Kinectics of particles. impact between 2 bodies 1 of which is stationary

[bobmarly12345]

Homework Statement

An engine is suddenly coupled to a rotating drum by a friction clutch. the moment of inertia of the engine is equivalent to a mass of 40kg acting with a radius of gyration of 111mm. the drum has a mass of 10kg and a radius of gyration of 100mm. the initial velocity of the engine before engagement is 12rev/s and the drum is initially at rest. find the velocity of engine and drum immediately after connection.

Homework Equations

total moment before impact = total moment after impact
(m1.u1) +(m2.u2) = (m1.v1)+(m2.v2)

The Attempt at a Solution

ive seen questions like this before without the gyration part in the question & i know that i'd just plug the weight into the equation along with the velocity for example
(40x12) +(10x0) = (40xV)+(10xv) and then just rearrange to find the end velocity. However the addition of the moment of inertia and radius gyration has thrown me. do i use the equation I=m.k² and use the answers for the engine and drum and replace the m1 and m2 with the I1 and I2 value id get from that equation?

 

[gneill]

Radius of gyration, r_g is related to the mass and moment of inertia of a rotating body:

$r_g^2 = \frac{I}{M}$

 

[bobmarly12345]

yeah I've got that part but how does the moment of inertia from that equation help me with finding the impact of the engine and drum? do i replace the mass in the impact before and after with I=M.R² so it would be (I1.u1) +(I2.u2) = (I1.v1)+(I2.v2) or do i use a different formula all together? I understand the theory of why the gyration is given in the question but how do i apply all the values I've been given into finding the answer?

 

[gneill]

You're dealing with an inelastic collision of the angular variety. When the bodies come together their moments of inertia combine...

 

[bobmarly12345]

so from that then would it follow that (I1.u1) +(I2.u2) = (I1+I2).V2,,,,,, V2=(I1.u1) +(I2.u2)/(I1+I2). therefore, I1=40X0.111m²=0.49284 & 10x0.1m²=0.1 then v2=9.98rev/s ? Would you agree with that or am i going in the wrong direction :)

 

[gneill]

so from that then would it follow that (I1.u1) +(I2.u2) = (I1+I2).V2,,,,,, V2=(I1.u1) +(I2.u2)/(I1+I2). therefore, I1=40X0.111m²=0.49284 & 10x0.1m²=0.1 then v2=9.98rev/s ? Would you agree with that or am i going in the wrong direction :)

Almost there... Remember that it's body 2, the motor, that's initially rotating.

 

[bobmarly12345]

oh of course yes! thank you for your help :)