625_Dynamics 11ed Manual

625_Dynamics 11ed Manual - The uniform stone (rectangular...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Engineering Mechanics - Dynamics Chapter 18 *Problem 18-48 The semicircular segment of mass M is released from rest in the position shown. Determine the velocity of point A when it has rotated counterclockwise 90°. Assume that the segment rolls without slipping on the surface. The moment of inertia about its mass center is I G . Given: M 15 kg = r 0.15 m = I G 0.25 kg m 2 = d 0.4 m = Solution: Guesses ω 1 rad s = v G 1 m s = Given Mgd 1 2 Mv G 2 1 2 I G 2 + Mg d r () + = v G ω 2 r = ω G Find ω G , () = ω rad s = v G 0.62 m s = v A 0 0 ω d 2 d 2 0 × = v A 2.48 2.48 0.00 m s = v A 3.50 m s = Problem 18-49 The uniform stone (rectangular block) of weight
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: The uniform stone (rectangular block) of weight W is being turned over on its side by pulling the vertical cable slowly upward until the stone begins to tip. If it then falls freely ( T = 0) from an essentially balanced at-rest position, determine the speed at which the corner A strikes the pad at B . The stone does not slip at its corner C as it falls. Given: W 150 lb = a 0.5 ft = b 2 ft = g 32.2 ft s 2 = Solution: Guess 1 rad s = 623...
View Full Document

Ask a homework question - tutors are online