Calculating Acceleration of a Falling Bucket Attached to a Pulley System

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Homework Help Overview

The problem involves a bucket connected to a frictionless pulley, where the objective is to determine the acceleration of the bucket as it falls. The discussion centers around applying Newton's laws, particularly for torques, to analyze the forces and motions involved in the system.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to identify the normal force and tension in the system, with some questioning how to derive these values. There are references to the relationship between gravitational force and tension, as well as the role of torque in the pulley system.

Discussion Status

Participants are actively exploring the relationships between forces acting on the bucket and the pulley. Some have provided hints and suggestions for approaching the problem, while others are clarifying concepts related to torque and tension. Multiple interpretations of the problem setup are being considered.

Contextual Notes

There is a mention of the frictionless nature of the pulley, which may influence the analysis of torques. The original poster expresses uncertainty regarding the normal force and tension, indicating potential gaps in the problem setup or assumptions.

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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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