Calculating Acceleration of a Falling Bucket Attached to a Pulley System

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To calculate the acceleration of a falling bucket in a pulley system, apply Newton's second law for both the bucket and the pulley. The net force on the bucket is given by the equation ma = mg - T, where T is the tension in the rope. For the pulley, the torque is expressed as T*R = I*alpha, linking tension and angular acceleration. By substituting the moment of inertia and solving the equations simultaneously, the acceleration of the bucket can be derived. This approach highlights the relationship between linear and rotational dynamics in the system.
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Homework Statement


1.A bucket of mass m is connected by a rope to a frictionless pulley of mass M. Use Newton’s 2nd law for torques and Newton’s 2nd law to find an equation for the acceleration of the bucket if it is released from rest and allowed to fall.



Homework Equations





The Attempt at a Solution



so i need we need to find Normal force in both direction. i know that

sum of Force in X direction is that o= N- Tcos(theta)

From the problem i am not sure how to find the Tension and where the normal force in the Y direction would come from.
 
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Isn't what you have is a F = m*g being retarded by an I*a/R where I is the moment of inertia?

Apparently the frictionless part refers to the rotation of the pulley and not the friction of the rope on the pulley because if the rope is free to slip, what torque would there be?
 
Take the tension T to be the reactant force

Hint...it is 90 degrees to the axis of rotation.
 
Forces acting on the bucket are T and mg. Since bucket is moving downwards, net force acting on it is
ma = mg - T...(1)
The tangential force actin on pulley is T.
Therefore The torque on pulley is T*R = I* alpha = I*Ra ...(2)
Substituting the expression for I and solving eq. 1 and 2 , find the equation for the acceleration.
 
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