In a friction clutch, a disc of moment of inertia I1 is rotating freely with an

Question

In a friction clutch, a disc of moment of inertia I1 is rotating freely with an angular speed of wi on a shaft whose moment of inertia is negligible. A second disc, with moment of inertia i2 and initially at rest, is coupled to the first disc such that a short time after the two discs are in contact, they rotate with a common angular speed wf:

If angular momentum is conserved, which of the following statements MUST be true during this process?

The net (resultant) torque acting on the pair of discs is zero at any time during this process.

B) The net (resultant) torque acting on the pair of discs is non-zero.

The torque on the disc with the larger moment of inertia is zero is always zero during this process.

The net (total) kinetic energy of the two discs remains constant during this process

If the initial KE is 1000 Joules, then which of the following isthe KE LOST in this process will be closest to (in Joules):
A)300, B)700, C)600, D)400

The net (resultant) torque acting on the pair of discs is zero at any time during this process.

Explanation / Answer

Statement is true

A)

The net (resultant) torque acting on the pair of discs is zero at any time during this process.

if angular momentum is conserved that means there is no net torque acting on the system, if it happend then angular momenum will not be conserved.

now,

conserve momentum

I1w1 = (I1+I2)w

w = I1w1/(I1+I2)

initial Rotational KE = 0.5I1w1^2

fnal KE = 0.5(I1+I2)w^2

Final KE = 0.5(I1+I2)[I1w1/(I1+I2)]^2

Final KE = 0.5(I1w1)^2 / [I1+I2]

so it would be around 33%

answer A is correct

A)

The net (resultant) torque acting on the pair of discs is zero at any time during this process.

if angular momentum is conserved that means there is no net torque acting on the system, if it happend then angular momenum will not be conserved.

now,

conserve momentum

I1w1 = (I1+I2)w

w = I1w1/(I1+I2)

initial Rotational KE = 0.5I1w1^2

fnal KE = 0.5(I1+I2)w^2

Final KE = 0.5(I1+I2)[I1w1/(I1+I2)]^2

Final KE = 0.5(I1w1)^2 / [I1+I2]

so it would be around 33%

answer A is correct

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