Find the magnitude of the initial acceleration of the rod's center of mass

  • Thread starter fruitl00p
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  • #1
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Homework Statement



A uniform rod of length 1.15m is attached to a frictionless pivot at one end. It is released from rest an angle theta=17.0 degrees above the horizontal.

Find the magnitude of the initial acceleration of the rod's center of mass

Homework Equations



torque=moment of inertia*angular acceleration
torque=length*force(perpendicular)

I=1/3 ML^2

The Attempt at a Solution



I set the torque about pivot equal to the moment of inertia about pivot times the angular acceleration to find the angular acceleration. Then I found the linear acceleration by multiplying the found angular acceleration by the length of rod. So my work was....

L/2*(Mg)cos17=1/3*ML^2*(angular acceleration)
3/2*(g/L)*cos17=(angular acceleration)
12.22 rad/s^2= (angular acceleration)

a=(angular acceleration)*L= 14.05 m/s^2

However the answer I got is wrong. (Unfortunately, I do not have access to the correct answer) What step did I miss?
 

Answers and Replies

  • #2
andrevdh
Homework Helper
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...... L/2*(Mg)cos17=1/3*ML^2*(angular acceleration)

This should be the sin not the cos.

The com will then experience only a tangential acceleration:

[tex]a_t = r \alpha[/tex]
 
  • #3
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I replaced the cos with the sin17 in my equation and I just tried to solve but I was told my answer was wrong. And I did exactly what you suggested. I wonder why that was wrong? Did I need to do something else?
 
  • #4
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Have you used the correct angle ? 17 deg above the horizontal is 90 + 17 from the vertical.
 
  • #5
hage567
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If you are trying to find the tangential acceleration of the centre of mass, wouldn't it be [tex]a_t = \alpha(L/2)[/tex], since the centre of mass will be at half the length of a uniform rod? Just a thought.
 
  • #6
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Mentz114, thank you for mentioning that! I looked through some notes and saw I need to add 90 degrees. However, I just redid the problem, using the info you told me and what andrevdh told me, and oddly enough, I got the same answer!

hage567, I will try to find the initial acceleration with L/2 instead of L.
 
  • #7
94
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Thanks everyone, I got the right answer now.
 

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