Rotational Motion: Force on Rod, Translation vs Rotation

In summary, when a force is exerted on a lever, such as a seesaw, it will cause both rotation and translation. The acceleration of rotation and translation can be determined by decomposing the external force vector into components. In the case of a seesaw with two children, the force exerted on the fulcrum would be the sum of their masses. However, if the seesaw were to rotate, the net torque would not be 0 and the force exerted on the fulcrum would be equal to the balanced force minus the rotational moment. This means that in the earlier experiment, the translation force would be 0 since the force was not balanced.
  • #1
Moose352
166
0
Imagine a rod in space. If I exert a force at one end, will the rod translate, rotate, or both? How do I determine what it will do?
 
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  • #2
It will rotate about it's centre of mass, and it's center of mass will move in a translation. Essentially, you are applying a moment and an unbalanced force to the body.
 
  • #3
But what will be the acceleration of its rotation and translation?
 
  • #4
Decompose the external force vector into components at the point of application. One componet which passes through the CM of the body the other perpendicular to it. The component through the CM will become a translational acceleration the other component times the distance to the CM will be the torque which cause rotation.
 
  • #5
So does that imply that when a force is acted upon a lever, like a seesaw, not all of the force acts in the rotation?

Okay, if I had a seesaw of 2 kg, which was 10 meters long, with a child (4kg) at each end, then what would be the force exerted on the fulcrum. Would it not be 2g + 4g + 4g?

What would be the force be if the seesaw became to rotate (that is, the net torque is not 0).
 
  • #6
Originally posted by Moose352
Okay, if I had a seesaw of 2 kg, which was 10 meters long, with a child (4kg) at each end, then what would be the force exerted on the fulcrum. Would it not be 2g + 4g + 4g?
Yes, it would be.
What would be the force be if the seesaw became to rotate (that is, the net torque is not 0).
Say for example one kid weighs 2kg (small kid). Then the balanced force acting on the fulcrum is 2+2+2 and the rotational force (moment) is 2.
 
  • #7
So the translation force is equal to: net force - rotational moment = balanced force? In that case, in the earlier experiment, would the translation force be 0 since the force is not balanced?
 

1. What is the difference between translation and rotation in rotational motion?

In rotational motion, translation refers to the linear movement of an object along a straight path, while rotation refers to the circular movement of an object around a fixed point.

2. How is the force on a rod calculated in rotational motion?

The force on a rod in rotational motion is calculated using the formula F = Iα, where F is the force applied to the rod, I is the moment of inertia of the rod, and α is the angular acceleration.

3. What is the relationship between torque and force in rotational motion?

Torque and force are directly proportional in rotational motion, meaning that an increase in torque will result in an increase in force, and vice versa.

4. How does the distribution of mass affect rotational motion?

The distribution of mass in an object affects its moment of inertia, which is a measure of the object's resistance to rotational motion. Objects with a greater mass distributed farther from the axis of rotation will have a higher moment of inertia and require more force to rotate.

5. What is the role of friction in rotational motion?

Friction plays a crucial role in rotational motion by providing a force that opposes the movement of an object, thus slowing down its rotation. This is important in controlling the speed and stability of rotating objects.

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