Finding Final Velocity of a Falling Rotating Rod

[blue5t1053]

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:
$$Rotational \ Inertia \ for \ a \ thin \ rod \ about \ central \ axis; \ \ I = \frac{1}{12} M L^{2}$$

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

$$Rotational \ Inertia \ for \ a \ sphere; \ \ I = \frac{2}{5} M r^{2}$$

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

 

[physixguru]

Show some attempt..

 

[tiny-tim]

$$Rotational \ Inertia \ for \ a \ thin \ rod \ about \ central \ axis; \ \ I = \frac{1}{12} M L^{2}$$

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

$$Rotational \ Inertia \ for \ a \ sphere; \ \ I = \frac{2}{5} M r^{2}$$

Hi blue5t1053!

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

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.

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

Now try it … !

 

[blue5t1053]

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.

 

[blue5t1053]

Would this be right?

$$m*g*h = \frac{1}{2} * m * v^{2}$$

$$9.8 \frac{m}{sec^{2}} * 1 m = \frac{1}{2} * v^{2}$$

$$\sqrt{\frac{9.8 \frac{m}{sec^{2}} * 1 m}{\frac{1}{2}}} = v$$

$$v = 4.427 \frac{m}{sec} \ ; \ then *2 m \ for \ top \ of \ rod$$

$$v = 8.85 \frac{m}{sec}$$

 

[tiny-tim]

Hi blue5t1053!

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!

 

[tiny-tim]

Would this be right? …

(I forgot you were only asked to find the final velocity! )

Yes, that looks fine!