Force exerted on the pivotal pin of an oscillating object.

In summary, the force and direction of the force on a frictionless pivot point near the end of a rod oscillating back and forth can be calculated by considering the acceleration and direction vectors caused by the oscillations in addition to the force of gravity. Ignoring torque is already included in the problem description due to the frictionless pivot point. The force of gravity will act downwards and the reaction forces at the pivot point will act through the pin. This problem is similar to calculating forces on a pin of a ball swinging in a circle in a vertical plane, but in this case, only part of a circle is involved.
  • #1
Ascendant78
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If you have a rod oscillating back and forth by a frictionless pivot point near the end of one side of the rod and the only external force is gravity, how would the magnitude and direction of the force of/on that pivot point be calculated? How does it change with the oscillations?

I am having difficulty with this since torque would be ignored at that point and gravity is always pushing directly downward. I'm thinking the way to figure it out would be calculating the direction and acceleration vectors caused by the oscillations and include those forces caused by the oscillations in addition to the gravitational acceleration, but I'm not 100% sure about this one? I couldn't find a problem that looked at this aspect of oscillations anywhere in any of my physics books, so any help would be greatly appreciated.
 
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  • #2
There is no torque at the frictionless pin so ignoring torque is already included in the desrciption of the problem.
And you also know that from the pin being frictionless, reaction forces at the pin will have to act through the pin.

Gravity does act downwards, and then for the rod, at what point on the rod can do we usually say the force of gravity is acting..

This problem is similar to calculating the forces on a pin of a ball swinging in a circle in a vertical plane, except instead of a full circle we have part of a circle.
 

Related to Force exerted on the pivotal pin of an oscillating object.

1. What is the force exerted on the pivotal pin of an oscillating object?

The force exerted on the pivotal pin of an oscillating object is dependent on the mass of the object, its velocity, and the distance of the pin from the center of mass. It is also affected by external forces such as friction and air resistance.

2. How is the force on the pivotal pin of an oscillating object calculated?

The force on the pivotal pin of an oscillating object can be calculated using the formula F = ma, where F is the force, m is the mass of the object, and a is the acceleration of the object. In the case of oscillating objects, the acceleration changes direction as the object moves back and forth, resulting in a periodic force on the pivotal pin.

3. What factors affect the force on the pivotal pin of an oscillating object?

The force on the pivotal pin of an oscillating object is affected by the mass of the object, its velocity, and the distance of the pin from the center of mass. It is also affected by external forces such as friction and air resistance.

4. How does the force on the pivotal pin change as the object oscillates?

The force on the pivotal pin of an oscillating object changes direction as the object moves back and forth. This is because the acceleration of the object changes direction, resulting in a periodic force on the pivotal pin. The force will also vary depending on the position of the object within its oscillation cycle.

5. How does the force on the pivotal pin of an oscillating object affect its motion?

The force on the pivotal pin of an oscillating object is responsible for the object's motion. It causes the object to accelerate and move in a back-and-forth motion. The magnitude and direction of the force will determine the behavior of the object, such as the amplitude and frequency of its oscillations.

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