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Question

The figure shows three rotating, uniform discs that are coupled by belts. One belt runs around the rims of discs A and C. Another belt runs around a central hub on the disc A and the rim of the disc B. The belts move smoothly without slippage on the rims and hub. Disc A has radius R; its hub has radius 0.500R; disc B has radius 0.250R and disc C has radius 2.00R. Discs B and C have the same density (mass per unit volume) and thickness. What is the ratio of the magnitude of the angular momentum of the disc C to that of the disc B ?

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Solution

The correct option is **B** 1024:1

Given:

Radius of disc A, rA=R

Radius of hub of disc A, rAh=0.500R

Radius of disc B, rB=0.250R

Radius of disc C, rC=2.00R

The motion of the all three discs can be related by linear velocity of the belts.

Let the angular velocities of the discs A, B and C be ωA, ωB and ωC respectively.

Then, for the hub of disc A and disc B,

⇒ωArAh=ωBrB

⇒ωA×0.500R=ωB×0.250R

⇒2ωA=ωB ……(1)

Now for disc A and C,

ωArA=ωCrC

⇒ωAR=ωC×2.00R

⇒ωA=2ωC ……(2)

From equation (1) and (2),

⇒ωB=4ωC ……(3)

Let the thickness and density [of each disc B and C] be t and ρ respectively.

Angular momentum of a disc is given by L=Iω=mr22ω

Now, Angular momentum of disk CAngular momentum of disk B=0.5(ρπr2Ct)×r2CωC0.5(ρπr2Bt)×r2BωB

⇒LCLB=r4CωCr4BωB

⇒LCLB=(2R)4ωC(0.25R)4×4ωC [from (3)]

⇒LCLB=1024:1

Given:

Radius of disc A, rA=R

Radius of hub of disc A, rAh=0.500R

Radius of disc B, rB=0.250R

Radius of disc C, rC=2.00R

The motion of the all three discs can be related by linear velocity of the belts.

Let the angular velocities of the discs A, B and C be ωA, ωB and ωC respectively.

Then, for the hub of disc A and disc B,

⇒ωArAh=ωBrB

⇒ωA×0.500R=ωB×0.250R

⇒2ωA=ωB ……(1)

Now for disc A and C,

ωArA=ωCrC

⇒ωAR=ωC×2.00R

⇒ωA=2ωC ……(2)

From equation (1) and (2),

⇒ωB=4ωC ……(3)

Let the thickness and density [of each disc B and C] be t and ρ respectively.

Angular momentum of a disc is given by L=Iω=mr22ω

Now, Angular momentum of disk CAngular momentum of disk B=0.5(ρπr2Ct)×r2CωC0.5(ρπr2Bt)×r2BωB

⇒LCLB=r4CωCr4BωB

⇒LCLB=(2R)4ωC(0.25R)4×4ωC [from (3)]

⇒LCLB=1024:1

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