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A \(2.00 \mathrm{~kg}\) textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is \(0.150 \mathrm{~m},\) to a hanging book with mass \(3.00 \mathrm{~kg} .\) The system is released from rest, and the books are observed to move \(1.20 \mathrm{~m}\) in \(0.800 \mathrm{~s}\) (a) What is the tension in each part of the cord? (b) What is the moment of inertia of the pulley about its rotation axis?

Short Answer

Expert verified
The tension in each part of the cord is 23.55 N and the moment of inertia of the pulley about its rotation axis is \(0.0677\, \mathrm{kg\, m^{2}}\).

Step by step solution

01

Find the acceleration of the books

First, find the net force acting on the two books. For the hanging book, the downward force is its weight, \(3.00 \, \mathrm{kg} \cdot 9.81 \, \mathrm{m/s^{2}} = 29.4 \, \mathrm{N}\). We'll treat the direction of this force as positive. The force on the book on the surface is just its weight, \(2.00 \, \mathrm{kg} \cdot 9.81 \, \mathrm{m/s^{2}} = 19.6 \, \mathrm{N}\), which we'll treat as negative because it opposes the motion of the hanging book. So the net force is \(29.4\, \mathrm{N} - 19.6\, \mathrm{N} = 9.8\, \mathrm{N}\). We also know that \(F=ma\), or \(a=F/m\). Our combined mass is \(2.00\, \mathrm{kg} + 3.00\, \mathrm{kg} = 5.00\, \mathrm{kg}\). Using these values, \(a = 9.8\, \mathrm{N} / 5.00\, \mathrm{kg} = 1.96\, \mathrm{m/s^{2}}\).
02

Find the tension in the cord

Now calculate the tension in the cord. The net force on the hanging book is the difference between the downward force (its weight) and the upward force (the tension in the cord). The downward force is \(3.00\, \mathrm{kg} \cdot 9.81\, \mathrm{m/s^{2}} = 29.43\, \mathrm{N}\), and the upward force is \(3.00\, \mathrm{kg} \cdot 1.96\, \mathrm{m/s^{2}} = 5.88\, \mathrm{N}\). Hence, the tension, \(T = 29.43\, \mathrm{N} - 5.88\, \mathrm{N} = 23.55\, \mathrm{N}\).
03

Find the moment of inertia of the pulley

First, find the angular acceleration of the pulley by dividing the linear acceleration of the books by the radius of the pulley. The acceleration is \(1.96\, \mathrm{m/s^{2}}\), and the radius of the pulley is \((0.150\, \mathrm{m}) / 2 = 0.075\, \mathrm{m}\), so the angular acceleration is \(1.96\, \mathrm{m/s^{2}} / 0.075\, \mathrm{m} = 26.13\, \mathrm{rad/s^{2}}\). Then use the Newton's second law in angular form: \(\tau = I \alpha\), where \(\tau\) is the torque, \(I\) is moment of inertia and \(\alpha\) is angular acceleration. The torque is given by the tension in the cord times the radius of the pulley, \(\tau = T \cdot r = 23.55\, \mathrm{N} \cdot 0.075\, \mathrm{m} = 1.77\, \mathrm{N \cdot m}\). Solving for \(I\) gives \(I = \tau / \alpha = 1.77\, \mathrm{N \cdot m} / 26.13\, \mathrm{rad/s^{2}} = 0.0677\, \mathrm{kg\, m^{2}}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tension in a Cord
Understanding the tension in a cord is essential when dealing with systems involving pulleys. Tension is a force that is transmitted through a string, rope, or cable when it is pulled tight by forces acting from opposite ends. In the context of the textbook and pulley problem, the tension represents the force which restrains the cord from free fall and enables the transfer of force from one book to the other.

In order to calculate the tension, one must consider the forces acting on the objects at either end of the cord. As we see in the solution, after calculating the net force acting on the system and the acceleration, we can determine the tension in the cord by accounting for the forces acting on the hanging book, including both its weight and the force due to acceleration.

It is crucial to remember that tension throughout the same cord or cable, assuming no friction in the pulley, is typically the same. This characteristic of tension is essential in solving many physics problems that include pulleys and cords.
Angular Acceleration
Angular acceleration plays a pivotal role in problems involving rotational motion, like that of a pulley in physics. It is defined as the rate at which the angular velocity of an object changes with respect to time. To find the angular acceleration \( \alpha \), you must consider the linear acceleration \( a \) and the radius of the rotation \( r \).

In the textbook exercise under scrutiny, angular acceleration is found by dividing the linear acceleration by the radius of the pulley. It is crucial because it connects the linear acceleration of the books to the rotational motion of the pulley. This concept is applied using the formula \( \alpha = \frac{a}{r} \), which is a simplified version of the more complex relationship between linear and angular quantities in rotational motion.

Understanding angular acceleration is not only critical for solving the problem at hand but also forms a foundational concept for understanding more complex systems involving rotational dynamics in physics.
Newton's Second Law for Rotation
Newton's second law for rotation is the rotational analogue to Newton's second law of motion for linear dynamics. It states that the torque \( \tau \) applied to an object is equal to the product of its moment of inertia \( I \) and its angular acceleration \( \alpha \) that is \( \tau = I \alpha \).

The moment of inertia represents the resistance of an object to changes in its rotational motion and is determined by both the mass of the object and the distribution of that mass relative to the axis of rotation.

By combining the concept of tension in the cord and angular acceleration, we can apply Newton's second law for rotation to find the moment of inertia for the pulley. The torque caused by the tension in the cord is countered by the pulley's resistance to change in its rotational speed, expressed through its moment of inertia. In the problem, finding the torque involved multiplying the tension in the cord by the radius of the pulley. This calculated torque, when set in the equation \( \tau = I \alpha \) along with the found angular acceleration, yields the moment of inertia of the pulley.

This law is a cornerstone in rotational dynamics, enabling the solution of a myriad of problems involving rotating objects, from simple pulleys to complex mechanical systems.

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Most popular questions from this chapter

A hollow, spherical shell with mass \(2.00 \mathrm{~kg}\) rolls without slipping down a \(38.0^{\circ}\) slope. (a) Find the acceleration, the friction force, and the minimum coefficient of static friction needed to prevent slipping. (b) How would your answers to part (a) change if the mass were doubled to \(4.00 \mathrm{~kg}\) ?

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A doubling of the torque produces a greater angular acceleration. Which of the following would do this, assuming that the tension in the rope doesn't change? (a) Increasing the pulley diameter by a factor of \(\sqrt{2} ;\) (b) increasing the pulley diameter by a factor of \(2 ;\) (c) increasing the pulley diameter by a factor of \(4 ;\) (d) decreasing the pulley diameter by a factor of \(\sqrt{2}\)

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