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Chapter 11: Problem 46

A \(2.85-\mathrm{kg}\) bucket is attached to a disk-shaped pulley of radius \(0.121 \mathrm{m}\) and \(\mathrm{mass} 0.742 \mathrm{kg} .\) If the bucket is allowed to fall, (a) what is its linear acceleration? (b) What is the angular acceleration of the pulley? (c) How far does the bucket drop in 1.50 s?

Short Answer

Expert verified
Linear acceleration: 8.01 m/s虏, angular acceleration: 66.20 rad/s虏, distance dropped: 9.01 m.

Step by step solution

01

Define Known Values

We have a bucket with mass \( m = 2.85 \, \text{kg} \), a pulley with radius \( R = 0.121 \, \text{m} \), and mass \( M = 0.742 \, \text{kg} \). We need to find linear acceleration of the bucket \( a \), angular acceleration of the pulley \( \alpha \), and distance the bucket drops in 1.50 s.
02

Use Moment of Inertia for the Pulley

The moment of inertia \( I \) for a disk-shaped pulley is \( I = \frac{1}{2} MR^2 \). Substituting the given values: \[ I = \frac{1}{2} \times 0.742 \, \text{kg} \times (0.121 \, \text{m})^2 = 0.00542851 \, \text{kg} \cdot \text{m}^2. \]
03

Establish Newton's Second Law for the System

Using Newton鈥檚 second law for the bucket: \[ T = m(g - a), \] where \( T \) is the tension and \( g = 9.81 \, \text{m/s}^2 \).
04

Relate Torque and Angular Acceleration

For the pulley, the torque \( \tau \) is \( \tau = I \alpha = TR \). Substitute the expression for \( T \) from Step 3 and the expression for the torque: \[ I \alpha = Rm(g - a). \]
05

Relate Linear and Angular Acceleration

The linear acceleration \( a \) and angular acceleration \( \alpha \) are related by \( \alpha = \frac{a}{R} \). Substitute this into the equation from Step 4: \[ I\frac{a}{R} = Rm(g - a). \]
06

Solve for Linear Acceleration

Rearrange the equation to solve for \( a \): \[ Ia = R^2m(g - a), \] leading to \[ a(I + R^2m) = R^2mg, \] thus, \[ a = \frac{R^2mg}{I + R^2m}. \] Substitute the known values: \[ a = \frac{(0.121)^2 \times 2.85 \times 9.81}{0.00542851 + (0.121)^2 \times 2.85} = 8.01 \, \text{m/s}^2. \]
07

Calculate Angular Acceleration

With linear acceleration \( a \), find angular acceleration \( \alpha \) using \( \alpha = \frac{a}{R} \): \[ \alpha = \frac{8.01}{0.121} = 66.20 \, \text{rad/s}^2. \]
08

Calculate Distance Dropped

Use the kinematic equation for distance with initial velocity \( v_0 = 0 \): \[ d = v_0 t + \frac{1}{2} at^2 = 0 + \frac{1}{2} \times 8.01 \times (1.50)^2. \] Calculate: \[ d = 0.5 \times 8.01 \times 2.25 = 9.01 \, \text{m}. \]

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Key Concepts

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

Moment of Inertia
The moment of inertia is a key concept in rotational dynamics, similar to mass in linear motion. It measures an object's resistance to changes in its rotation. For a disk-shaped pulley, like the one in our exercise, the formula to calculate the moment of inertia is \( I = \frac{1}{2} MR^2 \). Here, \( M \) is the mass of the disk and \( R \) is the radius.
  • In our exercise, with a mass \( M = 0.742 \, \text{kg} \) and radius \( R = 0.121 \, \text{m} \), the moment of inertia is calculated to be approximately \( 0.00542851 \, \text{kg} \cdot \text{m}^2 \).
  • This value helps determine how easily the pulley can start or stop spinning.
Understanding moment of inertia is crucial as it directly influences the angular behavior of rotating systems.
Linear Acceleration
Linear acceleration refers to the rate of change of velocity that an object experiences as it moves along a straight path. In the context of the falling bucket in this exercise, linear acceleration is directly influenced by both gravitational forces and the resistance provided by the pulley system.
  • The gravitational force acting downwards is countered by the tension in the rope, which relies on forces calculated using Newton's second law.
  • In our solution, we find that the linear acceleration \( a \) of the bucket is \( 8.01 \, \text{m/s}^2 \).
This value shows us how quickly the bucket is accelerating downward as it is released, factoring in not only gravity but also the rotational dynamics of the pulley.
Angular Acceleration
Angular acceleration describes how quickly an object's rotational speed changes over time. It depends on both the applied forces and the moment of inertia. In problems involving pulleys and gears, it's often found through the connection between linear and angular motions.
  • The relationship \( \alpha = \frac{a}{R} \) links linear acceleration \( a \) with the radius \( R \) of the object.
  • For our pulley, given that \( a = 8.01 \, \text{m/s}^2 \), angular acceleration \( \alpha \) becomes \( 66.20 \, \text{rad/s}^2 \).
Understanding angular acceleration is critical for analyzing how different parts of a system with rotational motion respond to forces over time.
Newton's Second Law
Newton's second law is foundational in understanding motion. It claims that the force acting on an object is equal to the mass of that object multiplied by its acceleration, expressed as \( F = ma \). This law is central when analyzing both linear and rotational motions.
  • In the problem, Newton鈥檚 second law helps in setting up the equations that govern both the motion of the falling bucket and the rotation of the pulley.
  • For linear motion: tension in the rope and gravitational pull on the bucket are balanced to find linear acceleration.
  • For rotational motion: the relationship between torque (caused by the tension) and angular acceleration of the pulley is elucidated.
This law builds a bridge between forces and motion, allowing us to predict how objects will move under various situations.

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Most popular questions from this chapter

A wheel on a game show is given an initial angular speed of \(1.22 \mathrm{rad} / \mathrm{s}\). It comes to rest after rotating through 0.75 of a turn. (a) Find the average torque exerted on the wheel given that it is a disk of radius \(0.71 \mathrm{m}\) and \(\mathrm{mass} 6.4 \mathrm{kg}\). (b) If the mass of the wheel is doubled and its radius is halved, will the angle through which it rotates before coming to rest increase, decrease, or stay the same? Explain. (Assume that the average torque exerted on the wheel is unchanged.)

A torque of \(0.12 \mathrm{N} \cdot \mathrm{m}\) is applied to an egg beater. (a) If the egg beater starts at rest, what is its angular momentum after \(0.65 \mathrm{s} ?\) (b) If the moment of inertia of the egg beater is \(2.5 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2},\) what is its angular speed after \(0.65 \mathrm{s} ?\)

To loosen the lid on a jar of jam \(8.9 \mathrm{cm}\) in diameter, a torque of \(8.5 \mathrm{N} \cdot \mathrm{m}\) must be applied to the circumference of the lid. If a jar wrench whose handle extends \(15 \mathrm{cm}\) from the center of the jar is attached to the lid, what is the minimum force required to open the jar?

A 0.34-kg meterstick balances at its center. If a necklace is suspended from one end of the stick, the balance point moves \(9.5 \mathrm{cm}\) toward that end. (a) Is the mass of the necklace more than, less than, or the same as that of the meterstick? Explain. (b) Find the mass of the necklace.

A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length \(3.15 \mathrm{m}\) and \(\mathrm{mass} 8.42 \mathrm{kg}\) find the torque the person must exert on the ladder to give it an angular acceleration of \(0.302 \mathrm{rad} / \mathrm{s}^{2}\)
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