# Rotational Motion Problem

1. Nov 30, 2007

### physucsc11

1. The problem statement, all variables and given/known data

A thin rod of length L stands vertically on a table. The rod begins to fall, but its lower end does no slide. (a) Determine the angular velocity of the rod as a function of the angle $$\phi$$ it makes with the tabletop. (b) What is the speed of the tip of the rod just before it strikes the table?

2. Relevant equations

1) Torque = Moment of Inertia * angular acceleration
T = I * $$\alpha$$

T = Fdsin$$\phi$$

3. The attempt at a solution

I am not sure if this is the correct solution because I'm not sure if I'm thinking about the problem correctly.

We know there is a force from gravity Fg. It will depend on the angle of the rod with the table. This relation is given by Fg($$\phi$$) = mgsin$$\phi$$.

so T($$\phi$$) = mgsin$$\phi$$d = mg(L/2)sin$$\phi$$.

now $$\alpha$$ = $$\frac{T}{I}$$, so

$$\alpha$$($$\phi$$) = (mgLsin$$\phi$$)/2I,

and we know I for a rod in this case is 1/3mL^2

so $$\alpha$$($$\phi$$) = $$\frac{3gsin\phi}{2L}$$])

then I integrate the ang. acceleration to find the angular velocity from $$\pi$$/2 to 0.

I found $$\omega$$ = $$\frac{3g}{2L}$$ rad/sec .

I have good feeling I am leaving something out in this solution, such as possibly something to do with friction with the surface. I have a feeling the words "falls without slipping" mean something I am not realizing.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 30, 2007

### Staff: Mentor

No, one correctly interpreted the "falls without slipping", which means that the rod pivots about the lower end, rather than at the center of gravity or point in between. One selected the appropriate from of the MI for this condition.

http://hyperphysics.phy-astr.gsu.edu/hbase/mi2.html#irod

3. Nov 30, 2007

### physucsc11

Ah, yes. Sometimes when thinking about more complex problems I start getting confused about the simple things and start questioning everything I normally know.

Thanks anyways.