Ball of putty hits rod (rotation)

[jonnyboy]

[SOLVED] Ball of putty hits rod (rotation)

I need help with three problems. For all of these topics we spent a small amount of class time on. And I'm having trouble knowing where to begin. I have posted each on their own thread.

A long thin rod of mass of .2 kg and length .04 m lie flat on a frictionless horizontal surface. A ball of putty of mass .05 kg traveling 3 m/s strikes the rod a distance (b) from the end of the stick. Treat the ball of putty as a point mass. Assume all

a. upon impact, determine the location of the center of mass
b. immediately after impact, what is the angular velocity about the axis through the center of mass of the system?
c. using differential calculus, determine the value of (b) that will cause the system to have the greatest angular velocity through the center mass of the system.

I'm just looking for the steps that I need to make to solve these questions. Like, how can I arrive to the formula to locate the center of mass of the system. Your help is greatly appreciated.

 

[jIyajbe]

I probably shouldn't try to answer questions when I'm tired, and without thinking it through, but hey...

For part a), define the system to be the rod + the putty. I would find the center of mass (COM) of this system by first noting (or assuming, depending on what the problem said) that the rod is uniform, and so its COM is at its geometric center; then use the standard equation for COM using the location of the rod's COM and the location of the putty:

Let R be the COM of the system, and M be the total mass of the system; i.e.,

M = \sum {m_i}

Then,

R = \frac{m_{1}r_{1}+m_{2}r_{2}}{M}

For (b)...If you have already studied conservation of energy, and the work-energy theorem, that would be the easiest way; if not, use conservation of angular momentum. (The putty's initial L = mvr.)

For part (c), sounds like you want to take the derivative of the expression for \omega, and set that derivative equal to zero.

HTH,

jIyajbe

 

[jonnyboy]

for part b, the rod inertia that I can use is I= (1/12)ml^2, right? Then I can use the parallel axis theorem, I = I_com + md^2

 

[jIyajbe]

Yup. Sounds good to me!