Classical mechanics-acc. of a rod using inertia

[indie452]

Homework Statement

two people are holding the ends of a rod length l and mass M, show that if one person let's go the initial acceleration of the free end is 3g/2

The Attempt at a Solution

i worked out the moment of inertia about centre mass (cm) and got = Ml²/12

because L=Iw
dL/dt = I*dw/dt
Torque (T) = I*ang.acc. (a)
so
a = T/I = mglsin[90] / Ml²/12
= 12g/l
this is wrong so i used l=l/2 as this is the distance to the cm.
so a = 24g/l also wrong

what am i doing wrong?

 

[Jebus_Chris]

Calculate the moment of inertia about the pivot point using the parallel axis theorem.

 

[kerimek]

I would suggest that you double check your calculations and make sure you are using the correct equations for rotational motion. In this scenario, the rod is being released at one end, which means the other end is rotating about a fixed point (the other person's hand). This requires the use of angular velocity and angular acceleration, instead of linear velocity and linear acceleration. Additionally, make sure you are using the correct values for the moment of inertia, which may differ depending on the axis of rotation.

Furthermore, it is important to consider the forces acting on the rod, such as the force of gravity and the force exerted by the person holding the other end. These forces may affect the acceleration of the free end of the rod and should be taken into account in your calculations.

I would also suggest breaking down the problem into smaller, simpler steps and using diagrams to visualize the situation. This can help in identifying any errors and understanding the concept better.

Overall, it is important to carefully analyze the problem and use the correct equations and values in order to arrive at the correct solution.