Rotational Motion of a Stick: Analyzing Accelerations

[Swany]

A thin, uniform stick of length 2.1 m and mass 4.2 kg is pinned through one end and is free to rotate. The stick is initially hanging vertically and at rest. You then rotate the stick so that you are holding it horizontally. You release the stick from that horizontal position. Remember that the moment of inertia for a stick of mass m and length L about its end is (1/3)m L2.

Also, using conservation of energy, it can be shown that the square of the angular speed as a function of angle is given by:

ω2 = 3 g sin(θ)/L

with θ the angle measured clockwise from horizontal and L the length of the stick.

What is the magnitude of the angular acceleration of the stick?
What is the magnitude of the tangential acceleration of the center of mass of the stick?
What is the magnitude of the centripetal acceleration of the center of mass of the stick?
What is the magnitude of the total acceleration of the center of mass of the stick?

I don't know know even where to begin, I am having a hard time with this subject area. It would be nice to get full drawn out answers. Thanks.

 

[apelling]

The rules of the forum mean that you have to have a go at the question before anyone can correct your answer.

To help you start, I like to compare angular motion to ordinary linear motion, but:-

Force becomes Torque (where Torque = force *perp. dist to pivot)
Mass m becomes moment of inertia I (youve got I from the question)
Initial velocity u becomes initial angular velocity w0[omega subscript 0] (u=r*w0)
Final velocity v becomes final ang. vel. w1
Acceleration a becomes ang. acc. alpha (a=r*alpha)
Time is still time t
Displacement s becomes ang. displacement theta. (s=r*theta)

You can then 'translate' all your linear physics equations to angular ones

e.g v=u+at becomes w1=w0+alpha*t
kinetic energy 1/2mv^2 becomes 1/2 I w^2
F=ma becomes Torque=I *alpha
etc.

Draw your stick at an angle theta to the horizon and see what you can come up with.

 

[m2003]

I understand your difficulty with this subject area and I will do my best to provide a clear and detailed explanation. Let's break down the questions one by one.

1. What is the magnitude of the angular acceleration of the stick?

Angular acceleration is defined as the rate of change of angular velocity, or how quickly the stick is rotating. In this scenario, the stick is initially at rest and then released from a horizontal position, meaning it will start rotating. The magnitude of the angular acceleration can be calculated using the equation:

α = ω/t

where α is the angular acceleration, ω is the angular velocity, and t is the time it takes for the stick to rotate.

Using the given equation for the angular speed as a function of angle, we can rearrange it to solve for the time t:

t = L/√(3g sin(θ))

Substituting this into the equation for angular acceleration, we get:

α = ω√(3g sin(θ))/L

Plugging in the values for the length of the stick (L = 2.1 m), mass of the stick (m = 4.2 kg), and acceleration due to gravity (g = 9.8 m/s^2), we can calculate the magnitude of the angular acceleration for different values of θ.

For example, at θ = 0 (when the stick is released from the horizontal position), the magnitude of the angular acceleration would be:

α = √(3g/L) = √(3*9.8/2.1) = 3.08 rad/s^2

2. What is the magnitude of the tangential acceleration of the center of mass of the stick?

Tangential acceleration is the component of acceleration that is tangent to the path of motion. In this case, as the stick rotates, the center of mass will also experience a tangential acceleration.

The magnitude of the tangential acceleration can be calculated using the equation:

at = αr

where at is the tangential acceleration, α is the angular acceleration (calculated in the previous question), and r is the distance from the center of mass to the pivot point (in this case, the pinned end of the stick).

Using the moment of inertia equation for a stick, we can calculate the distance r:

I = (1/3)mL^2 = (1/3)*4.2*(