Problem on Rotational mechanics / Conservation of angular momentum.

AI Thread Summary
The discussion centers on solving a problem involving two cylinders rotating and coming into contact, where slipping ceases due to friction. The user initially solved the problem using angular impulse and the change in angular momentum but seeks a solution based on the conservation of angular momentum. They derived an expression for the new angular speeds after slipping ceases, but their result differs from the expected answer. The correct formula incorporates the radii squared in the denominator, indicating a need for careful application of conservation principles. Clarification on the correct approach to using conservation of angular momentum is requested.
cantgetaname
Messages
41
Reaction score
0

Homework Statement


The figure shows two cylinders rotating about their axes, all the variables involved are shown.
The two cylinders are moved closer to touch each other keeping the axes parallel. The cylinders first slip over each other at the point of contact but slipping finally ceases due to friction between them.

Find the angular speeds of the cylinders after the slipping ceases.
2ztghkw.jpg

Homework Equations



  1. After slipping ceases, if w1' and w2' are new angular speeds then...
    w1' * r1 = w2' * r2
  2. Angular momentum = Iw.

The Attempt at a Solution


I have succeeding in solving the problem by equating the angular impulse due to friction to the change in angular momentums of both cylinders and eliminating friction force from the two equations.

But I'd like to solve this using conservation of angular momentum as well since it should be pretty straightforward and easy.

Since no external torque acts on the system, conserving angular momentum, I get:

I_{1}\omega_{1}+I_{2}\omega_{2}=I_{1}\omega_{1}' + I_{2}\omega_{2}'

Using this with w1' * r1 = w2' * r2, I get

\omega_{1}' = \frac{I_{1}\omega_{1}+I_{2}\omega_{2}}{I_{1}r_{2}+I_{2}r_{1}} r_{2}
and a similar expression for w2'.

The answer, however, is \omega_{1}' = \frac{I_{1}\omega_{1}r_{2}+I_{2}\omega_{2}r_{1}}{I_{1}r_{2}^{2}+I_{2}r_{1}^{2}} r_{2}
 
Last edited:
Physics news on Phys.org
Anyone?
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top