# Angular speed and acceleration of a rod

1. Jan 4, 2012

### songoku

1. The problem statement, all variables and given/known data
A rod with one end fixed on a smooth hinge is shown below.

The rod is released from horizontal position so that it rotates at the hinge. The mass of the rod is 800 gr and the length is 120 cm. Find:
a. the initial angular acceleration (at A)
b. the angular acceleration at B
c. angular speed at B
d. the speed of the edge of the rod at B

2. Relevant equations
KE rotation = 1/2 Iω2
KE translation = 1/2 mv2
PE = mgh
τ = I.α
τ = F.d

3. The attempt at a solution
The moment of inertia will be 1/3 ML2

a.
τ = I.α
F.d = 1/3 ML2
8 x 0.6 = 1/3 x 0.8 x (1.2)2 x α

b.
τ = I.α
8 x 0.6 sin 37° = 1/3 x 0.8 x (1.2)2 x α

c. don't know

d. My idea is using conservation of energy; the decrease of potential energy of the center of mass = the gain of rotational kinetic energy and translational kinetic energy

m.g.1/2 L cos 37° = 1/2 Iω2 + 1/2 mv2 → the question asks about v, right?

To solve for v, I have to get the ω from (c). Is my idea correct?

Thanks

2. Jan 4, 2012

### obafgkmrns

Try expressing ω as a function of v (or vice versa).

3. Jan 4, 2012

v = ωr ?

4. Jan 4, 2012

### BruceW

You've got parts a) and b) correct (although I don't know why you have used 8m/s^2 as the value of g, since it is roughly 9.8m/s^2)

For parts c) and d), you don't have to consider any translational KE. The moment of inertia you have given is around the hinge, so using this will mean that the rotational KE is the total KE.

5. Jan 4, 2012

### songoku

I used g = 10 m/s2. F = W = m.g = 0.8 x 10 = 8 N

So for (c), I just used:
m.g.1/2 L cos 37° = 1/2 Iω2
8 x 1/2 x 1.2 cos 37° = 1/2 x 1/3 x 0.8 x (1.2)2ω2

For (d), I used:
v = ωL = 4.47 x 1.2 = 5.36 m/s

Am I right? Thanks

6. Jan 5, 2012

### BruceW

Oh, yes that's right. I forgot that F=mg. Sorry about that!

Yep! that looks right to me.

7. Jan 5, 2012