Chapter 11, Problem 045 Your answer is partially correct. Try again. A man stand

Question

Chapter 11, Problem 045 Your answer is partially correct. Try again. A man stands on a platform that is rotating (without friction) with an angular speed of 2.24 rev/s; his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks, and platform about the central axis is 7.65 kgm2. If by moving the bricks the man decreases the rotational inertia of the system to 1.75 kg-m2, (a) what is the resulting angular speed of the platform and (b) what is the ratio of the new kinetic energy of the system to the original kinetic energy? (a) Numbe 9.79 (b) Number4 Click if you would like to Show Work for this question: UnitsT rad/s Units No units 4 Open Show Work

Explanation / Answer

here,

the initial angular speed , wi = 2.24 rev/s = 14.1 rad/s

the initial moment of inertia , Ii = 7.65 kg.m^2

the final moment of inertia , If = 1.75 kg.m^2

let the final angular speed be wf

a)

using conservation of angular momentum

Ii * wi = If * wf

7.65 * 14.1 = 1.75 * wf

wf = 61.5 rad/s

b)

the ratio of kinetic energy , R = (0.5 * If * wf^2)/( 0.5 * Ii * wi^2)

R = ( 1.75 * 61.5^2 )/(7.65 * 14.1^2)

R = 4.35

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