Angular Acceleration of A Rigid Rod

In summary: Now, how does this relate to the angular acceleration?In summary, a rigid rod with mass 3.50 kg and length of 2.70 m is rotating in a vertical plane about a frictionless pivot through its center. Two particles with masses 4.70 kg and 2.60 kg are attached to the ends of the rod. To determine the angular acceleration of the system when the rod makes an angle of 46.1o with the horizontal, the moment of inertia and torque equations are used. After calculating the moment of inertia and torque, it is determined that the torque is equal to r*m1g*cos(θ) - r*m2g*cos(θ). The direction of the force does not affect the
  • #1
tzacher
3
0

Homework Statement


A rigid rod of mass 3.50 kg and length of 2.70 m rotates in a vertical (x,y) plane about a frictionless pivot through its center. Particles m1 (mass=4.70 kg) and m2 (mass=2.60 kg) are attached at the ends of the rod. Determine the size of the angular acceleration of the system when the rod makes an angle of 46.1o with the horizontal.


Homework Equations


I = (1/12)ML^2 + MR^2 + MR^2
τ = Iα

The Attempt at a Solution


So I tried using the moment of inertia to find the I of the rod and each mass, then tried plugging that into the torque equation, but I can't seem to figure out what the torque equation is supposed to be. I tried to have τ=m1gL1sinθ-m2gL2sinθ, but that didn't work out. Can anyone help me?
 
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  • #2
Perhaps you could spell out more of your calculation details. Also, verify the location of the angle θ in the triangles you've constructed for the torque calculations.
 
  • #3
Alright so for moment of inertia, I had:

I = (1/12)(3.5)(2.7)^2 + 4.7(.675)^2 + 2.6(.675)^2

And then my torque was:

T = (.675)(9.81)(4.7)sin(46.1) + (.675)(9.81)(4.7)sin(46.1)

I'm not quite sure about the torque calculation being correct.
 
  • #4
tzacher said:
Alright so for moment of inertia, I had:

I = (1/12)(3.5)(2.7)^2 + 4.7(.675)^2 + 2.6(.675)^2

And then my torque was:

T = (.675)(9.81)(4.7)sin(46.1) + (.675)(9.81)(4.7)sin(46.1)

I'm not quite sure about the torque calculation being correct.

Yes, there's a bit of a problem with your torque calculation. Here's a diagram of the situation

attachment.php?attachmentid=40330&stc=1&d=1319593311.gif


Well, it's one of the possible situations; The problem doesn't specify which mass is on which end of the rod or which one is elevated above the pivot, or even on which side of the pivot the rod is elevated! So I suppose that you can assume they want the magnitude of the acceleration without specifying its direction.

Note where the angle θ appears in the force triangles relating the weights of the masses and the components perpendicular to rod. What trig function is applicable to find that component? Note also the directions that the torques have with respect to each other. Do they aid or oppose each other?
 

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  • #5
I don't think the direction matters, it rarely does.

So how would I find the torque? They're both going downwards, but I know that the larger mass is going to exert more of a torque. So would that make the torque:

T = r*m1g*cos(θ) - r*m2g*cos(θ)
 
  • #6
Yes, that looks better. Take the absolute value of the difference to indicate that direction is being ignored.
 

What is angular acceleration?

Angular acceleration is the rate of change of angular velocity of a rigid body. It is expressed in radians per second squared (rad/s^2).

How is angular acceleration different from linear acceleration?

Angular acceleration refers to the change in rotational speed of an object, while linear acceleration refers to the change in linear speed. Angular acceleration is measured in radians per second squared (rad/s^2) while linear acceleration is measured in meters per second squared (m/s^2).

How do you calculate angular acceleration?

Angular acceleration can be calculated by dividing the change in angular velocity by the change in time. It can also be calculated by taking the second derivative of the angular displacement with respect to time.

What factors affect the angular acceleration of a rigid rod?

The angular acceleration of a rigid rod can be affected by the magnitude and direction of the applied torque, the moment of inertia of the object, and the distribution of mass within the object.

Can angular acceleration be negative?

Yes, angular acceleration can be negative. This means that the object is decelerating or slowing down its rotational speed. A positive angular acceleration indicates an increase in rotational speed.

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