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Chapter 8: Problem 25

A flywheel of mass 182 kg has an effective radius of \(0.62 \mathrm{m}\) (assume the mass is concentrated along a circumference located at the effective radius of the flywheel). (a) How much work is done to bring this wheel from rest to a speed of 120 rpm in a time interval of 30.0 s? (b) What is the applied torque on the flywheel (assumed constant)?

Short Answer

Expert verified
(b) What is the constant applied torque on the flywheel? Answer: (a) The work done to bring the wheel from rest to 120 rpm in 30 seconds is approximately 2,192 J. (b) The constant applied torque on the flywheel is approximately 0.98 Nm.

Step by step solution

01

Convert angular speed to radians per second

First, we need to convert the given angular speed from rpm to radians per second. We are given 120 rpm and can use the following conversion factor: 1 rpm = \(\frac{2\pi}{60}\) rad/s Therefore, continuing the calculation: \(\omega = 120 \times \frac{2\pi}{60}= 2\pi\) rad/s
02

Calculate the flywheel's moment of inertia

Since the mass is concentrated along a circumference located at the effective radius of the flywheel, we can use the formula for the moment of inertia for a circular ring: \(I = m r^2\) Where \(m\) is the mass (182 kg), \(r\) is the effective radius (0.62 m) of the flywheel, and \(I\) is the moment of inertia. Plugging in the values: \(I = (182\,\text{kg}) (0.62\,\text{m})^2 = 69.93\, \text{kg} \cdot\text{m}^2\)
03

Calculate the angular acceleration

Now we have the moment of inertia and the final angular speed; we can find the angular acceleration (\(\alpha\)) using: \(\omega ^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)\) Since initially the flywheel is at rest, \(\omega_0 = 0\). Also, we are given the time taken to reach the final speed is 30s. So we can find the total angle covered as follows: \(\theta - \theta_0 = \omega_0t + \frac{1}{2}\alpha t^2\) After plugging in the values, we can solve for the angular acceleration: \(\alpha = \frac{2\pi}{(30\, s)^2} \approx 0.014 \, \text{rad} \cdot\text{s}^{-2}\)
04

Determine the work done

Now that we know the flywheel's angular acceleration, we can use the work-energy theorem to find the work done on the flywheel: \(W = \frac{1}{2} I (\omega^2 - \omega_0^2)\) The flywheel starts from rest, so \(\omega_0 = 0\): \(W = \frac{1}{2} (69.93\, \text{kg} \cdot\text{m}^2) (2\pi\, \text{rad} \cdot\text{s}^{-1})^2 \approx 2,192 \, \text{J}\)
05

Calculate the applied torque

Finally, we want to find the constant applied torque. Using the torque-angular acceleration relationship: \(\tau = I\alpha\) \(\tau = (69.93 \, \text{kg} \cdot\text{m}^2)(0.014 \, \text{rad} \cdot\text{s}^{-2}) \approx 0.98\, \text{N} \cdot\text{m}\) So, the applied torque on the flywheel is approximately 0.98 Nm. To summarize the answers: (a) The work done to bring the flywheel from rest to 120 rpm in 30 seconds is approximately 2,192 J. (b) The applied torque on the flywheel is approximately 0.98 Nm.

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Most popular questions from this chapter

The angular momentum of a spinning wheel is $240 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s} .$ After application of a constant braking torque for \(2.5 \mathrm{s}\), it slows and has a new angular momentum of $115 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s} .$ What is the torque applied?
An experimental flywheel, used to store energy and replace an automobile engine, is a solid disk of mass \(200.0 \mathrm{kg}\) and radius $0.40 \mathrm{m} .\( (a) What is its rotational inertia? (b) When driving at \)22.4 \mathrm{m} / \mathrm{s}(50 \mathrm{mph}),$ the fully energized flywheel is rotating at an angular speed of \(3160 \mathrm{rad} / \mathrm{s} .\) What is the initial rotational kinetic energy of the flywheel? (c) If the total mass of the car is \(1000.0 \mathrm{kg},\) find the ratio of the initial rotational kinetic energy of the flywheel to the translational kinetic energy of the car. (d) If the force of air resistance on the car is \(670.0 \mathrm{N},\) how far can the car travel at a speed of $22.4 \mathrm{m} / \mathrm{s}(50 \mathrm{mph})$ with the initial stored energy? Ignore losses of mechanical energy due to means other than air resistance.
A turntable of mass \(5.00 \mathrm{kg}\) has a radius of \(0.100 \mathrm{m}\) and spins with a frequency of 0.550 rev/s. What is its angular momentum? Assume the turntable is a uniform disk.
A centrifuge has a rotational inertia of $6.5 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2}$ How much energy must be supplied to bring it from rest to 420 rad/s \((4000 \text { rpm }) ?\)
A lawn sprinkler has three spouts that spray water, each \(15.0 \mathrm{cm}\) long. As the water is sprayed, the sprinkler turns around in a circle. The sprinkler has a total rotational inertia of $9.20 \times 10^{-2} \mathrm{kg} \cdot \mathrm{m}^{2} .\( If the sprinkler starts from rest and takes \)3.20 \mathrm{s}$ to reach its final speed of 2.2 rev/s, what force does each spout exert on the sprinkler?
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