Finding Acceleration and Reaction Forces in Rotating Thin Rod on a Hinge

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SUMMARY

The discussion focuses on calculating the acceleration and reaction forces of a rotating thin rod with a mass of 0.620 kg and length of 1.08 m, subjected to a horizontal impulsive force of 14.2 N. The key equations utilized include the moment of inertia \( I = \frac{1}{3} ML^2 \) and the relationship between angular and linear acceleration \( \alpha = \frac{a_T}{L} \). The center of mass acceleration is determined to be \( a_{com} = \frac{3F}{2M} \) when the force is applied at the bottom of the rod. Additionally, the discussion addresses the concept of the center of percussion, where no horizontal force is exerted by the hinge.

PREREQUISITES
  • Understanding of rotational dynamics and forces
  • Familiarity with moment of inertia calculations
  • Knowledge of angular acceleration and its relationship to linear acceleration
  • Basic principles of impulse and reaction forces
NEXT STEPS
  • Study the derivation of the center of mass acceleration in rotating systems
  • Learn about the center of percussion and its applications in mechanics
  • Explore the relationship between angular speed and linear speed in detail
  • Investigate the effects of different force application points on rigid body dynamics
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This discussion is beneficial for physics students, mechanical engineers, and anyone involved in dynamics and rotational motion analysis.

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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


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

The Attempt at a Solution


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

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!
 
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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)
 
Oops, thanks. Since α=atangentialL, α=acomL/2 right? That gives acom = 3F/2M
 

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