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Chapter 11: Problem 23

In the Olympic hammer throw, a contestant whirls a 7.3 -kg steel ball on the end of a 1.2 -m cable. If the contestant's arms reach an additional \(90 \mathrm{cm}\) from his rotation axis and if the ball's speed just prior to release is \(27 \mathrm{m} / \mathrm{s},\) what's the magnitude of the ball's angular momentum?

Short Answer

Expert verified
The magnitude of the ball's angular momentum is 412.87 Kg \cdot m^2/s

Step by step solution

01

Calculate the Moment of Inertia

Moment of Inertia is given by the formula \(I = m \cdot r^2\). Here, mass m=7.3Kg and radius r=1.2m+0.9m=2.1m. From the formula we find that \(I = 7.3Kg \cdot (2.1m)^2 = 32.13Kg \cdot m^2\)
02

Calculate the Angular Speed

The Angular Speed is given by the speed divided by the radius. Speed = 27m/s and radius = 2.1m. Hence, Angular Speed = 27m/s / 2.1m = 12.85 rad/s
03

Calculate the Angular Momentum

The Angular Momentum is given by the moment of inertia multiplied by the angular speed. From the previous steps we know that Moment of Inertia = 32.13Kg \cdot m^2 and Angular Speed = 12.85 rad/s. Hence, Angular Momentum = 32.13Kg \cdot m^2 \cdot 12.85 rad/s = 412.87 Kg \cdot m^2/s

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Most popular questions from this chapter

A 7.4 -cm-diameter baseball has mass \(145 \mathrm{g}\) and is spinning at 2000 rpm. Treating the baseball as a uniform solid sphere, what's its angular momentum?

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