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: Problem Set V Solutions 1. Consider masses m 1 ,m 2 ,m 3 at x 1 ,x 2 ,x 3 . Find X , the CM coordinate by finding X 12 , the CM of mass of 1 and 2, and combining it with m 3 . Show this is gives the same result as X = 3 i =1 m i x i 3 i =1 m i . The center of mass for masses 1 and 2 is given by X 12 = m 1 x 1 + m 2 x 2 m 1 + m 2 . The center of mass for all three masses is given by X = ( m 1 + m 2 ) X 12 + m 3 x 3 m 1 + m 2 + m 3 = ( m 1 + m 2 ) m 1 x 1 + m 2 x 2 m 1 + m 2 + m 3 x 3 m 1 + m 2 + m 3 = m 1 x 1 + m 2 x 2 + m 3 x 3 m 1 + m 2 + m 3 as we would expect. 2. Consider a square of mass 4 kg, side 2m, negligible thickness, with its sides oriented along the usual axes with its center at (0 , 0) . (i) Determine its CM using symmetry arguments. (ii) Imagine that the 1m 1m part of it in the fourth quadrant is chopped off. Where is the new CM? Do this using the extension of result in previous problem. Repeat using the following trick: view the chopped off shape as the full square plus a 1m 1m square of negative mass 1 kg in the fourth quadrant. (iii) A disk of radius R centered at the origin has a circular hole of radius R/ 2 centered at ( x = R/ 2 ,y = 0) . Where is its CM? (i) The center of mass is located in the middle of the square. This must be the case because flipping the square around the x or y axis does not alter the square, so the center of mass cannot move during these rotations. (ii) Using the method from problem 1, we can consider this shape as three 1 kg squares, one in each quadrent. Taking m i to be the square in quadrant i , ~ X 12 will be ( x = 0 ,y = 0 . 5 m ). Then, the x coordinate for the center of mass will be X = 2 kg (0) + 1 kg ( . 5 m ) 3 kg = 1 6 m and the y coordinate will be Y = 2 kg (0 . 5 m ) + 1 kg ( . 5 m ) 3 kg = 1 6 m. So, the center of mass is at ~ X = ( 1 / 6 m, 1 / 6 m ). For the second method, we can think of the square without the piece in the fourth quadrant as a mass of M = 4 kg at (0 , 0) plus a mass of 1 kg = M/ 4 at (0.5 m, 0.5 m). The x coordinate of the center of mass is x cm = ( M )(0) + ( M/ 4)(0 . 5 m ) M M/ 4 = 1 4 4 3 1 2 m = 1 6 m. 2 Similarly, the y coordinate is y cm = ( M )(0) + ( M/ 4)( . 5 m ) M M/ 4 = 1 6 m. As above, the center of mass of the square missing the piece in the fourth quadrant is ( 1 / 6 m, 1 / 6 m ). (iii) If the mass of the full disk is M , the mass of the missing piece will just be M times the ratio of the missing piece to the full piece. M missing = M ( R/ 2) 2 R 2 = M 4 Following the argument for the square, the center of mass of a circle must be in the middle of the circle. Therefore, we have a mass of M at (0,0) and a mass of M/ 4 at (R/2,0). Since both masses are centered at y = 0, y cm must equal 0. Solving for x cm x cm = M (0) + ( M/ 4)( R/ 2) M M/ 4 = R 6 The center of mass for the circle with a hole in it is at ( R/ 6 , 0)....
View
Full
Document
 Fall '08
 RAMAMURTISHANKAR
 Physics, Mass

Click to edit the document details