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Sagot :
If the bucket is allowed to fall, then its linear acceleration is 8.687 m/s^2.
From the given question,
We have to solve, if the bucket is allowed to fall, then its linear acceleration
Mass of Pulley(M)= 0.692 kg
Radius of pulley(r)= 0.131 m
Mass of bucket(m)= 2.70 kg
Net torque on pulley,
[tex]\Sigma[/tex] r = rT = Iα
T = Iα/r
T= tension on string(connecting pulley and bucket)
I= moment of inertia of pulley
α= angular acceleration of pulley
linear acceleration of bucket(a),
a = rα
α = a/r
Pulley is disc shaped so
I = 1/2 Mr^2
Now, T= Iα/r = (1/2 Mr^2) (a/r)/r
T = 1/2 Ma
Now considering the motion of bucket
ma = mg-T
ma = mg-1/2 Ma
ma+1/2 Ma = mg
a = m/(m+1/2 M) g
Now putting the value
a = (2.70)×9.8/ (2.70+1/2 (0.692))
After solving
a = 8.687 m/s^2
If the bucket is allowed to fall, then its linear acceleration is 8.687 m/s^2.
To learn more about linear acceleration link is here
brainly.com/question/13723307
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The right question is;
A 2.70 kg bucket is attached to a disk-shaped pulley of radius 0.131 mm and mass 0.692 kg . If the bucket is allowed to fall, What is its linear acceleration.
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