ps8b21 - Ph 111.01 October 22, 2008 PS 8b (21) Problem 1 =...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Ph 111.01 October 22, 2008 PS 8b (21) M R α Problem 1 = Walker P. 11-7. A torque of 1.14 N · m is applied to a bicycle wheel of radius 27 cm and mass 0.66 kg. Treating the wheel as a hoop, find its angular acceleration. rad/s 2 Solution. The equivalent to F = ma for rotational motion is I τ α = . We can solve for the angular acceleration. But first, we need the moment of inertia. We use the form that applies to a hoop: () ( ) 2 22 0.66 kg 0.27 m 0.0481 kg-m IM R == = Then, the angular acceleration is ( ) 2 2 1.14 N-m 23.70 radians/s 0.0481 kg-m I = (Sounds fast . . . .) Problem 2 = Walker P. 11-11. A wheel on a game show is given an initial angular speed of 1.31 rad/s. It comes to rest after rotating through 3/4 of a turn. M R α (a) Find the average torque exerted on the wheel given that it is a disk of radius 0.78 m and mass 6.4 kg. N · m (b) If the mass of the wheel is halved and its radius is doubled , will the angle through which it rotates before coming to rest increase, decrease, or stay the same? increase decrease stay the same Explain. (Assume that the average torque is unchanged.) Solution. (a)
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 4

ps8b21 - Ph 111.01 October 22, 2008 PS 8b (21) Problem 1 =...

This preview shows document pages 1 - 2. Sign up to view the full document.

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