141592653589793 thenbehaves like point particle in

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Unformatted text preview: enter of Mass=: 3.141592653589793.... Thenbehaves like= point particle in response Fnet,ext a M acm T∝ to the net external forces. L/g r Who wants to be an astronaut? Example: Astronauts & Rope • Two astronauts at rest in outer space are connected by a light rope. They begin to pull towards each other. Where do they meet? M = 1.5m Demo m Example: Astronauts & Rope... l l l l They start at rest, so VCM = 0. VCM remains zero because there are no external forces. So, the CM does not move! They will meet at the CM. m M = 1.5m CM L x=0 Finding the CM: If we take the astronaut on the left to be at x = 0: x cm M (0) + m( L) m( L) 2 = = =L M+m 2.5 m 5 x=L Ex: Center of Mass Motion • A man weighs exactly as much as his 20 foot long canoe. • Initially he stands in the center of the motionless canoe, a distance of 20 feet from shore. Next he walks toward the shore until he gets to the end of the canoe. –What is his new distance from the shore. (There no horizontal force on the canoe by the water). (transparency) 20 ft (a) 10 ft before 20 ft (b) 15 ft (c) 16.7 ft ? ft after Quick solution in this situation: Since the man and the canoe have the same mass, the CM of the man-canoe system will be halfway between the CM of the man and the CM of the canoe. l l Initially the CM of the system is 20 ft from shore. X X x 20 ft CM of system Solution l Since there is no force acting on the canoe in the x-direction, the location of the CM of the system can’t change! l Therefore, the man ends up 5 ft to the left of the system CM, and the center of the canoe ends up 5 ft to the right. l He ends up moving 5 ft toward the shore (15 ft away). 15 ft X 10 ft X 20 ft 5 ft CM of system x Measuring the center of mass If an object is suspended about an arbitrary point, the center of mass lies along the vertical line through that point. (due to torque, later.) Just suspend the object from two separate points and the CM at the intersection of the two vertical lines. Demo. suspend from here: center of mass: vertical line: net external force : Fnet,ext = Fi,ext. i Calculating Center of mass for continuous cm d r objects 2 acm = acceleration of center of mass = dt2 Breaking up a continuousnet,ext into infinitesimal pieces of Then : F object = M acm mass dm, the center of mass is expressed as a continuous integral: rcm 1 = M r dm Example worked out on transparency: Isosceles triangle T∝ Result: L/g π = 3.141592653589793.... h x h/3 r Then : Fnet,ext = M ￿ cm a ￿ Then : Fnet,ext = M ￿ cm a d2￿cm r ￿ cm = acceleration of center of mass = a dt2 ￿ 1 ￿￿ Then : 1Fnet,ext = M ￿ cm a ￿cm = r ￿ dm r ￿cmM r= ￿ dm r M ￿ Recall, that due to cancelation of Newton’s 3rd law pairs: ￿1 ￿ rcm ￿a r Fnet,ext...
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