Math Problem Statement
Find the magnitude of the force Fo exerted on the stick by the pivot immediately after it is released.
Solution
To determine the magnitude of the force exerted on the stick by the pivot immediately after it is released, we can break down the problem using the following physics principles:
Given Data
- Mass of stick,
- Length of stick,
- The stick is uniform, pivoted at one end, and initially held horizontally.
Approach to Solution
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Moment of Inertia Calculation: Since the stick is pivoted at one end, we use the moment of inertia for a rod of length pivoted at one end: Substituting the values:
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Torque Due to Gravity: The gravitational force acts at the center of mass, located at a distance of from the pivot. The torque due to gravity is given by: where is the acceleration due to gravity.
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Angular Acceleration: Using Newton's second law for rotation: where is the angular acceleration. Solving for :
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Force at the Pivot: Right after release, the stick experiences both translational and rotational accelerations. The force at the pivot can be found by considering the linear acceleration of the center of mass:
The linear acceleration .
Thus, the force exerted by the pivot is the resultant of the gravitational force and the force required to provide this linear acceleration.
Would you like to proceed with the calculations for each step in detail?
Here are some related questions:
- What is the significance of the moment of inertia in rotational motion?
- How does the pivot's frictionlessness affect the problem?
- Why does the gravitational force act at the center of mass?
- How would the force change if the rod were shorter or longer?
- How does this situation change if the stick were not uniform?
Tip: Understanding torque is essential in analyzing rotational motion, especially with objects pivoted at one end.
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Math Problem Analysis
Mathematical Concepts
Rotational Dynamics
Moment of Inertia
Torque
Angular Acceleration
Force Analysis
Formulas
Moment of Inertia for a rod pivoted at one end: I = (1/3)ML^2
Torque: τ = Mg(L/2)
Newton's second law for rotation: τ = Iα
Linear acceleration of the center of mass: a = α(L/2)
Theorems
Newton's Second Law for Rotation
Suitable Grade Level
Undergraduate Physics
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