
#1
Apr1105, 02:12 AM

P: 150

Please help.
Consider a turntable to be a circular disk of moment of inertia I_t rotating at a constant angular velocity omega_i around an axis through the center and perpendicular to the plane of the disk (the disk's "primary axis of symmetry"). The axis of the disk is vertical and the disk is supported by frictionless bearings. The motor of the turntable is off, so there is no external torque being applied to the axis. Another disk (a record) is dropped onto the first such that it lands coaxially (the axes coincide). The moment of inertia of the record is I_r. The initial angular velocity of the second disk is zero. There is friction between the two disks. After this "rotational collision," the disks will eventually rotate with the same angular velocity. What is the final angular velocity, omega_f, of the two disks? Express omega_f in terms of I_t, I_r, and omega_i. Because of friction, kinetic energy is not conserved while the disks' surfaces slip over each other. What is the final kinetic energy, K_f, of the two spinning disks? Express the final kinetic energy in terms of I_t, I_r, and the initial kinetic energy K_i of the twodisk system. No angular velocities should appear in your answer. Assume that the turntable deccelerated during time deltat before reaching the final angular velocity ( deltat is the time interval between the moment when the top disk is dropped and the time that the disks begin to spin at the same angular velocity). What was the average torque, \avg{\tau}, acting on the bottom disk due to friction with the record? Express the torque in terms of I_t, omega_i, omega_f, and deltat. Ok, I know the answer to the first question, which is (I_t*omega_i)/(I_t+I_r). But im lost on the whole KE thing. My answer was ((I_t+I_r)*(I_t*K_i))/(2*(I_t^2+2*I_t*I_r+I_r^2)), which was incorrect. Any help would be appreciated for the last question as well. Thanks in advance. 



#2
Apr1105, 05:02 AM

P: 453

[tex]K_i = \frac{1}{2}I_t\omega _i^2[/tex] So express [tex]\omega _i[/tex] in terms of K_i and I_t and substitute it in the equation for K_f: [tex]K_f = \frac{1}{2}(I_t + I_r)(\frac{I_t\omega _i}{I_t + I_r})^2[/tex] So show us whether your answer will be the same as the one in the textbook. 



#3
Apr1105, 05:36 AM

P: 150

I worked it out...the reason i got it wrong is because I put Ki instead of 2ki.
The answer is (I_t+I_r)*(I_t*2*K_i))/(2*(I_t^2+2*I_t*I_r+I_r^2) The answer is long because you need to eliminate any angular velocity variables. 



#4
Apr1105, 05:46 AM

P: 453

Conservation of angular momentum
You mean:
[tex]\frac{(I_t + I_r)(2I_tK_i)}{2I_t^2 + 2I_tI_r + I_r^2}[/tex] ? Then it is simplified to [tex]\frac{I_t}{I_t + I_r}K_i[/tex] 



#5
Apr1105, 05:48 AM

P: 150

Yeah. Just didnt simplify though. How would you determine the torque when the turntable decelerates?




#6
Apr1105, 05:53 AM

P: 453

Easy,
[tex]\tau_{av} = \frac{\Delta L}{\Delta t}[/tex], where L is the angular momentum. 



#7
Apr1105, 05:55 AM

P: 150

lol...ok. These questions are weird. I found the hardest question to be the easiest for me, and the easy ones to be hard.
Thanks 



#8
Apr2805, 01:09 AM

P: 13

heh i got the first 2 easy, but im totally lost on the third, how do u work out the angular momentum?




#10
Apr2805, 08:16 PM

P: 13

heh i got it, but i knew that, i jsut didnt realise i needed to use torque= I*alpha, nstead of momentum/time. then u solve alpha down into terms of intital and final angular velocity over time, and bingo



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