Problem Set 5 Solutions

# Problem Set 5 Solutions - Problem Set V Solutions 1...

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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 π ( R/ 2) 2 πR 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)....
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Problem Set 5 Solutions - Problem Set V Solutions 1...

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