Initial acceleration of two particles on a rod

AI Thread Summary
Two particles, A and B, each with a mass of 10 kg, are attached to a massless rod placed on a fulcrum, with A 10 cm from the fulcrum and B 70 cm away. The initial acceleration of particle A is derived from the net torque equation, where the difference in torque due to gravity is calculated as 0.6mg. The moment of inertia (I) is expressed as m(r_a^2 + r_b^2), leading to the angular acceleration (α) formula. The relationship between linear acceleration and angular acceleration is clarified, correcting the initial misunderstanding about the placement of r in the equations. The revised calculations for the accelerations of both particles are now based on the correct application of these principles.
DeviantSOP
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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:
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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.
 
Last edited:
Hi DeviantSOP,

DeviantSOP said:
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?
 
alphysicist said:
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.
 
DeviantSOP said:
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

<br /> a= r \alpha<br />

so I believe you had the r on the wrong side of the equation.
 
alphysicist said:
The relationship between linear (tangential) acceleration and angular acceleration is

<br /> a= r \alpha<br />

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
 

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