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Chapter 8: Problem 73

How long would a braking torque of \(4.00 \mathrm{N}\).m have to act to just stop a spinning wheel that has an initial angular momentum of $6.40 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s} ?$

Short Answer

Expert verified
Answer: The braking torque needs to act for 1.60 seconds to stop the spinning wheel.

Step by step solution

01

Identify the given values

In this exercise, we are given: - Braking Torque (T) = \(4.00 \mathrm{N} \cdot \mathrm{m}\) - Initial Angular Momentum (L_initial) = \(6.40 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}\)
02

Calculate the change in angular momentum

As the wheel must come to a complete halt, the final angular momentum will be zero. Hence, the change in angular momentum can be calculated as: Change in Angular Momentum (ΔL) = L_final - L_initial = \(0 - 6.40 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s} = -6.40 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}\)
03

Find the time required for braking torque to stop the wheel

Using the formula mentioned in the analysis, we can find the time required for the braking torque to stop the wheel. Rearrange the formula to find the time: Time (t) = Change in Angular Momentum / Torque Plug in the values: t = \(\frac{-6.40 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}}{4.00 \mathrm{N} \cdot \mathrm{m}}\) t = -1.60 s Since the time cannot be negative, we will take the absolute value: t = 1.60 s
04

Final Answer

The braking torque of \(4.00 \mathrm{N} \cdot \mathrm{m}\) must act for 1.60 seconds to stop the spinning wheel with an initial angular momentum of \(6.40 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}\).

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Most popular questions from this chapter

Two children standing on opposite sides of a merry-goround (see Fig. 8.5 ) are trying to rotate it. They each push in opposite directions with forces of magnitude \(10.0 \mathrm{N} .\) (a) If the merry-go-round has a mass of $180 \mathrm{kg}\( and a radius of \)2.0 \mathrm{m},$ what is the angular acceleration of the merry-go-round? (Assume the merry-go-round is a uniform disk.) (b) How fast is the merry-go-round rotating after \(4.0 \mathrm{s} ?\)
The Moon's distance from Earth varies between \(3.56 \times 10^{5} \mathrm{km}\) at perigee and \(4.07 \times 10^{5} \mathrm{km}\) at apogee. What is the ratio of its orbital speed around Earth at perigee to that at apogee?
A bicycle wheel, of radius \(0.30 \mathrm{m}\) and mass \(2 \mathrm{kg}\) (concentrated on the rim), is rotating at 4.00 rev/s. After 50 s the wheel comes to a stop because of friction. What is the magnitude of the average torque due to frictional forces?
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?
A weight of \(1200 \mathrm{N}\) rests on a lever at a point \(0.50 \mathrm{m}\) from a support. On the same side of the support, at a distance of $3.0 \mathrm{m}\( from it, an upward force with magnitude \)F$ is applied. Ignore the weight of the board itself. If the system is in equilibrium, what is \(F ?\)
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