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

In summary, the problem involves two blocks of equal mass suspended on a weightless rod, which is held horizontally and then released. The goal is to calculate the accelerations of the blocks as they start to move. Relevant formulas include the total torque, individual torques for each block, and the total moment of inertia. The tension in the string can also be calculated using the weight and mass of the blocks. The total moment of inertia for the system is equal to the sum of the individual inertial moments for each block. The angular acceleration can be found using the formula α = a/L, where a is the linear acceleration and L is the length of the rod. Since the rod is rigid, both blocks will have the same angular acceleration.
  • #1
Bedeirnur
18
0

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
KKmwl3Y.png



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
 
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  • #2
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.
 
  • #3
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?
 
  • #4
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
 
Last edited:
  • #5
Bedeirnur said:
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.
 
  • #6
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 (?)
 
  • #7
Bedeirnur said:
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?
 

1. What is torque?

Torque is a measure of the ability of a force to rotate an object around an axis. It is calculated by multiplying the force applied by the distance from the axis of rotation.

2. How is torque related to rotational motion?

Torque is directly related to rotational motion, as it is the force that causes an object to rotate around an axis. The greater the torque applied, the greater the rotational acceleration of the object will be.

3. What is Newton's third law as it applies to rotatory motion?

Newton's third law states that for every action, there is an equal and opposite reaction. In terms of rotational motion, this means that the force applied to rotate an object will result in an equal and opposite reaction force acting on the object, causing it to rotate in the opposite direction.

4. How can torque and Newton's third law be applied to everyday situations?

Many everyday situations involve the application of torque and Newton's third law. For example, riding a bike involves applying torque to the pedals to rotate the wheels, and the reaction force from the ground pushes the bike forward.

5. What are some real-life examples of torque and Newton's third law in action?

Some real-life examples of torque and Newton's third law include using a wrench to loosen a bolt, throwing a ball, and opening a door. In each of these situations, an applied force results in a reaction force, causing rotational motion.

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