Rotational Motion of a block of mass

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SUMMARY

The discussion focuses on the rotational motion of a block with a mass of 2.4 kg hanging from a massless cord wrapped around a pulley with a moment of inertia of 1.5 x 10-3 kg·m2. The key equations utilized include Newton's Second Law (F=ma) and the rotational equivalent (Torque=Iα). The user attempts to derive the angular acceleration of the pulley and the tension in the cord using the relationships between linear and angular motion, specifically noting the importance of the radius of the pulley (0.032 m) in these calculations.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Familiarity with rotational dynamics (Torque=Iα)
  • Knowledge of the relationship between linear acceleration and angular acceleration
  • Basic principles of pulleys and tension in cords
NEXT STEPS
  • Calculate angular acceleration using the formula α = a/radius of pulley
  • Determine the tension in the cord using the equation T = mg - ma
  • Explore the effects of varying the moment of inertia on angular acceleration
  • Investigate the implications of cord slip on pulley dynamics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and rotational motion, as well as educators seeking to clarify concepts related to pulleys and forces.

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!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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