/*! 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 12 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
Mass of stone: 2.0 kg; Tension: 14.0 N.

Step by step solution

01

Calculate the acceleration of the stone

The stone travels 12.6 m in 3.00 s from rest. Use the kinematic equation for distance: \[ \Delta s = v_0 t + \frac{1}{2} a t^2 \] where \( v_0 = 0 \). Therefore, \[ 12.6 = \frac{1}{2} a (3.00)^2. \] Solving for \( a \), we get \[ a = \frac{2 \times 12.6}{3.00^2} = 2.8 \text{ m/s}^2. \]
02

Use Newton's Second Law for the stone

Apply Newton's second law \( F = ma \) to find the mass of the stone. The forces acting on the stone are its weight \( mg \) and the tension \( T \) in the wire. The net force is thus: \[ mg - T = ma \] Where \( g = 9.8 \text{ m/s}^2 \).
03

Relate tension and torque on the pulley

The pulley experiences a torque due to the tension in the wire. For a disk, \[ \tau = I\alpha \] and the moment of inertia \( I \) for a disk is \( \frac{1}{2} m r^2 \). Since \( \tau = Tr \) and \( \alpha = \frac{a}{r} \):\[ Tr = \left( \frac{1}{2} M r^2 \right) \frac{a}{r} \] where \( M = 10.0 \text{ kg} \) is the mass of the pulley. This simplifies to \[ T = \frac{1}{2} M a \].
04

Solve for the mass of the stone

Substitute \( T = \frac{1}{2} M a \) back into the equation from Step 2: \[ mg - \frac{1}{2} M a = ma, \]\[ m(g - a) = \frac{1}{2} M a. \] Solve for \( m \):\[ m = \frac{\frac{1}{2} M a}{g - a} = \frac{\frac{1}{2} \times 10.0 \times 2.8}{9.8 - 2.8} = 2.0 \text{ kg}. \]
05

Calculate the tension in the wire

Use the equation from Step 3: \[ T = \frac{1}{2} M a = \frac{1}{2} \times 10.0 \times 2.8 = 14.0 \text{ N}. \]

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Key Concepts

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

Newton's Second Law
Understanding Newton's Second Law is fundamental in classical mechanics. This law states that the acceleration of an object depends on two things: the net force acting on that object and the object's mass. Mathematically, it can be expressed as
  • \[ F = ma \]
Here, \( F \) represents the net force exerted on the object, \( m \) is the mass, and \( a \) is the acceleration. The force and acceleration are vector quantities, meaning they have both magnitude and direction.
In our exercise, the stone is subjected to the gravitational force (its weight) and the tension in the wire. Therefore, the net force acting on the stone is the difference between the gravitational force \( mg \) and the tension \( T \). This is crucial for determining the mass of the stone:
  • \[ mg - T = ma \]
Thus, by rearranging this equation and incorporating the known acceleration of the stone, we can successfully solve for its mass. Remember: When objects accelerate, it's because there is an unbalanced force acting on them.
Kinematics
Kinematics is a branch of mechanics that describes the motion of objects without considering the causes of this motion. It focuses on parameters like displacement, speed, velocity, and acceleration. In our given problem, the stone's motion is analyzed using the kinematic equations.
One primary kinematic equation we use is:
  • \[ \Delta s = v_0 t + \frac{1}{2} a t^2 \]
Here, \( \Delta s \) is the displacement, \( v_0 \) is the initial velocity, \( t \) is the time, and \( a \) is the acceleration. As the stone starts from rest, \( v_0 = 0 \), simplifying our equation to:
  • \[ 12.6 = \frac{1}{2} a (3.00)^2 \]
Solving this gives us the stone's acceleration. Kinematics allows us to predict the future position and velocity of an object when its initial velocity, acceleration, and time duration of travel are known. It provides insights purely based on geometrical and time-based parameters of motion.
Torque and Rotational Motion
Torque plays a vital role in understanding rotational motion. It is similar to force in linear motion but applies to the rotational aspect. The torque \( \tau \) is calculated by
  • \[ \tau = Tr \]
where \( T \) is the force applied (tension in the wire) and \( r \) is the radius of the pulley.
Rotational motion is also characterized by angular acceleration \( \alpha \), which relates to the linear acceleration of the stone by the equation
  • \[ \alpha = \frac{a}{r} \]
The pulley's rotational inertia, \( I \), also influences its motion.
For a uniform disk like our pulley, the rotational inertia is given by
  • \[ I = \frac{1}{2} M r^2 \]
where \( M \) is the mass of the pulley. Using the relationship between torque and angular acceleration, \( \tau = I\alpha \), we can find the tension in the wire as
  • \[ T = \frac{1}{2} M a \]
This demonstrates how the linear and rotational properties of mechanics are intertwined, revealing the interconnected nature of motion.

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