Solve Acceleration of Rod's Centre of Mass

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SUMMARY

The discussion focuses on calculating the acceleration of the center of mass of a vertically standing rod when disturbed. The key equations involved include the net force equation (acom = Fnet/mnet) and the torque equation (tau = I * alpha). The user attempted to derive the acceleration using a combination of forces and torques but arrived at an incorrect expression: a = g - (l * alpha/(6 sin theta)). The correct expression for the acceleration of the center of mass is l * (alpha/2) * sin theta + (w^2 * l/2) * cos theta, as clarified by another participant.

PREREQUISITES
  • Understanding of angular motion concepts, including angular speed (omega) and angular acceleration (alpha).
  • Familiarity with torque calculations and the moment of inertia (I) for a rod.
  • Knowledge of basic mechanics, particularly Newton's laws of motion.
  • Ability to apply trigonometric functions in physics problems involving angles.
NEXT STEPS
  • Study the derivation of angular motion equations in classical mechanics.
  • Learn about the moment of inertia for various shapes, focusing on rods and their applications.
  • Explore the relationship between linear and angular acceleration in rigid body dynamics.
  • Investigate the effects of frictionless surfaces on the motion of objects.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and angular motion, as well as educators looking for examples of torque and acceleration calculations in rigid bodies.

Diganta28
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1. Homework Statement :

A rod of length l is vertically standing on a friction less surface.
It is slightly disturbed from this position. Let w(omega) and alpha be the angular speed and angular acceleration of the rod, when it turns through an angle theta, then find the value of acceleration of centre of mass of the rod.

2. Homework Equations :

acom = ( Fnet/mnet)

tau = i * alpha

tau= force*perpendicular distance from axis of rotation

3. The Attempt at a Solution :

Tried like this (but no luck):

(mg-N)=ma , where a is the required answer.N * (l/2) sin theta = (1/12 m (l)^2) * alpha

Solved to get : a = g-(l * alpha/(6 sin theta)).

But answer given :

l *(alpha/2)* sin theta + (w^2*l/2) cos theta

 
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Ok I understood what they are doing.

But is the answer :
g - (l*alpha/(6 sin theta)) wrong?
 

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