motion.
Part a
The rate of deceleration of the car is \(\mathrm{a}=-7.0 \mathrm{m} / \mathrm{s}^{2}\)
The radius of the tire is \(r=0.28m\)
Therefore, the angular acceleration \(\mathrm{a}=|\mathrm{a}| / \mathrm{r}\)
\(\begin{array}{l}\mathrm{a}=7.0\mathrm{\ m}/\mathrm{s}^2/0.28\mathrm{\ rad}/\mathrm{s}^2\\ \mathrm{a}=25.0\mathrm{\ rad}/\mathrm{s}^2\end{array}\)
Therefore, the angular acceleration of the tire is \(25.0\mathrm{\ rad}/\mathrm{s}^2\).
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Car Deceleration: Angular Acceleration, Revolutions, Time, Distance, a
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Explore the physics of a car's quick deceleration, including angular acceleration, revolutions, time, distance, initial velocity, and the realism of the scenario.