Rotational Motion of a block of mass

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Homework Help Overview

The problem involves a block of mass 2.4 kg hanging from a massless cord that is wrapped around a pulley with a specified moment of inertia. The scenario describes the block's descent and the resulting rotation of the pulley, with specific parameters provided, including the radius of the cord.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law for both linear and rotational motion. There is an exploration of the relationship between tension, gravitational force, and acceleration. Some participants question the relevance of the radius of the cord in the context of the problem.

Discussion Status

Participants are actively engaging with the problem, offering different equations and reasoning. Some have expressed uncertainty about their approaches, while others have indicated that previous suggestions were helpful in clarifying their understanding.

Contextual Notes

There is a mention of a missing figure that may provide additional context for the problem setup. Participants are also navigating the implications of the problem's constraints, such as the massless cord and the constant radius of the pulley.

simplygenuine07
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Homework Statement



A block (mass = 2.4 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1.5 x 10-3 kg·m2), as the figure shows. Initially the pulley is prevented from rotating and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of 0.032 m during the block's descent.
Find (a) the angular acceleration of the pulley and (b) the tension in the cord.


Homework Equations



Newtons Second law and Newtons second law of rotation
F=ma and Torque=Ialpha

The Attempt at a Solution


I tried using this equation, but i get the wrong answer no matter what I do.
T=mg+ma
a=Lalpha
 
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i think it must be

ma=mg-T

because the tension is opposite to the gravitational pull...
and where is the pic?
 
ma=mg-T => a=g-T/m (1)
also, a=alpha*radius of pulley and alpha= Torque/inertia = T*radius of pulley/inertia of pulley
=> a=radius^2*T/inertia (2)
from 1 and 2, hopefully we can find the answer. I'm not sure why they mention the radius of the cord?
Tell me if it works out, I didn't have time to actually solve it myself.
 
Yes thankyou, that helped a lot!
 

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