Initial acceleration of two particles on a rod

[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.

 

[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)) / 0.7Thats 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.

 

[alphysicist]

Hi 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

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?

 

[DeviantSOP]

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?

Not really sure what's 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.

 

[alphysicist]

Not really sure what's 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.

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.

 

[DeviantSOP]

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.

Ha, can't believe 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