# Homework Help: Initial acceleration of two particles on a rod

1. Jul 23, 2008

### DeviantSOP

Two particles(A and B) each of mass 10kg are attached to a massless rod on each on opposite ends of the rod. The rod is placed horizontally on a fulcrum. Particle A is 10 cm away from the fulcrum. And particle B is 70cm away from the fulcrum. The rod is now released.
What is the magnitude of the initial acceleration of particle A?
What is the magnitude of the initial acceleration of particle A?

My solution so far
B_torque - A_torque
= mgb - mga
= mg(0.7)-mg(0.1)
=0.6mg

Torque = Iα
I = m(a^2 + b ^2)

Thats the only thing I can think off right now. Totally stuck have no idea what to do.
Wasn't in class when this lesson was given so i have no idea what i am doing.

Last edited: Jul 23, 2008
2. Jul 23, 2008

### DeviantSOP

My solution so far
B_torque - A_torque
= mgb - mga
= mg(0.7)-mg(0.1)
=0.6mg

Torque = Iα
I = m(r_a^2 + r_b^2)

0.6mg = m(r_a^2 + r_b^2) * α
α = 0.6g / (r_a^2 + r_b^2)

α = a * r
a = α/r

Acceleration of a = (0.6g / (r_a^2 + r_b^2) ) / 0.1
Acceleration of b = (0.6g / (r_a^2 + r_b^2)) / 0.7

Thats the only thing I can think off right now. Totally stuck have no idea what to do.
Wasn't in class when this lesson was given so i have no idea what i am doing.

Last edited: Jul 23, 2008
3. Jul 23, 2008

### alphysicist

Hi DeviantSOP,

I don't think these two equations are correct; for example, check the units on each side. Do you see what they need to be?

4. Jul 24, 2008

### DeviantSOP

Not really sure whats wrong with them. I used net torque = Iα and then I = mr^2 and also α = a * r .
As I said earlier, i have no idea what I am doing.

5. Jul 24, 2008

### alphysicist

The relationship between linear (tangential) acceleration and angular acceleration is

$$a= r \alpha$$

so I believe you had the r on the wrong side of the equation.

6. Jul 24, 2008

### DeviantSOP

Ha, can't beleive I missed that. Thanks a lot. Amending my solution.

My solution so far
B_torque - A_torque
= mgb - mga
= mg(0.7)-mg(0.1)
=0.6mg

Torque = Iα
I = m(r_a^2 + r_b^2)

0.6mg = m(r_a^2 + r_b^2) * α
α = 0.6g / (r_a^2 + r_b^2)

a = α * r

Acceleration of a = (0.6g / (r_a^2 + r_b^2)) * 0.1
Acceleration of b = (0.6g / (r_a^2 + r_b^2)) * 0.7