An object of a mass 1.50 kg is suspended from a rough pulley of a radius 30.0 cm by light stirring as shown in Figure 1. This pulley has a moment of inertia 0.030 kg m² about the axis of the pulley. The object is released from rest and the pulley rotates without encountering frictional force. Assume that the string does not slip on the pulley, Determinea) the angular acceleration of the pulleyb) the tension in the string

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Question:

An object of a mass 1.50 kg is suspended from a rough pulley of a radius 30.0 cm by light stirring as shown in Figure 1. This pulley has a moment of inertia 0.030 kg m² about the axis of the pulley. The object is released from rest and the pulley rotates without encountering frictional force. Assume that the string does not slip on the pulley, Determine a) the angular acceleration of the pulley b) the tension in the string

Newton's second law allows us to find the results for the tension of the rope and the angular acceleration of the body are a) The angular acceleration is α = 26.7 rad / s²
b) The tension of the rope T = 2.67 N
Given parameters
The mass of the block m = 1.50 kg
The radius of the pulley R = 0.30 m
The moment of inertia I = 0.030 kg m²
To find
Angular acceleration
The tension of the rope
Newton's second law indicates that the net force is proportional to the product of the mass and the acceleration of the bodies
∑ F = m a
Where the bold letters indicate vectors, F is the external forces, m the mass and the acceleration of the body
A free body diagram is a diagram of the forces without the details of the bodies, in the attached we see a free body diagram of the system
Let's apply Newton's second law to the block
W - T = m a
Body weight is
W = mg
Let's substitute
mg - T = ma (1)
The pulley is a body that is in rotational motion so we use Newton's second law for rotation
Σ τ = I α
Where τ is the torque, I the moment of inertia and α the angular acceleration
In this case the reference system is located at the turning point T R = I α
T = [tex]\frac{I \alpha }{R}[/tex] Linear and angular variables are related
a = α R
We substitute in equation 1 and write the system of equations
mg - T = m α R
T = [tex]\frac{I \alpha }{R}[/tex]
We resolve
mg = α (m R + I / R)
α = [tex]\frac{g}{R + \frac{I}{mR} }[/tex] We calculate α = [tex]\frac{ 9.8}{ 0.30+\frac{0.030}{1.50 \ 0.30} }[/tex] α = 26.7 rad / s²
Let's calculate the tension of the rope
T = [tex]\frac{I \alpha }{R}[/tex] T = [tex]\frac{0.030 \ 26.7 }{0.30}[/tex]
T = 2.67 N
In conclusion with Newton's second law we can find the results for the tension of the rope and the angular acceleration of the body are a) The angular acceleration is alpha = 26.7 rad / s²
b) The tension of the rope T = 2.67 N
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