/*! 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 14 A stone is suspended from the fr... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(\mathrm{kg}\) and radius 50.0 \(\mathrm{cm}\) and turns on frictionless bearings. You measure that the stone travels 12.6 \(\mathrm{m}\) in the first 3.00 s starting from rest. Find (a) the mass of the stone and (b) the tension in the wire.

Short Answer

Expert verified
The stone's mass is 2 kg and the tension in the wire is 14 N.

Step by step solution

01

Understand the Problem

We have a disk-shaped pulley without friction and a stone attached to a wire wrapped around the pulley. The stone starts from rest and travels 12.6 m in 3 seconds. We need to find the mass of the stone and the tension in the wire.
02

Set Up Equations of Motion

Since the stone travels 12.6 m in 3 seconds starting from rest, we use the kinematic equation: \[ s = ut + \frac{1}{2}at^2 \] where \( s = 12.6 \) m, \( u = 0 \) \( m/s \) (initial velocity), and \( t = 3 \) s. Solving for the acceleration \( a \), we have:\[ 12.6 = 0 + \frac{1}{2}a(3)^2 \] \[ 12.6 = \frac{9}{2}a \] \[ a = \frac{12.6 \times 2}{9} = 2.8 \ \mathrm{m/s^2} \]
03

Relate Forces and Torque

For the stone, the force equation is: \[ mg - T = ma \]For the pulley (rotating disk), torque \( \tau \) is related to tension \( T \) by:\[ \tau = I \alpha \] Since \( \alpha \) (angular acceleration) is related to linear acceleration by \( a = \alpha R \), then:\[ I = \frac{1}{2}MR^2 = \frac{1}{2}(10)(0.5)^2 = 1.25 \ \mathrm{kg \, m^2} \]\[ T \cdot R = I \alpha = I \left( \frac{a}{R} \right) = 1.25 \times \frac{2.8}{0.5} \]
04

Solve for Tension and Mass

From torque equation, we have:\[ T \cdot 0.5 = 1.25 \times 5.6 = 7 \] \[ T = \frac{7}{0.5} = 14 \ \mathrm{N} \] Substituting back to the force equation:\[ mg - 14 = ma \] \[ m \times 9.8 - 14 = m \times 2.8 \] \[ m \times (9.8 - 2.8) = 14 \] \[ m \times 7 = 14 \] \[ m = \frac{14}{7} = 2 \ \mathrm{kg} \]
05

Conclusion

The mass of the stone is 2 kg and the tension in the wire is 14 N. We found these by using the linear motion equations, force equilibrium, and rotational dynamics principles.

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Key Concepts

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

Kinematic Equations
Kinematic equations are a set of equations that describe the motion of objects under constant acceleration. These equations allow us to calculate various parameters, such as distance, velocity, and time. In this exercise, since the stone starts from rest and travels a distance in a given time, we can use the kinematic equation:
  • \( s = ut + \frac{1}{2}at^2 \)
Here, \( s \) is the distance traveled, \( u \) is the initial velocity (which is 0 in this case), \( a \) is the acceleration, and \( t \) is the time.
By plugging in the values given in the problem—distance \( s = 12.6 \, \mathrm{m} \) and time \( t = 3 \, \mathrm{s} \)—we can solve for the linear acceleration \( a \).
These equations provide a way to understand motion and are crucial when objects are moving in a straight line with a uniform acceleration.
Torque
Torque is pivotal when it comes to rotational motion, similar to how force applies to linear motion. Torque is defined as the rotational equivalent of a force that causes an object to rotate around an axis. It is calculated as:
  • \( \tau = r \times F \)
where \( r \) is the distance from the pivot point to the point where the force is applied, and \( F \) is the force applied.
In the problem, the tension in the wire creates a torque on the pulley, forcing it to rotate. This is expressed as:
  • \( T \cdot R = I \alpha \)
where \( T \) is the tension in the wire, \( R \) is the pulley's radius, and \( \alpha \) is the angular acceleration. It’s essential to understand torque to analyze and predict the rotational dynamics of systems effectively.
Angular Acceleration
Angular acceleration is the rate of change of angular velocity over time. Just as linear acceleration measures changes in speed in a straight line, angular acceleration does the same for circular motion. It is related to the linear acceleration that we calculated from the kinematic equations by the formula:
  • \( a = \alpha R \)
where \( \alpha \) is the angular acceleration and \( R \) is the radius of the circular path.
For the pulley, the angular acceleration can be determined using:
  • \( \alpha = \frac{a}{R} \)
This relationship helps understand how forces applied tangentially on a rotating body affect its rotational speed, providing insight into how rotational and linear dynamics are interconnected.
Frictionless Pulley
A frictionless pulley is an idealized concept in physics, used to simplify calculations by assuming there is no friction resisting the rotation of the pulley. In real-life scenarios, pulleys do have friction, but by considering an ideal frictionless scenario, the focus can be placed solely on the forces and torques without additional complexities.
In this exercise, the frictionless pulley ensures that the only forces at play are the force of gravity on the stone and the tension in the wire. This simplification allows us to set up the equations more straightforwardly and directly relate the tension in the wire to the weight of the stone and its acceleration.
Using a frictionless pulley model helps students understand the core principles of rotational dynamics without being sidetracked by the influences of friction.

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Most popular questions from this chapter

The rotor (flywheel) of a toy gyroscope has mass 0.140 kg. Its moment of inertia about its axis is \(1.20 \times 10^{-4} \mathrm{kg} \cdot \mathrm{m}^{2} .\) The mass of the frame is 0.0250 \(\mathrm{kg}\) . The gyroscope is supported on a single pivot (Fig. E10.53) with its center of mass a horizontal distance of 4.00 \(\mathrm{cm}\) from the pivot. The gyroscope is precessing in a horizontal plane at the rate of one revolution in 2.20 \(\mathrm{s}\) . (a) Find the upward force exerted by the pivot. (b) Find the angular speed with which the rotor is spinning about its axis, expressed in rev/min. (c) Copy the diagram and draw vectors to show the angular momentum of the rotor and the torque acting on it.

A large 16.0 -kg roll of paper with radius \(R=18.0 \mathrm{cm}\) rests against the wall and is held in place by a bracket attached to a rod through the center of the roll (Fig. P10.69). The rod turns without friction in the bracket, and the moment of inertia of the paper and rod about the axis is 0.260 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) The other end of the bracket is attached by a frictionless hinge to the wall such that the bracket makes an angle of \(30.0^{\circ}\) with the wall. The weight of the bracket is negligible. The coefficient of kinetic friction between the paper and the wall is \(\mu_{\mathrm{k}}=0.25 .\) A constant vertical force \(F=60.0 \mathrm{N}\) is applied to the paper, and the paper unrolls. (a) What is the magnitude of the force that the rod exerts on the paper as it unrolls? (b) What is the magnitude of the angular acceleration of the roll?

A uniform solid cylinder with mass \(M\) and radius 2\(R\) rests on a horizontal tabletop. A string is attached by a yoke to a frictionless axle through the center of the cylinder so that the cylinder can rotate about the axle. The string runs over a disk-shaped pulley with mass \(M\) and radius \(R\) that is mounted on a frictionless axle through its center. A block of mass \(M\) is suspended from the free end of the string (Fig. P10.87). The string doesn't slip over the pulley surface, and the cylinder rolls without slipping on the tabletop. Find the magnitude of the acceleration of the block after the system is released from rest.

The Yo-yo. A yo-yo is made from two uniform disks, each with mass \(m\) and radius \(R\) , connected by a light axle of radius \(b .\) A light, thin string is wound several times around the axle and then held stationary while the yo-yo is released from rest, dropping as the string unwinds. Find the linear acceleration and angular acceleration of the yo-yo and the tension in the string.

In a lab experiment you let a uniform ball roll down a curved track. The ball starts from rest and rolls without slipping.While on the track, the ball descends a vertical distance \(h .\) The lower end of the track is horizontal and extends over the edge of the lab table; the ball leaves the track traveling horizontally. While free-falling after leaving the track, the ball moves a horizontal distance \(x\) and a vertical distance \(y\) . (a) Calculate \(x\) in terms of \(h\) and \(y\) ignoring the work done by friction. (b) Would the answer to part (a) be any different on the moon? (c) Although you do the experiment very carefully, your measured value of \(x\) is consistently a bit smaller than the value calculated in part (a). Why? (d) What would \(x\) be for the same \(h\) and \(y\) as in part (a) if you let a silver dollar roll down the track? You can ignore the work done by friction.

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