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Falling Rod, Rotation?

  • Thread starter blue5t1053
  • Start date
23
1
Problem:
A rigid assembly is made of a light weight rod 2 m long and a heavy gold ball which is attached at the middle of the rod. The rigid assembly initially rests with one end on the ground in a vertical position. If released what speed does the top of the rod have when it strikes the ground?

Equations:
[tex]Rotational \ Inertia \ for \ a \ thin \ rod \ about \ central \ axis; \ \ I = \frac{1}{12} M L^{2}[/tex]

[tex]Rotational \ Inertia \ for \ a \ thin \ rod \ about \ non-central \ axis; \ \ I = \frac{1}{12} M L^{2} + M R(from \ center)^{2}[/tex]

[tex]Rotational \ Inertia \ for \ a \ sphere; \ \ I = \frac{2}{5} M r^{2}[/tex]

My Work:
Any help where to start? I am lost. I know that the final answer is 8.85 m/s.
 

Answers and Replies

335
0
Show some attempt..
 
tiny-tim
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[tex]Rotational \ Inertia \ for \ a \ thin \ rod \ about \ central \ axis; \ \ I = \frac{1}{12} M L^{2}[/tex]

[tex]Rotational \ Inertia \ for \ a \ thin \ rod \ about \ non-central \ axis; \ \ I = \frac{1}{12} M L^{2} + M R(from \ center)^{2}[/tex]

[tex]Rotational \ Inertia \ for \ a \ sphere; \ \ I = \frac{2}{5} M r^{2}[/tex]
Hi blue5t1053! :smile:

You're making this too complicated … no wonder you don't know where to start. :rolleyes:

When a question uses words like "lightweight" and "heavy", it means that you can ignore the lightweight one.

So you don't need to consider moments of inertia … the rod is lightweight, so it counts as 0, and the ball is heavy, so you can regard it as a point.

So this is just a point mass. :smile:

(But is the ground frictionless, or is the rod attached to the ground? :confused:)

Now try it … ! :smile:
 
23
1
I am to assume that the ground is frictionless and that it will rotate from where the rod had made contact with the ground.

I can't find any way to calculate it without mass. What principle would I use to find it? I tried using angular velocity, but I couldn't make it work.
 
23
1
Would this be right?

[tex]m*g*h = \frac{1}{2} * m * v^{2}[/tex]

[tex]9.8 \frac{m}{sec^{2}} * 1 m = \frac{1}{2} * v^{2}[/tex]

[tex]\sqrt{\frac{9.8 \frac{m}{sec^{2}} * 1 m}{\frac{1}{2}}} = v[/tex]

[tex]v = 4.427 \frac{m}{sec} \ ; \ then *2 m \ for \ top \ of \ rod[/tex]

[tex]v = 8.85 \frac{m}{sec}[/tex]
 
tiny-tim
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Hi blue5t1053! :smile:

You don't need the mass … it will be the same on both sides of any equation … it always cancels out.

Draw a diagram showing the rod at a typical angle θ.

Since you don't know the value of the normal force… and you'd rather avoid calculating it … get an angular acceleration by taking moments about the bottom of the rod! :smile:
 
tiny-tim
Science Advisor
Homework Helper
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