/*! 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 74 A block (mass \(=2.0 \mathrm{kg}... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 74

A block (mass \(=2.0 \mathrm{kg}\) ) is hanging from a massless cord that is wrapped around a pulley (moment of inertia \(=1.1 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2}\) ), as the drawing shows. Initially the pulley is prevented from rotating and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of \(0.040 \mathrm{m}\) during the block's descent. Find the angular acceleration of the pulley and the tension in the cord.

Short Answer

Expert verified
Angular acceleration is \( 9.66 \mathrm{\,rad/s^2} \); tension is \( 18.83 \mathrm{\,N} \).

Step by step solution

01

Identify Known Values

We have the mass of the block, \( m = 2.0 \mathrm{\,kg} \), the moment of inertia of the pulley, \( I = 1.1 \times 10^{-3} \mathrm{\,kg \cdot m^2} \), and the radius of the pulley, \( r = 0.040 \mathrm{\,m} \). We need to find the angular acceleration (\( \alpha \)) and the tension (\( T \)).
02

Apply Newton's Second Law for Translation

For the falling block, apply \( F = ma \), where the net force is equal to gravitational force minus the tension: \[ mg - T = ma \]Thus, \[ 2.0 \times 9.8 - T = 2.0a \]Which simplifies to \[ 19.6 - T = 2.0a \]
03

Relate Linear Acceleration and Angular Acceleration

Since the cord does not slip, the linear acceleration \( a \) of the block is related to the angular acceleration \( \alpha \) by \[ a = r\alpha \]
04

Apply Newton's Second Law for Rotation

Using the rotational form of Newton's second law on the pulley, \[ \tau = I\alpha \]where torque (\( \tau \)) is given by \( Tr \): \[ Tr = I\alpha \]Substitute \( T \) from earlier equation and \( a = r\alpha \).
05

Solve for Angular Acceleration

Substitute \( T = 19.6 - 2.0r\alpha \) in \( Tr = I\alpha \): \[ (19.6 - 2.0 \times 0.040\alpha) \times 0.040 = 1.1 \times 10^{-3} \alpha \]Solve for \( \alpha \):\[ 0.784 - 0.08\alpha = 1.1 \times 10^{-3} \alpha \]\[ 0.784 = 0.0811\alpha \]\[ \alpha \approx 9.66 \mathrm{\,rad/s^2} \]
06

Calculate Tension in the Cord

Substitute \( \alpha \) back into the equation for \( T \):\[ T = 19.6 - 2.0 \times 0.040 \times 9.66 \]\[ T = 19.6 - 0.7728 \]\[ T \approx 18.83 \mathrm{\,N} \]

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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 fundamental concept in rotational motion. It's often compared to mass in linear motion, as it represents how much an object resists changes to its rotational motion. For a pulley, the moment of inertia (\( I \)) determines how easily it can be set into motion by a torque. In our exercise, the moment of inertia of the pulley is given as \( 1.1 \times 10^{-3} \mathrm{kg \cdot m^2} \).
  • A smaller moment of inertia means the object will accelerate more easily.
  • The larger the moment of inertia, the harder it is to start rotating.
Understanding this concept is crucial, as it connects directly with the torque required to rotate the pulley and subsequently with the angular acceleration of the system. The moment of inertia is a key player in rotational dynamics, offering a parallel to the mass we use in linear systems.
Tension in the Cord
Tension is a force that is transmitted through a string, rope, cable, or cord when it is pulled tight by forces acting from opposite ends. In this problem, tension in the cord plays an integral role in facilitating rotational motion without slipping. It affects both the linear motion of the block and the rotation of the pulley.
  • Tension opposes the gravitational pull on the block, leading to net force calculation.
  • It also provides the necessary torque to rotate the pulley.
To find the tension, we consider the balance of forces on the block. As shown in the detailed solution, the force due to gravity (\( mg \)) minus the tension (\( T \)) dictates the linear acceleration of the block. This relationship is crucial in translating the block's motion into rotational motion around the pulley, as dictated through both Newton's second laws for translation and rotation.
Newton's Second Law
Newton's second law is a cornerstone of physics, relating the net force on an object to its mass and acceleration. In its rotational form, it applies to torques and angular accelerations. For our scenario, two forms of Newton's second law are applied.
  • Translational: \( F = ma \), where the net force on the block is gravity minus tension (\( mg - T = ma \)).
  • Rotational: \( \tau = I\alpha \), where torque (\( \tau \)) is caused by tension applied at the pulley's radius (\( Tr \)).
Both forms allow us to relate the block's linear properties to the pulley's rotational characteristics. The linear acceleration is linked to the angular acceleration by the pulley's radius, transforming force equations into ones involving torque and angular motion. Combining these laws paints a full picture of the dual motion occurring and bridges the gap between translational and rotational dynamics.

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Most popular questions from this chapter

A man drags a 72 -kg crate across the floor at a constant velocity by pulling on a strap attached to the bottom of the crate. The crate is tilted \(25^{\circ}\) above the horizontal, and the strap is inclined \(61^{\circ}\) above the horizontal. The center of gravity of the crate coincides with its geometrical center, as indicated in the drawing. Find the magnitude of the tension in the strap.

The wheel of a car has a radius of \(0.350 \mathrm{m}\). The engine of the car applies a torque of \(295 \mathrm{N} \cdot \mathrm{m}\) to this wheel, which does not slip against the road surface. since the wheel does not slip, the road must be applying a force of static friction to the wheel that produces a countertorque. Moreover, the car has a constant velocity, so this countertorque balances the applied torque. What is the magnitude of the static frictional force?

Multiple-Concept Example 10 offers useful background for problems like this. A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of \(0.0830 \mathrm{m}\), an angular speed of \(76.0 \mathrm{rad} / \mathrm{s},\) and a moment of inertia of \(0.615 \mathrm{kg} \cdot \mathrm{m}^{2} . \mathrm{A}\) brake shoe presses against the surface of the cylinder and applies a tangential frictional force to it. The frictional force reduces the angular speed of the cylinder by a factor of two during a time of 6.40 s. (a) Find the magnitude of the angular deceleration of the cylinder. (b) Find the magnitude of the force of friction applied by the brake shoe.

The drawing shows a rectangular piece of wood. The forces applied to corners \(\mathrm{B}\) and \(\mathrm{D}\) have the same magnitude of \(12 \mathrm{N}\) and are directed parallel to the long and short sides of the rectangle. The long side of the rectangle is twice as long as the short side. An axis of rotation is shown perpendicular to the plane of the rectangle at its center. A third force (not shown in the drawing) is applied to corner A, directed along the short side of the rectangle (either toward \(\mathrm{B}\) or away from \(\mathrm{B}\) ), such that the piece of wood is at equilibrium. Find the magnitude and direction of the force applied to corner A.

A 15.0 -m length of hose is wound around a reel, which is initially at rest. The moment of inertia of the reel is \(0.44 \mathrm{kg} \cdot \mathrm{m}^{2},\) and its radius is \(0.160 \mathrm{m} .\) When the reel is turning, friction at the axle exerts a torque of magnitude \(3.40 \mathrm{N} \cdot \mathrm{m}\) on the reel. If the hose is pulled so that the tension in it remains a constant \(25.0 \mathrm{N},\) how long does it take to completely unwind the hose from the reel? Neglect the mass and thickness of the hose on the reel, and assume that the hose unwinds without slipping.
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