- #1
daveamal
- 15
- 0
Stuck in this question -- 2 spinning disks brought together...
Two disks are spinning freely about axes that run through their respective centres. The larger disk
(R1 = 1.42 m)
has a moment of inertia of 1180 kg · m2 and an angular speed of 4.0 rad/s. The smaller disk
(R2 = 0.60 m)
has a moment of inertia of 906 kg · m2 and an angular speed of 8.0 rad/s. The smaller disk is rotating in a direction that is opposite to the larger disk. The edges of the two disks are brought into contact with each other while keeping their axes parallel. They initially slip against each other until the friction between the two disks eventually stops the slipping. How much energy is lost to friction? (Assume that the disks continue to spin after the disks stop slipping.)
What I did was,
First I found w(angular velocity when the speed becomes same) by:
w=(w1I1+w2I2)/(I1+I2)
The the Kf=1/2w^2(I1+I2)
And Ki=1/2w1^2I1+1/2w2^2I2
Then I found the energy lost by Kf-Ki but The answer is wrong...
Two disks are spinning freely about axes that run through their respective centres. The larger disk
(R1 = 1.42 m)
has a moment of inertia of 1180 kg · m2 and an angular speed of 4.0 rad/s. The smaller disk
(R2 = 0.60 m)
has a moment of inertia of 906 kg · m2 and an angular speed of 8.0 rad/s. The smaller disk is rotating in a direction that is opposite to the larger disk. The edges of the two disks are brought into contact with each other while keeping their axes parallel. They initially slip against each other until the friction between the two disks eventually stops the slipping. How much energy is lost to friction? (Assume that the disks continue to spin after the disks stop slipping.)
What I did was,
First I found w(angular velocity when the speed becomes same) by:
w=(w1I1+w2I2)/(I1+I2)
The the Kf=1/2w^2(I1+I2)
And Ki=1/2w1^2I1+1/2w2^2I2
Then I found the energy lost by Kf-Ki but The answer is wrong...