How Do You Calculate the Initial Acceleration of a Rod's Center of Mass?

AI Thread Summary
To calculate the initial acceleration of a uniform rod's center of mass, the relevant equations involve the rotational dynamics equation Rmg(sinθ) = Iα, where R is the distance from the pivot to the center of mass, g is gravitational acceleration, θ is the angle, I is the moment of inertia, and α is angular acceleration. The moment of inertia for a uniform rod pivoted at one end is I = (1/3)mL^2. After substituting the known values, the calculated angular acceleration α is found to be 4.58 rad/s², leading to a linear acceleration a of 2.63 m/s². A mistake in using sin instead of cos for the angle was noted, which could affect the final result. The calculations appear correct if the initial values are verified.
WillP
Messages
8
Reaction score
0
Pivot point on a Uniform Rod and Acceleration

I'm stuck on this question, which seems like it should be fairly simple:

A uniform rod of length 1.15 m is attached to a frictionless pivot at one end. It is released from rest from an angle theta = 21.0° above the horizontal. Find the magnitude of the initial acceleration of the rod's CM.
theres a graphic of it here:
http://capaserv.physics.mun.ca/msuphysicslib/Graphics/Gtype20/prob28_1012rod.gif

ok so what I've done so far is:
Rmg(sinθ) = Iα
Rmg(sinθ) = (1/3)mL^2α

getting the expression I = (1/3)mL^2 from my textbook for a uniform rod with a pivot at its end.

working that out I get:
(0.575)(9.8)(sin21) = (1/3)(1.15)^2 α
α = 4.58 rad/s

then converting α to a with:
a = Rα
a = (.575)(4.58)
a = 2.63 m/s^2

but apparently this isn't right.
can anyone help me out?
 
Last edited by a moderator:
Physics news on Phys.org
nevermind... my stupid mistake should have used cos instead of sin :redface:
 


The rotational moment of inertia is a measure of an object's resistance to rotational motion, similar to how mass is a measure of an object's resistance to linear motion. It is dependent on the object's mass distribution and the distance of the mass from the axis of rotation. In this case, the uniform rod has a moment of inertia given by I = (1/3)mL^2, where m is the mass of the rod and L is its length.

To find the initial acceleration of the rod's center of mass, we can use the rotational dynamics equation Rmg(sinθ) = Iα, where R is the distance from the pivot point to the center of mass, g is the acceleration due to gravity, θ is the initial angle above the horizontal, I is the moment of inertia, and α is the angular acceleration.

Substituting in the known values, we get (0.575)(9.8)(sin21) = (1/3)(1.15)^2α. Solving for α, we get α = 4.58 rad/s^2.

To find the linear acceleration of the center of mass, we can use the relation a = Rα, where a is the linear acceleration and R is the distance from the pivot point to the center of mass. Substituting in the values, we get a = (0.575)(4.58) = 2.63 m/s^2.

Your calculations look correct, so it's possible that there may be a mistake in the given values or the problem statement. Double check the given values and make sure they are correct. If they are, then your solution is correct.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top