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Chapter 8: Problem 72

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?

Short Answer

Expert verified
Answer: 50 N·m

Step by step solution

01

Identify the known quantities

We are given the following information: - Initial angular momentum \(L_i = 240 \, \mathrm{kg} \cdot \mathrm{m}^2 / \mathrm{s}\) - Final angular momentum \(L_f = 115 \, \mathrm{kg} \cdot \mathrm{m}^2 / \mathrm{s}\) - Time \(t = 2.5 \, \mathrm{s}\) We know that torque is related to the change in angular momentum and the time duration. Our objective is to find the magnitude of the constant braking torque applied to the spinning wheel.
02

Apply formula for torque

The formula for torque is given by: $$ \tau = \frac{\Delta L}{\Delta t}$$ Where \(\tau\) is the torque, \(\Delta L\) is the change in angular momentum, and \(\Delta t\) is the time duration.
03

Calculate the change in angular momentum

We can find the change in angular momentum by subtracting the initial angular momentum from the final angular momentum: $$\Delta L = L_f - L_i$$ Plugging in the values: $$\Delta L = 115 \, \mathrm{kg} \cdot \mathrm{m}^2 / \mathrm{s} - 240 \, \mathrm{kg} \cdot \mathrm{m}^2 / \mathrm{s} = -125\, \mathrm{kg} \cdot \mathrm{m}^2 / \mathrm{s}$$
04

Calculate the torque

Now that we have found the change in angular momentum, we can use the formula for torque to find the magnitude of the constant braking torque. $$ \tau = \frac{-125 \, \mathrm{kg} \cdot \mathrm{m}^2 / \mathrm{s}}{2.5 \, \mathrm{s}}$$ $$ \tau = -50 \, \mathrm{N} \cdot \mathrm{m}$$ Since we are only interested in the magnitude of the torque, we can take the absolute value: $$|\tau| = 50 \, \mathrm{N} \cdot \mathrm{m}$$ So, the magnitude of the constant braking torque applied to the spinning wheel is \(50\, \mathrm{N} \cdot \mathrm{m}\).

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Most popular questions from this chapter

A solid sphere of mass 0.600 kg rolls without slipping along a horizontal surface with a transnational speed of \(5.00 \mathrm{m} / \mathrm{s} .\) It comes to an incline that makes an angle of \(30^{\circ}\) with the horizontal surface. Ignoring energy losses due to friction, to what vertical height above the horizontal surface does the sphere rise on the incline?
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