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Conservation of Angular Kinetic Energy 
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#1
Jun2012, 12:48 AM

P: 916

1. The problem statement, all variables and given/known data
I have some difficulty understanding how they have obtained the equation for conservation of kinetic energy in the following solved problem: There is a 2 kg disk traveling at 3.0 m/s strikes a 1.0 kg stick of length 4.0 m that is lying flat on nearly frictionless ice. The disk strikes the endpoint of the stick, at a distance r = 2.0 m from the stick's center. Suppose the collision is elastic and the disk does not deviate from its original line of motion. The moment of inertia of the stick about its center of mass is 1.33 kg.m^{2}. If we apply the conservation of kinetic energy to the system we get: [itex]\frac{1}{2}m_d v_{d i}^2 = \frac{1}{2}m_d v_{d f}^2 + \frac{1}{2}m_s v_{s}^2 + \frac{1}{2} I \omega^2[/itex] [itex]\frac{1}{2} (2)(3)^2 = \frac{1}{2}(2) v_{d f}^2 + \frac{1}{2}(1) v_{s}^2 + \frac{1}{2} (1.33) \omega^2[/itex] Here is a diagram: 3. The attempt at a solution I'm confused because the only rotational kinetic energy considered was the rotational kinetic energy of the stick after the collision. But the disk has a rotational kinetic energy too both before and after the collision, right? Shouldn't we model the motion of the rolling disk as a combination of both translation and rotation? If so this is what we would get: [itex]\frac{1}{2}m_d v_{d i}^2 + \frac{1}{2}I_{d} \omega_{d, i} = \frac{1}{2}m_d v_{d f}^2 + \frac{1}{2}m_s v_{s}^2 + \frac{1}{2} I_s \omega_{s}^2 + \frac{1}{2}I_{d} \omega_{d, f}[/itex] So, why have they ignored the rotational kinetic energy of the disk? Any help is greatly appreciated. 


#2
Jun2012, 12:56 AM

Emeritus
Sci Advisor
PF Gold
P: 5,196

I think the answer is simply that your assumption that the disc is rolling is wrong. It does not say anywhere in the problem that the disc is rotating. It's probably just sliding across the surface of the ice like a hockey puck. Therefore, it has only translational kinetic energy.



#3
Jun2012, 01:58 AM

PF Gold
P: 456

A simple advice in problems like these  Since the radius of disc was not given, it was not possible to determine the Moment of Inertia of the disc about its centre, so it was safe to assume that we need not consider its angular kinetic energy as far as this problem goes.



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