Question

A textbook of mass 1.95 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.100 m , to a hanging book with mass 3.08 kg . The system is released from rest, and the books are observed to move a distance 1.27 m over a time interval of 0.830 s .

Part A:What is the tension in the part of the cord attached to the textbook?

Part B:What is the tension in the part of the cord attached to the book?
Take the free fall acceleration to be g = 9.80 m/s2 .

Part C:What is the moment of inertia of the pulley about its rotation axis?
Take the free fall acceleration to be g = 9.80 m/s2 .

Answer 1

potential energy loss = m g h
= 3.08 * 9.8 * 1.27

= kinetic energy gain
= (1/2) (1.95)v^2 + (1/2)I (v/r)^2

so what is v at the end of this experiment?
find acceleration
1.27 = (1/2)a t^2
1.27 = (1/2)a (.83)^2
so
a = 3.69 m/s^2
and
v = a t = 3.69(.83) = 3.06 m/s at the end
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well that should give you what you need
mg - Tension = m a
Tension = 3.08 (9.8-3.69)

You have I in that kinetic energy formula above