A textbook of mass 1.95 kg rests on a frictionless, horizontal surface. A cord a

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 0.200 m, to a hanging book with mass 2.95 kg. The system is released from rest, and the books are observed to move a distance 1.22 m over a time interval of 0.850 s. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of What is the tension in the part of the cord attached to the textbook? 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/s^2. What is the moment of inertia of the pulley about its rotation axis? Take the free fall acceleration to be g = 9.80 m/s^2.

Explanation / Answer

Let’s use the following equation to determine the acceleration.

d = vi * t + ½ * a * t^2, vi = 0
1.22= ½ * a * 0.850^2
a = 2.44÷ 0.7255
This is approximately 3.363m/s^2. Let’s use the following equation to determine the tension in the part of the cord attached to the textbook.

T = m * a = 1.95 * 3.363= 6.55785 N

Let’s determine the weight of the book.

Weight = 2.95 * 9.8 = 28. 91N
The net force on the book is its weight minus the tension.

28.91 – T = 2.95 * 3.363
T = 28. 91 - 9.92085 = 18.98915 N

C.) What is the moment of inertia of the pulley about its rotation axis?
over a pulley whose diameter is 0.200 m

These two forces produce the torque that causes the pulley to accelerate as it rotates.

T2 * r – T1 * r = I *
= a/r

T2 * r – T1 * r = I * a/r
(T2– T1) * r = I * a/r
(T2 – T1) * r^2 = I * a
I = (T2 – T1) * r^2 ÷ a

Net force = 18.98915 – 6.55785 = 12.43130 N
Radius = 0.100 meter

I = 12.43130* 0.100^2 ÷ 3.363 =. 036964
I is approximately 0.0369

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