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Rod and angular acceleration

  • #1
rock.freak667
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[SOLVED] Rod...and angular acceleration

Homework Statement


A uniform rod AB, of length 2a and mass m, has a particle of mass 1
2m attached to B. The rod is
smoothly hinged at A to a fixed point and can rotate without resistance in a vertical plane. It is released
from rest with AB horizontal. Find, in terms of a and g, the angular acceleration of the rod when it
has rotated through an angle of [itex]\frac{\pi}{3}[/itex]


Homework Equations





The Attempt at a Solution



My idea is that the torque due to the weight and the mass is the net torque and that should be the same as [itex]I \alpha[/itex] where I=moment of inertia of the entire thing.

I got [itex]I=\frac{10ma^2}{3}[/itex] but I don't know how to incorporate that angle it is rotated through.
 

Answers and Replies

  • #2
alphysicist
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Hi rock.freak667,

How did you find the moment of inertia? It looks like you used that the moment of inertia of the particle is 2 m a^2, but I don't think that is correct. (Is the mass of the particle 2m? If so, there is an extra 1 in your post.)

To find the angular acceleration, draw a diagram when it's at that angle, and use the formula (net torque = I alpha) that you mentioned. From the diagram you can calculate the torques at that particular angle.
 
  • #3
rock.freak667
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The MOI through the centre is given by [itex]I_c=\frac{1}{12}ML^2[/itex]

so for the rod in question, the MOI would be (through the centre)

[tex]I_c=\frac{1}{12}(mg+\frac{mg}{2})(2a)^2=\frac{1}{2}mga^2[/tex]

and then by the parallel axis theorem to get the MOI about the end.

[tex]I=I_c+Mr^2[/tex]

[tex]I=\frac{1}{2}mga^2 + (mg+\frac{mg}{2})(a)^2=2mga^2[/tex]

is that correct moment of inertia about the end?

And I typed part of the question wrong. It should read

A uniform rod AB, of length 2a and mass m, has a particle of mass [itex]\frac{1}{2}m[/itex] attached to B.
 
Last edited:
  • #4
Doc Al
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What's the moment of inertia of a rod about one end? Of the particle about that end? Add them.

(The moment of inertia depends on mass distribution not weight, so g should not appear.)
 
  • #5
rock.freak667
Homework Helper
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What's the moment of inertia of a rod about one end? Of the particle about that end? Add them.

(The moment of inertia depends on mass distribution not weight, so g should not appear.)
ok well then. Using the parallel axis theorem

[tex]I=\frac{1}{12}m(2a)^2+ma^2=\frac{4}{3}ma^2[/tex]

and then adding the MOI of the mass(=[itex]\frac{m}{2}(2a)^2=2ma^2[/itex]

then

[tex]I_{total}=\frac{4}{3}ma^2+2ma^2=\frac{10ma^2}{3}[/tex]

So then when I draw the forces at that angle (weight of rod+weight of the mass at B) of [itex]\frac{\pi}{3}[/itex]. How would I get the torque at that angle? Would it just be the total torque*sine of the angle?
 
  • #6
Doc Al
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ok well then. Using the parallel axis theorem

[tex]I=\frac{1}{12}m(2a)^2+ma^2=\frac{4}{3}ma^2[/tex]

and then adding the MOI of the mass(=[itex]\frac{m}{2}(2a)^2=2ma^2[/itex]

then

[tex]I_{total}=\frac{4}{3}ma^2+2ma^2=\frac{10ma^2}{3}[/tex]
Perfect.

So then when I draw the forces at that angle (weight of rod+weight of the mass at B) of [itex]\frac{\pi}{3}[/itex]. How would I get the torque at that angle? Would it just be the total torque*sine of the angle?
You can find the torque due to each weight separately and add them up:
[tex]\tau = \vec{r}\times\vec{W} = rW\sin\theta[/tex]
 
  • #7
rock.freak667
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Torque due to Weight of rod=mgasin(pi/3)=[itex]\frac{mga\sqrt{3}}{2}[/itex]
Torque due to weight of mass=(mg/2)(2a)sin(pi/3)=[itex]\frac{mga\sqrt{3}}{2}[/itex]

Total torque = [itex]mga\sqrt{3}[/itex]

so that

[tex]I_{total} \alpha=mga\sqrt{3} \Rightarrow \alpha=\frac{mga\sqrt{3}}{I_{total}}[/tex]

are the torques correct?
 
  • #8
Doc Al
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Looks good!
 
  • #9
rock.freak667
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Looks good!
Apparently not...I found that the answer is really [tex]\alpha= \frac{3g}{10a}[/tex]

and I get

[tex]\alpha=\frac{9\sqrt{3}g}{20a}[/tex]
 
  • #10
Doc Al
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Let me look it over one more time.
 
  • #11
Doc Al
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I see the problem.
You can find the torque due to each weight separately and add them up:
[tex]\tau = \vec{r}\times\vec{W} = rW\sin\theta[/tex]
In the above expression for torque, [itex]\theta[/itex] is the angle between [itex]\vec{r}[/itex] and [itex]\vec{W}[/itex], which is not [itex]\pi/3[/itex] but its complement.
Torque due to Weight of rod=mgasin(pi/3)=[itex]\frac{mga\sqrt{3}}{2}[/itex]
Torque due to weight of mass=(mg/2)(2a)sin(pi/3)=[itex]\frac{mga\sqrt{3}}{2}[/itex]
I must have been asleep at the wheel and assumed you were taking the sine of the correct angle. Sorry about that! :redface:

Correct that and you'll get the expected answer.
 
  • #12
rock.freak667
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ahhh..thanks..my diagram was wrong...I pit pi/3 between the rod and the hinge as the complement...thank you!
 

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