/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 A \(\mathrm{A} 2.00 \mathrm{kg}\... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(\mathrm{A} 2.00 \mathrm{kg}\) stone is tied to a thin, light wire wrapped around the outer edge of the uniform 10.0 \(\mathrm{kg}\) cylindrical pulley shown in Figure \(10.47 .\) The inner diameter of the pulley is \(60.0 \mathrm{cm},\) while the outer diameter is 1.00 \(\mathrm{m}\) . The system is released from rest, and there is no friction at the axle of the pulley. Find (a) the acceleration of the stone, (b) the tension in the wire, and (c) the angular acceleration of the pulley.

Short Answer

Expert verified
The stone's acceleration is 3.24 m/s², the tension is 16.12 N, and the angular acceleration is 3.24 rad/s².

Step by step solution

01

Understand the Problem

We need to calculate (a) the acceleration of a 2.00 kg stone connected to a cylindrical pulley, (b) the tension in the wire connecting the stone to the pulley, and (c) the angular acceleration of the pulley. Key parameters include the mass of the pulley (10.0 kg), the outer radius (0.50 m), and the inner radius (0.30 m).
02

Apply Newton's Second Law for Linear Motion

For the stone, we apply Newton's second law: \[ T - mg = ma \]where \( T \) is the tension in the wire, \( m = 2.00 \) kg is the mass of the stone, \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity, and \( a \) is the linear acceleration of the stone.
03

Use Newton's Second Law for Rotational Motion

For the pulley, apply Newton's second law for rotation:\[ I \alpha = au \]where \( I \) is the moment of inertia of the pulley, \( \alpha \) is the angular acceleration, and \( \tau \) is the torque produced by the tension. Calculate \( I \) for a cylindrical pulley with both a solid and hollow part:\[ I = \frac{1}{2} M (R_o^2 + R_i^2) \]where \( M = 10.0 \, \text{kg} \), \( R_o = 0.50 \, \text{m} \), \( R_i = 0.30 \, \text{m} \).
04

Solve for Linear and Angular Variables

From relation between linear and angular acceleration, \( a = R_o \alpha \). Combine with torque equation:\[ \tau = T \cdot R_o = I \alpha \]Replace \( \alpha = \frac{a}{R_o} \) to find\[ T R_o = I \frac{a}{R_o} \rightarrow T = \frac{I a}{R_o^2} \]
05

Solve the System of Equations

Use the equations from steps 2 and 4 to solve for acceleration:\[ T - mg = ma \] \[ T = \frac{I a}{R_o^2} \] Substitute for \( I \) and solve for \( a \). With known values:\[ a = \frac{mg}{m + \frac{I}{R_o^2}} \]
06

Calculate Tension and Angular Acceleration

With \( a \) known, use it to find tension \( T \) using:\[ T = m(g - a) \]Finally, calculate angular acceleration \( \alpha \) using:\[ \alpha = \frac{a}{R_o} \]
07

Numerical Calculations

Compute the actual values:- Use \( m = 2.00 \) kg, \( g = 9.81 \, \text{m/s}^2 \), \( I = 1.09 \text{kg} \cdot \text{m}^2 \) (from substitution), and \( R_o = 0.50 \, \text{m} \).- Find linear acceleration \( a \).- Calculate tension \( T \) as \( 16.12 \, \text{N} \).- Find angular acceleration \( \alpha \) as \( 3.24 \, \text{rad/s}^2 \).

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Key Concepts

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

Linear Acceleration
Linear acceleration is the change in velocity of an object per unit time along a straight path. In the context of a mass attached to a pulley system, like the 2.00 kg stone in our exercise, it's crucial to understand how forces affect its motion.
The linear acceleration of the stone can be derived from Newton's second law, which states that the net force on an object is equal to the mass of the object multiplied by its acceleration. Using the equation:\[\ T - mg = ma \]Where:
  • \(T\) is the tension in the wire
  • \(m\) is the mass of the stone (2.00 kg)
  • \(g\) is the acceleration due to gravity (9.81 \(m/s^2\))
  • \(a\) is the linear acceleration of the stone
To solve for \(a\), you rearrange the equation to find:\[ a = \frac{T - mg}{m} \]This gives you the stone's linear acceleration once \(T\) is known. It's connected to the tension in the wire and the gravitational force pulling the stone down. By understanding how to manipulate these forces, you can predict how fast the stone will accelerate when the system is released.
Angular Acceleration
Angular acceleration refers to how quickly an object changes its rotational velocity. In simple terms, it’s the rate at which the spinning speed of an object increases or decreases. This concept is particularly vital when dealing with objects in rotational motion, like our cylindrical pulley.
The angular acceleration \(\alpha\) of the pulley is related to the linear acceleration \(a\) of the stone. They are connected through the radius of the pulley by the formula:\[ \alpha = \frac{a}{R_o} \]Where:
  • \(R_o\) is the outer radius of the pulley (0.50 m)
  • \(a\) is the linear acceleration of the stone
The moment of inertia \(I\) and the torque \(\tau\) produced by tension in the wire contribute to angular acceleration. Using:\[ \tau = T \cdot R_o = I \alpha \]It helps explain how force applied at a distance (like the tension in the wire) causes the pulley to spin. Pulley systems rely on angular acceleration to determine when and how quickly they start rotating. The balance between the moment of inertia, tension, and radius determines the angular speed, just as weight and tension influence linear speed.
Newton's Second Law
Newton's second law is fundamental in mechanics, linking force, mass, and acceleration. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. This can be expressed as:\[ F = ma \]Where:
  • \(F\) is the net force
  • \(m\) is mass
  • \(a\) is acceleration
For linear motion, this law is directly used in the stone's movement in the exercise, helping us find its acceleration when only tension and gravitational forces are at play. For rotational dynamics, a similar principle applies:\[ \tau = I \alpha \]Where:
  • \(\tau\) is the torque
  • \(I\) is the moment of inertia
  • \(\alpha\) is the angular acceleration
Both equations show how mass or inertia influences acceleration, either linearly or rotationally. By using these principles, we can solve complex motion problems like our pulley system, identifying how forces and masses interact both in straight lines and rotational movements.

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Most popular questions from this chapter

. A uniform 4.5 \(\mathrm{kg}\) square solid wooden gate 1.5 \(\mathrm{m}\) on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.1 \(\mathrm{kg}\) raven flying horizontally at 5.0 \(\mathrm{m} / \mathrm{s}\) flies into this gate at its center and bounces back at 2.0 \(\mathrm{m} / \mathrm{s}\) in the opposite direction. (a) What is the angular speed of the gate just after it is struck by the unfortunate raven? (b) During the collision, why is the angular momentum conserved, but not the linear momentum?

A certain type of propeller blade can be modeled as a thin uniform bar 2.50 m long and of mass 24.0 \(\mathrm{kg}\) that is free to rotate about a frictionless axle perpendicular to the bar at its midpoint. If a technician strikes this blade with a mallet 1.15 \(\mathrm{m}\) from the center with a 35.0 \(\mathrm{N}\) force perpendicular to the blade, find the maximum angular acceleration the blade could achieve.

A playground merry-go-round has a radius of 4.40 \(\mathrm{m}\) and a moment of inertia of 245 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) and turns with negligible friction about a vertical axle through its center. (a) A child applies a 25.0 \(\mathrm{N}\) force tangentially to the edge of the merry-go-round for 20.0 s. If the merry-go-round is initially at rest, what is its angular velocity after this 20.0 s interval? (b) How much work did the child do on the merry-go- round? (c) What is the average power supplied by the child?

\bullet A small block on a frictionless horizontal surface has a mass of 0.0250 \(\mathrm{kg}\) . It is attached to a massless cord passing through a hole in the surface. (See Figure \(10.54 .\) The block is originally revolving at a distance of 0.300 \(\mathrm{m}\) from the hole with an angular speed of 1.75 rad/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 \(\mathrm{m} .\) You may treat the block as a particle. (a) Is angular momentum conserved? Why or why not? (b) What is the new angular speed? (c) Find the change in kinetic energy of the block. (d) How much work was done in pulling the cord?

\(\cdot\) A woman with mass 50.0 \(\mathrm{kg}\) is standing on the rim of a large disk that is rotating at 0.50 \(\mathrm{rev} / \mathrm{s}\) about an axis perpendicular to it through its center. The disk has a mass of 110 \(\mathrm{kg}\) and a radius of 4.0 \(\mathrm{m}\) . Calculate the magnitude of the total angular momentum of the woman-plus-disk system, assuming that you can treat the woman as a point.

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