Torque and Newton third's law applied to rotatory motion.

[Bedeirnur]

Homework Statement

Two blocks of mass m are suspended on the ends of a rigid weightless rod of length L1 + L2; With L1=20 cm and L2= 80 cm. The rod is held horizontally on the fulcrum shown in the diagram and then released. Calculate the accelerations of the two blocks as they start to move

Homework Equations

Mtot(total torque) = M1 + M2
M1 = l1*T
M2 = -l2*T
Mtot = Itot (total moment of inertia) * α (angular acceleration)
(T is the tension)

The Attempt at a Solution

I've really thought on it but i can think of nothing if not that i will have to use those formulaes to resolve it...

I thought that as the two masses are equal, even the two tensions are equal so T1=T2=T

The forces that causes the rod to rotate are T1 and T2 and act on the ends but i don't know how to find those and actually can't understand if that's what really happens.

I am really looking forward help as I'm having serious problems with that, i hope that someone will be able to help.

Thank you very much

 

[NascentOxygen]

This has a resemblance to those weapons once used to hurl a rock or a jar of burning oil at the defences of the enemy.

The tension in a string = weight + m.a

I don't see moment of inertia comes into the picture, the beam is weightless.

 

[PhanthomJay]

looks like you have the correct relevant equations, so in order to solve for the initial (angular) acceleration, you need to calculate the Moment of Inertia (I) of the system. Thoughts?

 

[Bedeirnur]

The total moment of inertia should be I1+I1 --> m*L1^2+m*L2^2= m*(L1^2+L2^2) right?

PhantomJay isn't it asking the linear acceleration? because as α = a * L it is different for the 2 blocks?

EDIT it's a = α * L so it might be α as it's equal for both

 

[PhanthomJay]

The total moment of inertia should be I1+I1 --> m*L1^2+m*L2^2= m*(L1^2+L2^2) right?

PhantomJay isn't it asking the linear acceleration? because as α = a * L it is different for the 2 blocks?

Yes, your calc for I is correct. I guess you are also correct about finding the linear (tangential) initial acceleration of each block, as opposed to their angular acceleration.

 

[Bedeirnur]

But there's something i can't understand.

When using the Torque formula Mtot = Iα... I have that Mtot is the sum of the 2 torques, I is the sum of the 2 inertial moments. But i can't understand at all what that α is here...as the two have 2 different angular velocities i can't understand how to use it.

Even if i see it as a = α*r so α=a/L we have to L's (?)

 

[PhanthomJay]

But there's something i can't understand.

When using the Torque formula Mtot = Iα... I have that Mtot is the sum of the 2 torques, I is the sum of the 2 inertial moments. But i can't understand at all what that α is here...as the two have 2 different angular velocities i can't understand how to use it.

Even if i see it as a = α*r so α=a/L we have to L's (?)

Why do you say that each mass has a different angular velocity...or more to the point, why do you say that each mass has a different angular acceleration? The rod is rigid, so if one mass rotates one degree in so many seconds, does not the other mass have to rotate at the same rate?