- #1
Naeem
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Q. A disk of mass M1 = 350 g and radius R1 = 10 cm rotates about its symmetry axis at finitial = 198 rpm. A second disk of mass M2 = 198 g and radius R2 = 5 cm, initially not rotating, is dropped on top of the first. Frictional forces act to bring the two disks to a common rotational speed ffinal.
a) What is ffinal? Please give your answer in units of rpm, but do not enter the units.
Ans. Since there are no external torques on the system, Angular momentum must be conserved.
Finally, the total angular momentum is due to both disks spinning:
Lf = I1w1 + I2w2 = MR2Wf
Since, Li = Lf
1/2 MR2wi = MR2Wf
So, Wf = 1/2Wi
I tried to put Wi as 198 rpm and found out Wf to be 99, which the computer says is wrong.
b) In the process, how much kinetic energy is lost due to friction?
Ans. The initial kinetic energy is 1/2 Iiw12, and the final Kinetic energy is
K2 = 1/2( I1 + 12 )w2
There fore the fraction lost is :
| Delta K | / K1
which is ,
(1/2( I1 + 12 )w2 - 1/2 Iiw12)/ 1/2Iw12, which is 2/3 , which the computer says is wrong.
Can anybody help!
a) What is ffinal? Please give your answer in units of rpm, but do not enter the units.
Ans. Since there are no external torques on the system, Angular momentum must be conserved.
Finally, the total angular momentum is due to both disks spinning:
Lf = I1w1 + I2w2 = MR2Wf
Since, Li = Lf
1/2 MR2wi = MR2Wf
So, Wf = 1/2Wi
I tried to put Wi as 198 rpm and found out Wf to be 99, which the computer says is wrong.
b) In the process, how much kinetic energy is lost due to friction?
Ans. The initial kinetic energy is 1/2 Iiw12, and the final Kinetic energy is
K2 = 1/2( I1 + 12 )w2
There fore the fraction lost is :
| Delta K | / K1
which is ,
(1/2( I1 + 12 )w2 - 1/2 Iiw12)/ 1/2Iw12, which is 2/3 , which the computer says is wrong.
Can anybody help!