A bar subject to a rolling disk which is released on an inclined plane

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SUMMARY

The discussion focuses on calculating the initial acceleration of a rod of mass m and length l, released from a vertical position on an inclined plane. The correct formula for the acceleration at point A of the rod is given as a_A = g sin θ / (1 - (3/4) cos² θ). Key equations utilized include the moment of inertia I_cm = mL²/12 and the relationship between linear and angular acceleration. The user attempts to derive the solution using Newton's laws and rotational dynamics but encounters difficulties in the calculations.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with rotational dynamics and moment of inertia
  • Knowledge of trigonometric functions in physics
  • Basic principles of inclined plane mechanics
NEXT STEPS
  • Review the derivation of angular acceleration in rigid body dynamics
  • Study the effects of friction on inclined planes in physics problems
  • Explore advanced topics in rotational motion, such as torque and angular momentum
  • Practice similar problems involving rigid bodies on inclined planes
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of rigid bodies on inclined planes, particularly in the context of rotational motion and acceleration calculations.

Telemachus
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Homework Statement


I have this other exercise rigid in the plane, with which I am having problems.

The rod of mass m and length l, is released based on the vertical position of rest with the small roller end A resting on the slope. Determine the initial acceleration A.
attachment.php?attachmentid=32766&stc=1&d=1299177876.png

(neglect friction and mass of the roller A)

The answer given by the book is [tex]a_A=\displaystyle\frac{g\sin \theta}{1-\frac{3}{4}\cos^2\theta}[/tex]

Homework Equations


I try to raise the moment equation, and Newton. But not me, not that I'm doing wrong. For this consider that the bar rotates about its center of mass.
[tex]I_{cm}=\displaystyle\frac{mL^2}{12}[/tex]
[tex]N-mg\cos\theta=0[/tex]
[tex]mg\sin\theta=ma_{cm}[/tex]
[tex]I_{cm}\alpha=\displaystyle\frac{L}{2}mg\sin\theta\cos\theta[/tex]

The Attempt at a Solution



[tex]\alpha=\displaystyle\frac{6g\sin\theta\cos\theta}{L}[/tex]
[tex]a_cm=g\sin\theta[/tex]

Then: [tex]a_A=a_{cm}+\displaystyle\frac{L}{2}\alpha=g\sin\theta+3g\cos\theta\sin\theta[/tex]

Greetings and thanks for posting.
 

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