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Lecture_13

# Lecture_13 - 7.5 Center of Mass In analyzing the collisions...

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7.5 Center of Mass In analyzing the collisions of two objects, the mass of each object before the ollision is located at different points in space collision is located at different points in space. We define an average location in space for the total mass of the system, whether it be for 2 objects or n objects: This location in space is called the Center of Mass of the system. The Center of Mass is a position relative to some fixed coordinate system, which ay or may not be moving with the object(s) may or may not be moving with the object(s). Units? [displacement] = [ m ] For example, let’s say I have a solid sphere of mass m . Where is the center of mass of the sphere? Clearly, it’s at the center of the sphere. Now let’s say I have two spheres, m 1 and m 2 , where m 2 = m 1 . The pheres are separated by some distance Where is the center spheres are separated by some distance x . Where is the center of mass? m 1 m 2 CM x x CM x The center of mass is now located exactly between the two spheres.

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Now let’s say I have two spheres, m 1 and m 2 , where m 2 > m 1 . The spheres are separated by some distance x . Where is the center of mass? M m 1 m 2 x x CM The CM is no longer exactly between the two masses, since more of the system’s mass is located to the right, since m 2 > m 1 . Thus, the CM lies closer to the mass on the right. Note: In the last two examples, the CM was not located within either object of the ystem but was located out in space between them system, but was located out in space between them. Here we looked at just two objects, but we can generalize this to n objects. The location of the CM of n objects is given by: = = n n i i i x m x 1 CM n i i i x m x = = 1 CM = i i m 1 tot M
Example : A 5-kg mass is located 2 m from the origin along the x-axis. A 12-kg mass is located 10 m from the origin along the x -axis. What is the position of the CM of this system? = 12 kg x x = 0 m 1 12 kg m 1 = 5 kg x = 10 m x = 2 m x CM x = 7.65 m n i i x m x m x m + ) g)(10 2 ) g)(2 tot i M x = = 1 CM 2 1 2 2 1 1 m m + = kg 12 kg 5 m) kg)(10 (12 m) kg)(2 5 ( + + = m 65 . 7 = the CM were exactly between the two masses then its position would have been If the CM were exactly between the two masses, then its position would have been at x = 6 m. But, since the mass on the right is larger, the CM is shifted farther to the right at =765m x 7.65 m.

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Example : Where is the CM of the Earth-Moon system? Assume the Earth is located at the origin. 598 0 4 g m m = 7.35 × 10 22 kg m e = 5.98 × 10 24 kg x = 0 x = 3.85 × 10 8 m t n i i i x m x = = 1 CM m m e e m m x m x m + + = 0 m m m m x m = 22 24 8 22 0 5 0 8 ) 10 85 . 3 )( 10 35 . 7 ( × × × = m 10 67 . 4 6 × = tot m e m e 10 35 . 7 10 98 . 5 + The radius of the Earth is 6.38 × 10 6 m, and the location of the CM is less than this. Thus, the CM of the Earth-Moon x CM system, actually exists within the mass of the Earth!
Center of Mass and Conservation of Linear Momentum Let’s look at our collision example again: 25m/s v o1 = 25 m/s v o2 = 0 x CM v f = ?

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Lecture_13 - 7.5 Center of Mass In analyzing the collisions...

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