Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Question
Chapter 9, Problem 40P
To determine
The moment of inertia of a ball about its diameter, modelling the ball as a spherical shell.
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Physics for Scientists and Engineers
Ch. 9 - Prob. 1PCh. 9 - Prob. 2PCh. 9 - Prob. 3PCh. 9 - Prob. 4PCh. 9 - Prob. 5PCh. 9 - Prob. 6PCh. 9 - Prob. 7PCh. 9 - Prob. 8PCh. 9 - Prob. 9PCh. 9 - Prob. 10P
Ch. 9 - Prob. 11PCh. 9 - Prob. 12PCh. 9 - Prob. 13PCh. 9 - Prob. 14PCh. 9 - Prob. 15PCh. 9 - Prob. 16PCh. 9 - Prob. 17PCh. 9 - Prob. 18PCh. 9 - Prob. 19PCh. 9 - Prob. 20PCh. 9 - Prob. 21PCh. 9 - Prob. 22PCh. 9 - Prob. 23PCh. 9 - Prob. 24PCh. 9 - Prob. 25PCh. 9 - Prob. 26PCh. 9 - Prob. 27PCh. 9 - Prob. 28PCh. 9 - Prob. 29PCh. 9 - Prob. 30PCh. 9 - Prob. 31PCh. 9 - Prob. 32PCh. 9 - Prob. 33PCh. 9 - Prob. 34PCh. 9 - Prob. 35PCh. 9 - Prob. 36PCh. 9 - Prob. 37PCh. 9 - Prob. 38PCh. 9 - Prob. 39PCh. 9 - Prob. 40PCh. 9 - Prob. 41PCh. 9 - Prob. 42PCh. 9 - Prob. 43PCh. 9 - Prob. 44PCh. 9 - Prob. 45PCh. 9 - Prob. 46PCh. 9 - Prob. 47PCh. 9 - Prob. 48PCh. 9 - Prob. 49PCh. 9 - Prob. 50PCh. 9 - Prob. 51PCh. 9 - Prob. 52PCh. 9 - Prob. 53PCh. 9 - Prob. 54PCh. 9 - Prob. 55PCh. 9 - Prob. 56PCh. 9 - Prob. 57PCh. 9 - Prob. 58PCh. 9 - Prob. 59PCh. 9 - Prob. 60PCh. 9 - Prob. 61PCh. 9 - Prob. 62PCh. 9 - Prob. 63PCh. 9 - Prob. 64PCh. 9 - Prob. 65PCh. 9 - Prob. 66PCh. 9 - Prob. 67PCh. 9 - Prob. 68PCh. 9 - Prob. 69PCh. 9 - Prob. 70PCh. 9 - Prob. 71PCh. 9 - Prob. 72PCh. 9 - Prob. 73PCh. 9 - Prob. 74PCh. 9 - Prob. 75PCh. 9 - Prob. 76PCh. 9 - Prob. 77PCh. 9 - Prob. 78PCh. 9 - Prob. 79PCh. 9 - Prob. 80PCh. 9 - Prob. 81PCh. 9 - Prob. 82PCh. 9 - Prob. 83PCh. 9 - Prob. 84PCh. 9 - Prob. 85PCh. 9 - Prob. 86PCh. 9 - Prob. 87PCh. 9 - Prob. 88PCh. 9 - Prob. 89PCh. 9 - Prob. 90PCh. 9 - Prob. 91PCh. 9 - Prob. 92PCh. 9 - Prob. 93PCh. 9 - Prob. 94PCh. 9 - Prob. 95PCh. 9 - Prob. 96PCh. 9 - Prob. 97PCh. 9 - Prob. 98PCh. 9 - Prob. 99PCh. 9 - Prob. 100PCh. 9 - Prob. 101PCh. 9 - Prob. 102PCh. 9 - Prob. 103PCh. 9 - Prob. 104PCh. 9 - Prob. 105PCh. 9 - Prob. 106PCh. 9 - Prob. 107PCh. 9 - Prob. 108PCh. 9 - Prob. 109PCh. 9 - Prob. 110PCh. 9 - Prob. 111PCh. 9 - Prob. 112PCh. 9 - Prob. 113PCh. 9 - Prob. 114PCh. 9 - Prob. 115PCh. 9 - Prob. 116PCh. 9 - Prob. 117PCh. 9 - Prob. 118PCh. 9 - Prob. 119PCh. 9 - Prob. 120PCh. 9 - Prob. 121PCh. 9 - Prob. 122PCh. 9 - Prob. 123PCh. 9 - Prob. 124PCh. 9 - Prob. 126PCh. 9 - Prob. 127PCh. 9 - Prob. 128PCh. 9 - Prob. 129P
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- A playground merry-go-round of radius R = 2.00 m has a moment of inertia I = 250 kg m2 and is rotating at 10.0 rev/min about a frictionless, vertical axle. Facing the axle, a 25.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round?arrow_forwardA rigid body with a cylindrical cross-section is released from the top of a 30 incline. It rolls 10.0 m to the bottom in 2.60 s. Find the moment of inertia of the body In terms of its mass m and radius r.arrow_forwardCalculate the moment of inertia by direct integration of a thin rod of mass M and length L about an axis through the rod L/3 , as shown below. Check your answer with the parallel-axis theorem.arrow_forward
- Calculate the moment of inertia of a skater given the following information. (a) The 60.0-kg skater is approximated as a cylinder that has a 0.110-m radius. b) The skater with arms extended is approximated by a cylinder that is 52.5 kg, has a 0.110-m radius, and has two 0.900-m-long arms which are 3.75 kg each and extend straight out from the cylinder like rods rotated about their ends.arrow_forwardA buzzard (m = 9.29 kg) is flying in circular motion with aspeed of 8.44 m/s while viewing its meal below. If the radius ofthe buzzards circular motion is 8.00 m, what is the angularmomentum of the buzzardaround the center of its motion?arrow_forwardCheck Your Understanding What is the moment of inertia of a cylinder of radius R and mass m about an axis through a point on the surface, as shown below? A uniform thin disk about an axis through the center Integrating to find the moment of inertia of a two-dimensional object is a little bit trickier, but one shape is commonly done at this level of study—a uniform thin disk about an axis through its center (Figure 10.27). Figure 10.27 Calculating the moment of inertia for a thin disk about an axis through its center.arrow_forward
- A solid cylinder of mass 2.0 kg and radius 20 cm is rotating counterclockwise around a vertical axis through its center at 600 rev/min. A second solid cylinder of the same mass and radius is rotating clockwise around the same vertical axis at 900 rev/min. If the cylinders couple so that they rotate about the same vertical axis, what is the angular velocity of the combination?arrow_forwardFind the moment of inertia of the rod in the previous problem by direct integration.arrow_forwardA system consists of a disk of mass 2.0 kg and radius 50 cm upon which is mounted an annular cylinder of mass 1.0 kg with inner radius 20 cm and outer radius 30 cm (see below). The system rotates about an axis through the center of the disk and annular cylinder at 10 rev/s. (a) What is the moment of inertia of the system? (b) What is its rotational kinetic energy?arrow_forward
- Big Ben (Fig. P10.17), the Parliament tower clock in London, has hour and minute hands with lengths of 2.70 m and 4.50 m and masses of 60.0 kg and 100 kg, respectively. Calculate the total angular momentum of these hands about the center point. (You may model the hands as long, thin rods rotating about one end. Assume the hour and minute hands are rotating at a constant rate of one revolution per 12 hours and 60 minutes, respectively.)arrow_forwardFigure P10.41 shows a side view of a car tire before it is mounted on a wheel. Model it as having two side-walls of uniform thickness 0.635 cm and a tread wall of uniform thickness 2.50 cm and width 20.0 cm. Assume the rubber has uniform density 1.10 103 kg/m3. Find its moment of inertia about an axis perpendicular to the page through its center. Figure P10.41arrow_forwardThe velocity of a particle of mass m = 2.00 kg is given by v= 5.10 + 2.40 m /s. What is the angular momentumof the particle around the origin when it is located atr= 8.60 3.70 m?arrow_forward
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Moment of Inertia; Author: Physics with Professor Matt Anderson;https://www.youtube.com/watch?v=ZrGhUTeIlWs;License: Standard Youtube License