Rotational Motion of a block of mass

AI Thread Summary
The discussion focuses on a physics problem involving a block hanging from a cord wrapped around a pulley, requiring the calculation of the pulley’s angular acceleration and the tension in the cord. The user attempts to apply Newton's second law and rotational dynamics equations but struggles to reach the correct answers. Key equations mentioned include the relationship between linear acceleration, gravitational force, and tension, as well as the connection between linear and angular acceleration through the radius of the pulley. The radius is significant as it relates to the torque and angular motion of the pulley. Overall, the conversation highlights the complexities of applying rotational motion principles in this scenario.
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 alot!
 
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