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The rotor (flywheel) of a toy gyroscope has mass 0.140 kg.

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 53E Chapter 10

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 53E

The rotor (flywheel) of a toy gyroscope has mass 0.140 kg. Its moment of inertia about its axis is The mass of the frame is 0.0250 kg. The gyroscope is supported on a single pivot (?Fig. E10.51?) with its center of mass a horizontal distance of 4.00 cm from the pivot. The gyroscope is precessing in a horizontal plane at the rate of one revolution in 2.20 s. (a) Find the upward force exerted by the pivot. (b) Find the angular speed with which the rotor is spinning about its axis, expressed in rev/min. (c) Copy the diagram and draw vectors to show the angular momentum of the rotor and the torque acting on it.

Step-by-Step Solution:

Solution 53E Step 1: a) The upward force is nothing but torque on the axis, = Force × Lever distance = mg r Mass of the system, m = mass of the gyroscope + mass of the frame = 0.140 kg + 0.0250 kg m = 0.165 kg Acceleration due to gravity, g = 9.8 m/s 2 Lever distance, r = 4 cm = 0.04 m 2 Therefore, torque, = 0.165 kg × 9.8 m/s × 0.04 m = 0.065 Nm The upward force exerted on the axis, = 0.065 Nm Step 2: b) The precession rate, = Torque / Angular momentum = mgr / I Where, I = moment of inertia of the gyroscope = Angular velocity of the rotor about its own axis. Provided, I = 1.20 × 10 kg m 2 Rate of precession, = 1 rev / 2.20 s = 0.454 rev/ s 1 rev = 2 rad Therefore, = 0.454 × 2 rad /s = 2.851 rad/s Substituting all these values in the above equation, -4 2 2.851 rad/s = 0.065 Nm / ( 1.20 × 10 kg m × ) -4 2 Rearranging the equation, we get, = 0.065 Nm / ( 1.20 × 10 kg m × 2.851 rad/s ) The angular velocity of the rotor, = 190 rad/s 1 rad = 1/2 revolutions 1 s = 1/60 min Therefore, 1 rad/s = (1/2) / ( 1/60) rev/min = 9.554 rev/min Therefore, angular velocity of the rotor, = 190× 9.554 rev/min = 1815.26 rev/ min

Step 3 of 3

Chapter 10, Problem 53E is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

This full solution covers the following key subjects: its, mass, gyroscope, rotor, pivot. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. This textbook survival guide was created for the textbook: University Physics, edition: 13. University Physics was written by and is associated to the ISBN: 9780321675460. The full step-by-step solution to problem: 53E from chapter: 10 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. Since the solution to 53E from 10 chapter was answered, more than 465 students have viewed the full step-by-step answer. The answer to “The rotor (flywheel) of a toy gyroscope has mass 0.140 kg. Its moment of inertia about its axis is The mass of the frame is 0.0250 kg. The gyroscope is supported on a single pivot (?Fig. E10.51?) with its center of mass a horizontal distance of 4.00 cm from the pivot. The gyroscope is precessing in a horizontal plane at the rate of one revolution in 2.20 s. (a) Find the upward force exerted by the pivot. (b) Find the angular speed with which the rotor is spinning about its axis, expressed in rev/min. (c) Copy the diagram and draw vectors to show the angular momentum of the rotor and the torque acting on it.” is broken down into a number of easy to follow steps, and 115 words.

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