This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 7A Midterm II Sample Questions - Solutions Kevin Young This is a list of sample problems that I will be solving for the midterm review on Saturday, 12-2 in 1 Pimentel. I may not get to solve all of these in the time allotted, and I will most likely be doing them out of order. Problem 1 - Semicircular rod A rod of mass, m, length, r, is bent into a semicircle. 1. Calculate the center of mass of the rod after it has been bent. Use symme- try considerations to reduce the problem to one-dimension. I would suggest polar coordinates for this integral. This problem is, like it suggests, best solved in polar coordinates. Be- cause of the symmetry of the problem, the axis must lie along the y-axis. Now, this reduces the problem to finding the center of mass in the y-direction. The formula for the center of mass is: y = R ydm M tot Where here dm = ds = r d , and = M tot /( l eng th ) = M tot /( r ) is the lin- ear mass density. And in polar coordinates, y = r sin . The integral becomes: y = R y dm M tot = R r 2 sin d M tot = r 2 M tot Z sin d = M tot r 2 M tot r (- cos ) = 2 r 2. Calculate the moment of inertia about an axis perpendicular to the plane of the semicircle and passing through the point midway between the ends of the rod. (This is like a ring spun about its center of mass, but with the bottom half cut away) 1 No calculation needs to be done here, as all of the mass is located the same distance ( r ) from the axis of rotation. So the moment of inertia is just I = mr 2 3. Calculate the moment of inertia about and axis perpendicular to the plane of the semicircle and through the center of mass. Dont do any integrals for this part. This is occasion for the parallel axis theorem. We know where the center of mass is, and we know what the moment of inertia through a different point is, so the parallel axis theorem tells us that: I = I cm + md 2 Or, cast into a more useful form: I cm = I- md 2 Where here, d is the distance from the center of mass to the center of curvature of the ring, d = 2 r . Then the moment of inertia is: I = mr 2- m ( 2 r ) 2 = (1- 4/ 2 ) mr 2 . Problem 2 - Mid-space Collision Two identical satellites are in orbit around the Earth. The Earth has mass 6 10 24 kg and radius 6400 km . The gravitational constant is G = 6.67 10- 11 N m 2 kg- 2 ....
View Full Document
This note was uploaded on 04/06/2009 for the course PHYSICS 7A taught by Professor Lanzara during the Spring '08 term at University of California, Berkeley.
- Spring '08