PHYSICS HIGH SCHOOL

A textbook of mass 2.09kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.120m , to a hanging book with mass 2.99kg . The system is released from rest, and the books are observed to move a distance 1.30m over a time interval of 0.750s . Part A What is the tension in the part of the cord attached to the textbook? =9.66N.
Part B What is the tension in the part of the cord attached to the book? Take the free fall acceleration to be = 9.80m/s^2 =15.5N.
Part C What is the moment of inertia of the pulley about its rotation axis? Take the free fall acceleration to be = 9.80 m/s^2????

Answers

Answer 1
Answer:

Answer:

(A) 9.7 N

(B) 15.4 N

 (C) I = 0.0045 kg m^{2}

Explanation:

mass of text book (M1) = 2.09 kg

mass of book (M2) = 2.99 kg

diameter of the pulley (d) = 0.12 m

radius (r) = 0.06 m

distance moved (s) = 1.30 m

time (t) = 0.75 s

acceleration due to gravity (g) = 9.8 m/s^[2}

(a) what is the tension in the part of the cord attached to the text book?

the text book is moving horizontally, so the tension in this case becomes

tension = mass x acceleration

we can get the acceleration from s = ut + 0.5 at^{2}

since the books are initially at rest u = 0

s = 0.5 at^{2}

1.3 = 0.5 x a x 0.75^{2}

a = 4.643 m/s^[2}

 

tension (T1) = 2.09 x 4.643 = 9.7 N

(b) what is the tension in the part of the cord attached to the book?

   the book is hanging vertically, so the tension in this case becomes

tension = m x ( g - a )

(g-a) is the net acceleration of the first book

tension (T2) = 2.99 x (9.8 - 4.643) = 15.4 N

(c) What is the moment of inertia of the pulley?

    if the books were to move in the direction of the book, it will cause the pulley to rotate clockwise and if they were to move in the direction of the text book on the table the pulley will rotate in an anticlockwise direction. Taking clockwise rotation of the pulley to be negative while anticlockwise to be positive, we can say  ( T2 - T1 )r = I∝

      where ∝ is the angular acceleration of the pulley relative to its radial  

      acceleration, ∝ = \frac{a}{r}

      ( T2 - T1 )r = I\frac{a}{r}

      I = \frac{(T2 - T1)r^{2}}{a}

      I = \frac{(15.5 - 9.7)0.060^{2}}{4.643}

      I = 0.0045 kg m^{2}


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Answer

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