Chapter_11_Problems

Chapter_11_Problems - Problem 11-41 A man stands on a...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem 11-41 I 1 ω 1 = I 2 ω 2 ω 2 = (I 1 /I 2 1 ω 2 = (6.0/2.0)1.2 ω 2 = 3.6 rev/s K 2 /K 1 =( ½ I 2 ω 2 2 )/ (½ I 1 ω 1 2 ) K 2 /K 1 = 3.0 The extra kinetic energy came from the work done by the man in decreasing the rotational inertia by pulling the bricks closer. This energy came from the man’s store of internal energy. A man stands on a platform that is rotating (without friction) with an angular speed of 1.2 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 vertical axis of the platform is 6.0 kg · m 2 . If by moving the bricks the man decreases the rotational inertia of the system to 2.0 kg · m 2 , what are (a) the resulting angular speed of the platform and (b) the ratio of the new kinetic energy of the system to the original kinetic energy? (c) What source provided the added kinetic energy?
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Problem 11-43 I A ω 1 = (I A + I B 2 ω 2 = [(I A
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

Chapter_11_Problems - Problem 11-41 A man stands on a...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online