Skip to main content
Ch 10: Dynamics of Rotational Motion
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 10, Problem 16b

A 12.0-kg box resting on a horizontal, frictionless surface is attached to a 5.00-kg weight by a thin, light wire that passes over a frictionless pulley (Fig. E10.16). The pulley has the shape of a uniform solid disk of mass 2.00 kg and diameter 0.500 m. After the system is released, find the acceleration of the box.

Verified step by step guidance
1
Identify the forces acting on the system: The 5.00-kg weight experiences a gravitational force downward, which is the driving force for the system. The tension in the wire acts upward on the 5.00-kg weight and horizontally on the 12.0-kg box.
Apply Newton's second law to the 5.00-kg weight: The net force on the weight is the difference between the gravitational force and the tension in the wire. This can be expressed as: \( m_2 g - T = m_2 a \), where \( m_2 = 5.00 \text{ kg} \), \( g = 9.81 \text{ m/s}^2 \), and \( a \) is the acceleration.
Apply Newton's second law to the 12.0-kg box: The only horizontal force on the box is the tension in the wire, so \( T = m_1 a \), where \( m_1 = 12.0 \text{ kg} \).
Consider the rotational motion of the pulley: The torque on the pulley due to the tension is \( \tau = T R \), where \( R = 0.250 \text{ m} \) is the radius of the pulley. The moment of inertia of the pulley is \( I = \frac{1}{2} m R^2 \), where \( m = 2.00 \text{ kg} \). The angular acceleration \( \alpha \) is related to the linear acceleration by \( \alpha = \frac{a}{R} \).
Combine the equations: Use the relationship between torque and angular acceleration \( \tau = I \alpha \) to find \( T \) in terms of \( a \). Substitute this expression into the equations for the 5.00-kg weight and the 12.0-kg box to solve for the acceleration \( a \).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
17m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Newton's Second Law

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass (F = ma). In this problem, it helps determine the acceleration of the box by considering the forces acting on both the box and the weight, including gravitational force and tension in the wire.
Recommended video:
Guided course
06:54
Intro to Forces & Newton's Second Law

Rotational Dynamics

Rotational dynamics involves the study of objects in rotational motion, where torque and angular acceleration play key roles. The pulley in this problem is a solid disk, and its rotational inertia affects the system's dynamics. Understanding how the torque due to the tension in the wire causes angular acceleration in the pulley is crucial for solving the problem.
Recommended video:
Guided course
08:12
Torque & Acceleration (Rotational Dynamics)

Conservation of Energy

Conservation of energy states that energy in a closed system remains constant. In this scenario, the potential energy of the weight is converted into kinetic energy of the box and rotational kinetic energy of the pulley. Analyzing energy transformations helps in understanding the system's behavior and calculating the acceleration of the box.
Recommended video:
Guided course
06:24
Conservation Of Mechanical Energy
Related Practice
Textbook Question

A 2.20-kg hoop 1.20 m in diameter is rolling to the right without slipping on a horizontal floor at a steady 2.60 rad/s. Find the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop; (iii) a point on the right side of the hoop, midway between the top and the bottom.

2922
views
Textbook Question

A 2.00-kg textbook 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 3.00 kg. The system is released from rest, and the books are observed to move 1.20 m in 0.800 s. What is the tension in each part of the cord?

2780
views
2
rank
Textbook Question

A solid ball is released from rest and slides down a hillside that slopes downward at 65.0° from the horizontal. In part (a), why did we use the coefficient of static friction and not the coefficient of kinetic friction?

1302
views
Textbook Question

A stone is suspended from the free end of a wire that is wrapped around the outer rim of a pulley, similar to what is shown in Fig. 10.10. The pulley is a uniform disk with mass 10.0 kg and radius 30.0 cm and turns on frictionless bearings. You measure that the stone travels 12.6 m in the first 3.00 s starting from rest. Find the tension in the wire.

4414
views
1
rank
1
comments
Textbook Question

A 2.20-kg hoop 1.20 m in diameter is rolling to the right without slipping on a horizontal floor at a steady 2.60 rad/s. Find the velocity vector for each of the points in part (c), but this time as viewed by someone moving along with the same velocity as the hoop.

1667
views
Textbook Question

A 15.0-kg bucket of water is suspended by a very light rope wrapped around a solid uniform cylinder 0.300 m in diameter with mass 12.0 kg. The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 10.0 m to the water. With what speed does the bucket strike the water?

3068
views