exam3_2004S_sol

exam3_2004S_sol - Mark answers in spaces 53-75 on the...

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PHYSICS 221 Spring 2004 EXAM 3: April 15 2004 8:00pm—9:30pm SOLUTIONS [53] Disk #1 of uniform density of mass 1kg and radius 2m is spinning centered on a massless turntable with angular velocity ω =12 rad/s. Disk #2 of uniform density of mass 2kg and radius 1m is dropped on it and due to friction between the disks, the system of two disks eventually rotate at a common angular velocity about the axis of the turntable. What is the final angular velocity of the system? (A) 2 rad/s (B) 4 rad/s (C) 6 rad/s (D) 8 rad/s (E) 10 rad/s The moment of inertia of disk 1 is I 1 =(M 1 r 1 ²)/2=2 kg m². The moment of inertia of disk 2 is I 2 =(M 2 r 2 ²)/2=1 kg m² . The initial angular momentum is L 1 =I 1 ω 1 =(2 kg m²)(12 rad/s)=24 J s. Since the situation applies no torque to the system, the angular momentum will be preserved so the final angular momentum will be the same. The final moment of inertia is I f =I 1 +I 2 =3 kg m². The final angular velocity is therefore ω f =L f / I f =(24Js)/(2 kg m²)=8rad/s. Correct response=60% Mark answers in spaces 53-75 on the answer sheet No class Friday April 16 No class Friday April 16 Massless Turntable Disk 1 r 1 =2m; M 1 =1kg Disk 2 r 2 =1m; M 2 =2kg
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[54] What is the x-component of the angular momentum about the origin of a particle located at the point m j r ) ˆ 2 ( = G with mass 2kg and a velocity of s m k j i v / ) ˆ 3 ˆ 2 ˆ ( + + = G (A) +6 kg m²/s (B) -6 kg m²/s (C) +12 kg m²/s (D) -12 kg m²/s (E) 0 kg m²/s Angular momentum is given by s m kg s m k j i m j kg v m r L / 12 / ) ˆ 3 ˆ 2 ˆ )( ˆ 2 )( 2 ( 2 + = + + = × = G G G Correct response=44% The most popular wrong answer was A which I guess results from forgetting to multiply by the mass.
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[55] Three balls are rolled down an incline ramp and roll without slipping. Ball X is a solid ball of radius 5cm and mass 1kg. Ball Y is another solid ball of radius 10cm and mass 0.5kg. Ball Z is a hollow ball of radius 5cm and mass 1kg. Neglecting drag, kinetic and rolling friction, if these three balls are released simultaneously from the top of the ramp, in which order do they arrive at the bottom? (A) X, Y, Z (B) Z, Y, X (C) X tied with Y followed by Z (D) X tied with Z followed by Y (E) All three tied. As these balls roll down the ramp, they should undergo uniform acceleration so that if v f is the final velocity at the bottom of the ramp and L is the length of the ramp, the time to get to the bottom is: f v L t / 2 = Thus, the ordering of the objects is the objects in time is the reverse of the ordering in terms of final velocity. The fastest final velocity is the ball which gets to the bottom first (hopefully this intuitively obvious). The problem is thus reduced to determining the velocity of the balls at the bottom of the ramp. Consider now a ball of radius r, moment of inertia I and mass m. Denote the height of the ramp by h. The total kinetic energy which the ball will have at the bottom of the ramp will be K=mgh.
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This note was uploaded on 03/26/2008 for the course PHYS 221 taught by Professor Herrera-siklody during the Spring '08 term at Iowa State.

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exam3_2004S_sol - Mark answers in spaces 53-75 on the...

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