Solving Rotational Motion Problems: Inertia & Angular Acceleration

[YoungBuddhist]

Thanks in advance for the help. Here are the problems:

A uniform rod is free to rotate about a frictionless pivot at one end. The rod is released from rest in the horizontal position. What is the magnitude of the angular acceleration of the rod at the instant it is 60 degrees below the horizontal?

A uniform meter stick is pivoted to rotate about a horizontal axis through the 25 cm mark on the stick. The stick is released from rest in a horizontal position. Determine the inertia(calculus) and the magnitude of the initial angular acceleration of the stick.

I am a first time poster and would greatly appreciate someone's detailed analysis on how to solve these and other similar types of problems.

Thanks once again.

 

[krab]

Please read one of the top Stickys: FAQ: Why hasn't anybody answered my question?

 

[kyle8921]

I am happy to provide a response to your question about solving rotational motion problems involving inertia and angular acceleration.

First, let's define some key concepts. Inertia is the property of an object to resist changes in its state of motion. In rotational motion, inertia is related to an object's mass and its distribution around an axis of rotation. Angular acceleration, on the other hand, is the rate at which an object's angular velocity changes, and is related to the net torque acting on the object.

For the first problem, we have a uniform rod rotating about a frictionless pivot at one end. The rod is released from rest in the horizontal position and we are asked to find the magnitude of the angular acceleration at the instant it is 60 degrees below the horizontal.

To solve this problem, we can use the equation for angular acceleration, which states that the angular acceleration (α) is equal to the net torque (τ) divided by the moment of inertia (I) of the object. In this case, since the rod is uniform, we can use the formula for the moment of inertia of a rod rotating about one end, which is 1/3 * mass * length^2.

We also know that at the instant the rod is 60 degrees below the horizontal, the torque acting on the rod is equal to the weight of the rod multiplied by the perpendicular distance from the pivot to the center of mass of the rod (since the rod is rotating about its center of mass). Therefore, we can set up the equation: α = τ/I = (mg * L/2) / (1/3 * mL^2) = 3g/2L.

Solving for α, we get an angular acceleration of 3g/2L. This means that the rod will accelerate at a rate of 3g/2L radians per second squared, where g is the acceleration due to gravity and L is the length of the rod.

For the second problem, we have a uniform meter stick pivoted at the 25 cm mark and released from rest in a horizontal position. We are asked to determine the inertia using calculus and the magnitude of the initial angular acceleration.

To solve this problem, we can use the same equation for angular acceleration as before. However, since the stick is pivoted at a point other than one end, we cannot use the formula for the moment of inertia of a rod rotating about one end. Instead, we