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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 (螖L) = 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

Find the force exerted by the biceps muscle in holding a 1-L milk carton (weight \(9.9 \mathrm{N}\) ) with the forearm parallel to the floor. Assume that the hand is \(35.0 \mathrm{cm}\) from the elbow and that the upper arm is $30.0 \mathrm{cm}$ long. The elbow is bent at a right angle and one tendon of the biceps is attached to the forearm at a position \(5.00 \mathrm{cm}\) from the elbow, while the other tendon is attached at \(30.0 \mathrm{cm}\) from the elbow. The weight of the forearm and empty hand is \(18.0 \mathrm{N}\) and the center of gravity of the forearm is at a distance of \(16.5 \mathrm{cm}\) from the elbow.
A planet moves around the Sun in an elliptical orbit (see Fig. 8.39 ). (a) Show that the external torque acting on the planet about an axis through the Sun is zero. (b) since the torque is zero, the planet's angular momentum is constant. Write an expression for the planet's angular momentum in terms of its mass \(m,\) its distance \(r\) from the Sun, and its angular velocity \(\omega\) (c) Given \(r\) and a, how much area is swept out during a short time At? [Hint: Think of the area as a fraction of the area of a circle, like a slice of pie; if \(\Delta t\) is short enough, the radius of the orbit during that time is nearly constant.] (d) Show that the area swept out per unit time is constant. You have just proved Kepler's second law!
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?

Assume the Earth is a uniform solid sphere with radius of \(6.37 \times 10^{6} \mathrm{m}\) and mass of \(5.97 \times 10^{24} \mathrm{kg} .\) Find the magnitude of the angular momentum of the Earth due to rotation about its axis.
A uniform rod of length \(L\) is free to pivot around an axis through its upper end. If it is released from rest when horizontal, at what speed is the lower end moving at its lowest point? [Hint: The gravitational potential energy change is determined by the change in height of the center of gravity.]
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