Power, Torque, and Velocity in a Pulley System

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SUMMARY

The discussion focuses on calculating the tension in the cord attached to two books and the moment of inertia of a pulley in a frictionless pulley system. The system consists of a 1.99 kg textbook and a 2.97 kg hanging book, with a pulley diameter of 0.150 m. Key equations used include F=ma, Torque= Iα, and I=mR². The solution involves determining the acceleration of the system and the relationship between tension and the moment of inertia of the pulley.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of torque and rotational dynamics (Torque= Iα)
  • Familiarity with kinematics equations for uniformly accelerated motion
  • Basic concepts of moment of inertia (I=mR²)
NEXT STEPS
  • Calculate the acceleration of the system using kinematic equations.
  • Determine the tension in the cord for both books using F=ma.
  • Explore the relationship between linear acceleration and angular acceleration in pulley systems.
  • Investigate the effects of pulley mass on tension and acceleration in similar systems.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators looking for practical examples of pulley systems in action.

David112234
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Homework Statement


A textbook of mass 1.99 kg 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 2.97 kg . The system is released from rest, and the books are observed to move a distance 1.20 m over a time interval of 0.790 s .

A.What is the tension in the part of the cord attached to the textbook?
B.What is the tension in the part of the cord attached to the book?
C.What is the moment of inertia of the pulley about its rotation axis?

2. Homework Equations

F=ma
Torque= Iα
a=Rα
I=mR2

The Attempt at a Solution


book.jpg

Forces on book 1 (on top/textbook)

mg and Normal force cancel each other so the only force is to the right, Tension

t1=m1a

Forces on book 2 (hanging one)

-t2 + m2g= -m2a

Forces on wheel pulley

The string will create a torque in the clockwise direction

T = Iα
a = Rα
a/R = α

T = I (a/R)

I am stumped as what to do, I can find the t1 by using the second equation and solving for a in terms of t2 and plugging it back into the first one but then I would have another unknown variable.
 
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Before the books start to move the tension in the cable is the hanging mass X the acceleration of gravity. The tension in the cable is the same at all points in the cable when the books are at rest. Assume that the books accelerate at a constant rate. The distance traveled is equal to 1/2 the acceleration times the square of the elapsed time. Rearrange terms to find the acceleration. Compare that acceleration to the acceleration due to the masses and forces involved. Acceleration equals force divided by mass. Any difference between the two acceleration rates could be assumed to be the effects of the inertia of the pulley. That's as far as you can go without knowing either the mass or the radius of the pulley (or both mass and radius).

Once things start moving the forces on the cable will be different on each side of the pulley. You can calculate that using the observed acceleration and the mass of the books.

That should get you started.
 

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