Thin rod rotating on a hinge

  • Thread starter mli
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  • #1
mli
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Homework Statement


A thin rod of mass 0.620 kg and length 1.08 m is at rest, hanging vertically from a strong, fixed hinge at its top end. Suddenly, a horizontal impulsive force (14.2 i ) N is applied to it.
(a) Suppose the force acts at the bottom end of the rod. Find the acceleration of its center of mass.
(b) Find the horizontal force the hinge exerts.
(c) Suppose the force acts at the midpoint of the rod. Find the acceleration of this point.
(d) Find the horizontal hinge reaction force.
e) Where can the impulse be applied so that the hinge will exert no horizontal force? This point is called the center of percussion.

Homework Equations


[tex]\Sigma\tau = FL =I\alpha \\ I=1/3 ML^2 \\ \alpha=a_{T}/L[/tex]

The Attempt at a Solution


[tex]FL=1/3 ML^2\alpha[/tex]
Rearranging for alpha gives
[tex]\alpha=3F/ML[/tex]

I'm stuck on part a. I can work out the tangential acceleration but it's looking for the acceleration at the com. I can't recall any equations about the relationship between tangential and center of mass acceleration and can't seem to find anything on google either. Thank you in advance!
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
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Do you know how
angle is related to position ?
angular speed is related to velocity ?
angular acceleration is related to acceleration ?

Perhaps you can find yourself a hint by looking at the dimensions ?

google (angular speed linear speed)
 
  • #3
mli
2
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Oops, thanks. Since α=atangentialL, α=acomL/2 right? That gives acom = 3F/2M
 

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