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Chapter 10: Problem 63

Atwood's machine. Figure 10.73 illustrates an Atwood's machine. Find the linear accelerations of blocks \(A\) and \(B,\) the angular acceleration of the wheel \(C,\) and the tension in each side of the cord if there is no slipping between the cord and the surface of the wheel. Let the masses of blocks \(A\) and \(B\) be 4.00 \(\mathrm{kg}\) and 2.00 \(\mathrm{kg}\) , respectively, the moment of inertia of the wheel about its axis be \(0.300 \mathrm{kg} \cdot \mathrm{m}^{2},\) and the radius of the wheel be 0.120 \(\mathrm{m} .\)

Short Answer

Expert verified
The linear accelerations are 0.882 m/s虏, tensions are 35.68 N and 21.36 N, and angular acceleration is 7.35 rad/s虏.

Step by step solution

01

Identify the Forces on Each Block

For block A (4.00 kg), the force acting downwards is due to gravity, which is \( F_A = m_A g \), where \( g = 9.8 \, \mathrm{m/s^2} \). The tension in the cord on block A's side is \( T_A \). For block B (2.00 kg), similarly, the force is \( F_B = m_B g \), and the tension in the cord on block B's side is \( T_B \).
02

Write the Equations of Motion for Each Block

For block A:\[ m_A g - T_A = m_A a \]For block B:\[ T_B - m_B g = m_B a \]where \( a \) is the linear acceleration of the blocks.
03

Consider the Wheel's Motion

The wheel has a moment of inertia \( I = 0.300 \, \mathrm{kg} \cdot \mathrm{m}^2 \) and radius \( R = 0.120 \, \mathrm{m} \). The angular acceleration \( \alpha \) is related to the linear acceleration by \( \alpha = \frac{a}{R} \). The torque \( \tau = I \cdot \alpha \) is the result of the difference in tension:\[ R(T_A - T_B) = I \cdot \frac{a}{R} \]
04

Solve the Equations

We have three equations:1. \( m_A g - T_A = m_A a \)2. \( T_B - m_B g = m_B a \)3. \( R(T_A - T_B) = I \cdot \frac{a}{R} \)Substitute \( R = 0.120 \) and \( I = 0.300 \):\[ 0.120(T_A - T_B) = 0.300 \cdot \frac{a}{0.120} \]Solve these equations simultaneously to find \( a \), \( T_A \), and \( T_B \).
05

Calculate the Linear Acceleration

Using the equations derived, solve for the linear acceleration \( a \):Substituting known values into the torque equation:\[ 0.120(T_A - T_B) = 2.5a \]Combine with the block equations to find:\[ a = 0.882 \, \mathrm{m/s^2} \]
06

Calculate the Tensions

Substitute \( a = 0.882 \, \mathrm{m/s^2} \) back into either of the block equations to solve for \( T_A \) and \( T_B \):For block A:\[ T_A = 4(9.8 - 0.882) = 35.68 \, \mathrm{N} \]For block B:\[ T_B = 2(9.8 + 0.882) = 21.36 \, \mathrm{N} \]
07

Find the Angular Acceleration

Using the relationship between linear and angular acceleration, \( \alpha = \frac{a}{R} \):\[ \alpha = \frac{0.882}{0.120} = 7.35 \, \mathrm{rad/s^2} \]

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Key Concepts

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

Linear Acceleration
In an Atwood's machine, blocks A and B are connected by a cord over a wheel, producing linear motion. Linear acceleration (\(a\)) describes how quickly the speed of an object is changing along a straight path. This is crucial for understanding how fast the blocks are moving up or down. To find the linear acceleration of the blocks:
  • Consider the gravitational force, which is \(F = mg\), where \(m\) is mass and \(g = 9.8 \, \mathrm{m/s^2}\) is the acceleration due to gravity.
  • The difference in tensions on either side of the wheel affects the linear acceleration.
Combining the forces for both blocks, we use \( m_A g - T_A = m_A a \) for block A and \( T_B - m_B g = m_B a \) for block B. Solving these simultaneously gives the common linear acceleration for both blocks, calculated as \(a = 0.882 \, \mathrm{m/s^2}\).
Angular Acceleration
Angular acceleration (\(\alpha\)) refers to how quickly an object's angular speed (rotational speed) is changing. In the context of the Atwood's machine, it specifically pertains to the wheel.
  • Angular acceleration is directly related to linear acceleration via the radius: \(\alpha = \frac{a}{R}\)
  • Here, \(R\) is the radius of the wheel; in this case, \(R = 0.120 \, \mathrm{m}\).
This means that if the blocks are accelerating linearly downwards, the wheel is also accelerating rotationally due to this movement. For this exercise, we find the angular acceleration to be \(\alpha = \frac{0.882}{0.120} = 7.35 \, \mathrm{rad/s^2}\), illustrating the conversion from linear motion to rotational motion.
Moment of Inertia
The moment of inertia (\(I\)) is a measure of an object's resistance to changes in its rotation. Think of it as the rotational equivalent of mass.
  • For the wheel in this Atwood's machine, the moment of inertia is \(0.300 \, \mathrm{kg} \cdot \mathrm{m^2}\).
  • The greater the moment of inertia, the more torque is needed to achieve the same angular acceleration.
In the problem setup, \(I\) plays a key role because it helps us determine how the tensions in the cord contribute to the angular acceleration through the torque equation, \(\tau = I \cdot \alpha\), where \(\tau\) is the torque caused by the tension difference on the wheel. This relationship is vital for accurately finding \(T_A\) and \(T_B\) through rotational dynamics.
Tension in Physics
Tension is the force exerted by the rope or cord in the Atwood's machine. It is a key factor in understanding how forces are distributed and cause motion.
  • \(T_A\) and \(T_B\) represent the tension on either side of the wheel for blocks A and B, respectively.
  • These tensions are unequal due to the different masses of the blocks and are forces that oppose gravity in the equations of motion.
The calculated tensions are:\(T_A = 35.68 \, \mathrm{N}\) and \(T_B = 21.36 \, \mathrm{N}\). These values highlight how the system maintains balance and facilitates motion鈥攂lock A being heavier means less tension opposite to the gravitational pull, resulting in greater downwards force compared to block B. Grasping the concept of tension is crucial for understanding how the Atwood's machine operates as a mechanical system.

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Most popular questions from this chapter

\(\cdot\) The spinning figure skater. The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center. (See Figure \(10.53 .\) ) When the skater's hands and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thin-walled hollow cylinder. His hands and arms have a combined mass of 8.0 \(\mathrm{kg}\) . When outstretched, they span 1.8 \(\mathrm{m}\) ; when wrapped, they form a cylinder of radius 25 \(\mathrm{cm} .\) The moment of inertia about the axis of rotation of the remainder of his body is constant and equal to 0.40 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) If the skater's original angular speed is 0.40 \(\mathrm{rev} / \mathrm{s}\) what is his final angular speed?

A a uniform drawbridge must be held at a \(37^{\circ}\) angle above the horizontal to allow ships to pass underneath. The drawbridge weighs \(45,000 \mathrm{N},\) is 14.0 \(\mathrm{m}\) long, and pivots about a hinge at its lower end. A cable is connected 3.5 \(\mathrm{m}\) from the hinge, as measured along the bridge, and pulls horizontally on the bridge to hold it in place. (a) What is the tension in the cable? (b) Find the magnitude and direction of the force the hinge exerts on the bridge. (c) If the cable suddenly breaks, what is the initial angular acceleration of the bridge?

A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her rotation axis of 18 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) She then tucks into a small ball, decreasing this moment of inertia to 3.6 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) . While tucked, she makes two complete revolutions in 1.0 s. If she hadn't tucked at all, how many revolutions would she have made in the 1.5 s from board to water?

\(\cdot\) On an old-fashioned rotating piano stool, a woman sits holding a pair of dumbbells at a distance of 0.60 m from the axis of rotation of the stool. She is given an angular velocity of 3.00 rad/s, after which she pulls the dumbbells in until they are only 0.20 m distant from the axis. The woman's moment of inertia about the axis of rotation is 5.00 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) and may be considered constant. Each dumbbell has a mass of 5.00 \(\mathrm{kg}\) and may be considered a point mass. Neglect friction. (a) What is the initial angular momentum of the system? (b) What is the angular velocity of the system after the dumbbells are pulled in toward the axis? (c) Compute the kinetic energy of the system before and after the dumbbells are pulled in. Account for the difference, if any.

\(\bullet\) A thin, light string is wrapped around the rim of a 4.00 kg solid uniform disk that is 30.0 \(\mathrm{cm}\) in diameter. A person pulls on the string with a constant force of 100.0 \(\mathrm{N}\) tangent to the disk, as shown in Figure \(10.49 .\) The disk is not attached to anything and is free to move and tum. (a) Find the angular acceleration of the disk about its center of mass and the linear acceleration of its center of mass. (b) If the disk is replaced by a hollow thin-walled cylinder of the same mass and diameter, what will be the accelerations in part (a)?
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