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Advanced Physics THP#4 Rotational Dynamics point mass m = 1kg ...

Advanced Physics THP#4 Rotational Dynamics point mass m = 1kg ...

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<strong>Advanced</strong> <strong>Physics</strong> <strong>THP#4</strong> <strong>Rotational</strong> <strong>Dynamics</strong>These problems are open book and open notes. You may not consult or confer with anyone other thanMr. Burns. Full credit will be awarded for each problem only if the correct answer is accompanied bysufficient work that is presented in such a manner that the logic and mathematical operations are clearand easy to follow. All answers must be boxed, labeled and accompanied by the appropriate units.frictionless hingeuniform rodM =10 kgL = 2 m<strong>mass</strong>less string<strong>point</strong> <strong>mass</strong>m = <strong>1kg</strong>0.5 m1. (10) A 10 kg uniform rod with a length of 2 meters is connected to a frictionless hinge. Therod is released from rest and it swings down and strikes a 1 kg <strong>point</strong> <strong>mass</strong> (initially at rest) thathangs from the ceiling at a <strong>point</strong> that is 0.5 m from the end of the rod.a. Calculate the initial angular acceleration of the rod. (I rod = 1 / 3 ML 2 ).b. Find the angular acceleration of the rod when the rod is at a 36.87° angle relative to its starting<strong>point</strong> (below horizontal).c. Find the angular speed of the rod just before impact.d. The angular speed of the rod after impact is 1 rad/sec. Find the linear speed of the <strong>point</strong> <strong>mass</strong>after impact.


2. (10) A space shuttle astronaut in circular orbitaround the Earth has an assembly consisting of two smalldense spheres (pt. <strong>mass</strong>es) each of <strong>mass</strong> M, whose centersare connected by a rigid rod of length L, and negligible<strong>mass</strong>. The astronaut also has a device that will launch asmall lump of clay of <strong>mass</strong> M at speed v 0 . Express youranswers in terms of M, v 0 , L and fundamental constants.MLMv 0Initially the assembly is “floating” freely at rest relative tothe cabin. The astronaut launches the clay lump so that itperpendicularly strikes and sticks to one of the spheres ofthe assembly.Ma. (2) Determine the distance from the left end of the rod to the center of <strong>mass</strong> of thesystem (assembly and clay lump) immediately after the collision. (Assume that the radii of the spheresand the clay lump are much smaller than the separation of the spheres.)b. (1) On the figure above, indicate the direction of motion of the center of <strong>mass</strong>immediately after the collision. (Hint: Momentum is a vector quantity.)c. (2) Determine the linear speed of the center of <strong>mass</strong> immediately after the collision.d. (3) Determine the angular speed of the system (assembly and clay lump) around(relative to) the center of <strong>mass</strong> immediately after the collision.e. (2) Determine the change in kinetic energy as a result of the collision.3. (10) A <strong>mass</strong>less rope is wrapped around a 30 kg turntable (I = 3.75 kg-m 2 ) that is free to spinon a frictionless axle. The rope passes over an ideal pulley and is attached to a 10 kg <strong>mass</strong>. The systemis released from rest.a. (1) The tension in the rope is less than 100 N. Why?b. (3) Determine linear acceleration of the <strong>mass</strong>.c. (1) Determine the angular acceleration, α, of the turntable.d. (2) The rope detaches from the turntable after 5 seconds. Find the final angular speed, ω, ofthe table.e. (3) A 4 kg lump of clay (a <strong>point</strong> <strong>mass</strong>) is now dropped (straight down) onto the turntable ata <strong>point</strong> halfway between the center and the edge of the turntable. Calculate the new angular speed ofthe turntable.M (turntable) = 30 kgR = 0.5 m<strong>mass</strong>less ropeidealpulleyfrictionless axle10 kg

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