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g 1.90 kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.150 m, to a hanging book with mass 3.10 kg. The system is released from rest, and the books are observed to move 1.20 m in 0.900 s . Part A What is the tension in the part of the cord attached to the textbook? Express your answer with the appropriate units. T1 = nothing nothing Request Answer Part B What is the tension in the part of the cord attached to the hanging book? Express your answer with the appropriate units. T2 = nothing nothing Request Answer Part C What is the moment of inertia of the pulley about its rotation axis? Express your answer with the appropriate units. I = nothing nothing Request Answer Provide Feedback

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Answer:

Explanation:

First of all we shall calculate the acceleration of the system . Let it be a .

s = ut + 1/2 a t²

1.2 = 1/2 x a x .9²

a = 2.963

A ) Applying 2nd law of motion on textbook

T₁  = ma , m is mass of the textbook a is its acceleration

T₁ = 1.9 x 2.963

= 5.63 N .

B ) Applying 2nd law of motion on hanging book

mg - T₂ = ma , m is mass of hanging book , a is its acceleration

T₂ = mg - ma

= m ( g - a )

= 3.1 x ( 9.8 - 2.963 )

= 21.2 N

C )

Torque on the pulley

= ( T₂ - T₁ )  x R

R is radius of the pulley

= (21.2 - 5.63 ) x .075

= 1.16775  Nm

angular acceleration of pulley

ι = linear acceleration /  radius

= 2.963 /  .075

= 39.5 rad / s²

torque on pulley = moment of inertia x angular acceleration

1.16775 = moment of inertia x 39.5

moment of inertia = 29.56 x 10⁝³ kg m² .

onyebuchinnaji

(a) The tension of the chord attached to the textbook is 5.624 N.

(b) The tension of the chord attached to the hanging book is 21.2 N.

(c) The moment of inertia of the pulley about its rotation axis is 0.0296 kgm².

The given parameters:

  • Mass of the text book, m = 1.90 kg
  • Diameter of the pulley, d = 0.15 m
  • Mass of hanging book = 3.10 kg
  • distance traveled by the book, s = 1.2 m
  • time of motion of the book, t = 0.9 s

The acceleration of the book at the given displacement and time is calculated as;

[tex]s = ut \ + \ \frac{1}{2} at^2\\\\s = 0 + \frac{1}{2} at^2\\\\s = \frac{1}{2} at^2\\\\a = \frac{2s}{t^2} \\\\a = \frac{2 \times 1.2 }{0.9^2} \\\\a = 2.96 \ m/s^2[/tex]

The tension of the chord attached to the textbook is calculated as follows;

T = ma

T = 1.9 x 2.96

T = 5.624 N

The tension of the chord attached to the hanging book is calculated as follows;

T = mg - ma

T = m(g - a)

T = 3.1(9.8 - 2.96)

T = 21.2 N

The torque on the pulley is calculated as;

[tex]\tau = \Delta T R\\\\ \tau = (T_2-T_1)R\\\\ \tau = (21.2 - 5.624) \times 0.075\\\\ \tau = 1.168 \ Nm[/tex]

The angular acceleration of the books is calculated as;

[tex]\alpha = \frac{a}{R} \\\\\alpha = \frac{2.964}{0.075}\\\\\alpha =39.52 \ rad/s^2[/tex]

The moment of inertia of the pulley about its rotation axis is calculated as follows;

[tex]\tau = I \alpha\\\\I = \frac{\tau }{\alpha} \\\\I = \frac{1.168}{39.52} \\\\I = 0.0296 \ kg m^2[/tex]

Learn more about moment of inertia of pulley here: https://brainly.com/question/25044518

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