/*! 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 27 An air puck of mass \(m_{1}=0.25... [FREE SOLUTION] | 91Ó°ÊÓ

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An air puck of mass \(m_{1}=0.25 \mathrm{~kg}\) is tied to a string and allowed to revolve in a circle of radius \(R=1.0 \mathrm{~m}\) on a frictionless horizontal table. The other end of the string passes through a hole in the center of the table, and a mass of \(m_{2}=1.0 \mathrm{~kg}\) is tied to it (Fig. P7.27). The suspended mass remains in equilibrium while the puck on the tabletop revolves. (a) What is the tension in the string? (b) What is the horizontal force acting on the puck? (c) What is the speed of the puck? the horizontal force acting on the puck? (c) What is the the horizontal force acting on the puck? (c) What is the speed of the puck?

Short Answer

Expert verified
The tension in the string is 9.8 N, the horizontal force acting on the puck is 9.8 N, and the speed of the puck is 6.26 m/s.

Step by step solution

01

Understand the forces acting on the mass m1 and m2

Firstly, the mass \(m_2\) is stationary. The forces acting on it are the tension, T in the upward direction and the gravitational force \(m_2 * g\) in downward direction. Since the mass is stationary, these forces are balanced. So, \(T = m_2 * g\). Where \(g\) is the acceleration due to gravity. For the puck of mass \(m_1\), the forces acting on it will be the tension (T) directed towards the centre of the circle providing the necessary centripetal force for circular motion.
02

Find the tension in the string

We know from Newton’s Second Law that, \(T = m_2 * g\), where T is the tension in the string, \(m_2\) is the mass of the object at the other end of the string, and \(g\) is the acceleration due to gravity. So, \(T = 1.0 kg * 9.8 m/s^2 = 9.8 N\). So, the tension in the string is 9.8 N.
03

Find the horizontal force acting on the puck

The horizontal force acting on the puck in the circular path is the centripetal force which is provided by the tension in the string. So, the horizontal force is the same as the tension in the string. Therefore, the horizontal force acting on the puck is also 9.8 N.
04

Find the speed of the puck

The centripetal force acting on the puck of mass \(m_1\) moving with speed v is given by \(F_c = m_1 * v^2 / R\). This is equal to the tension in the rope. We can solve this for v to find that \(v=sqrt(T * R/m_1)\). Substituting in the values for the tension, radius and mass of the puck, we find \(v=sqrt(9.8 N * 1.0 m / 0.25 kg) = sqrt(39.2) = 6.26 m/s\).

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Key Concepts

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

Centripetal Force
In the realm of circular motion physics, centripetal force plays a central role, quite literally. This force is necessary for any object to move in a circular path. It acts towards the center of the circle, constantly pulling the object inward to maintain its circular motion. In our textbook problem, the air puck is revolved in a circle, and the force that keeps it from flying away is the centripetal force, provided here by the tension in the string.

Imagine tying a ball to a string and swirling it above your head. The ball doesn't fly off because the string’s tension acts as the centripetal force. Without this force, the puck would continue in a straight line due to inertia, as per Newton's first law of motion. The formula to find the centripetal force (\(F_c\)) is given by \( F_c = \frac{m_1 v^2}{R} \), where \(m_1\) is the mass of the puck, \(v\) is the velocity, and \(R\) is the radius of the circle.
Tension in Physics
Tension is a force that is transmitted through a string, rope, or wire when it is pulled tight by forces acting from opposite ends. It's a pulling force that always acts along the length of the object. In our exercise, the string connecting the air puck and the suspended mass creates tension. This tension has two important roles: it supports the weight of the hanging mass (\(m_2\)) and simultaneously provides the centripetal force needed for the air puck's circular motion.

Tension can vary in magnitude along the length of the material depending on various factors such as weight, friction, or other forces acting on it. However, in an idealized scenario (as in this problem), tension is considered uniform throughout. It is important to note that tension also follows Newton’s third law of motion, for every action, there is an equal and opposite reaction, indicating the mutual forces of action and reaction between two objects.
Newton's Second Law
Newton's second law is best described by the relationship \( F = ma \), where \( F \) is the net force applied to an object, \( m \) is the mass of the object, and \( a \) is the acceleration. This fundamental principle of physics tells us that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass.

In the context of the air puck problem, this law helps us to understand how the net force acting on the puck and the suspended mass determines their acceleration. For the suspended mass \( m_2 \) in the problem, Newton's second law explains why the tension \( T \) equals the gravitational force \( m_2 g \) when in equilibrium, since the acceleration is zero. For the puck moving in a circle, this law ties together the tension (acting as a centripetal force) and the resulting circular motion of the puck.
Gravitational Force
Gravitational force is a force of attraction that exists between all objects with mass. It is one of the four fundamental forces in the universe. On the surface of the Earth, this force causes objects to have weight and it is the force that keeps our planet in orbit around the Sun. The gravitational force that an object experiences near the Earth's surface is the product of its mass (\( m \)) and the acceleration due to gravity (\( g \), approximately \(9.8 m/s^2\) on Earth).

In our textbook problem, the gravitational force is what's pulling the suspended mass (\( m_2 \) of \(1.0 kg\)) downwards, which can be calculated by \( F = m_2 g \). This is exactly balanced by the tension in the string when the system is in equilibrium. Understanding gravitational force is key to understanding why objects fall at the same rate regardless of their mass, provided there is no air resistance, as the acceleration \( g \) is constant.

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Most popular questions from this chapter

A \(50.0\)-kg child stands at the rim of a merry-go-round of radius \(2.00 \mathrm{~m}\), rotating with an angular speed of \(3.00 \mathrm{rad} / \mathrm{s}\). (a) What is the child's centripetal acceleration? (b) What is the minimum force between her feet and the floor of the carousel that is required to keep her in the circular path? (c) What minimum coefficient of static friction is required? Is the answer you found reasonable? In other words, is she likely to stay on the merry-go-round?

A pail of water is rotated in a vertical circle of radius \(1.00 \mathrm{~m}\). (a) What two external forces act on the water in the pail? (b) Which of the two forces is most important in causing the water to move in a circle? (c) What is the pail's minimum speed at the top of the circle if no water is to spill out? (d) If the pail with the speed found in part (c) were to suddenly disappear at the top of the circle, describe the subsequent motion of the water. Would it differ from the motion of a projectile?

A car initially traveling at \(29.0 \mathrm{~m} / \mathrm{s}\) undergoes a constant negative acceleration of magnitude \(1.75 \mathrm{~m} / \mathrm{s}^{2}\) after its brakes are applied. (a) How many revolutions does each tire make before the car comes to a stop, assuming the car does not skid and the tires have radii of \(0.330 \mathrm{~m}\) ? (b) What is the angular speed of the wheels when the car has traveled half the total distance?

A \(0.400-\mathrm{kg}\) pendulum bob passes through the lowest part of its path at a speed of \(3.00 \mathrm{~m} / \mathrm{s}\). (a) What is the tension in the pendulum cable at this point if the pendulum is \(80.0 \mathrm{~cm}\) long? (b) When the pendulum reaches its highest point, what angle does the cable make with the vertical? (c) What is the tension in the pendulum cable when the pendulum reaches its highest point?

A digital audio compact disc carries data along a continuous spiral track from the inner circumference of the disc to the outside edge. Each bit occupies \(0.6 \mu \mathrm{m}\) of the track. A CD player turns the disc to carry the track counterclockwise above a lens at a constant speed of \(1.30 \mathrm{~m} / \mathrm{s}\). Find the required angular speed (a) at the beginning of the recording, where the spiral has a radius of \(2.30 \mathrm{~cm}\), and (b) at the end of the recording, where the spiral has a radius of \(5.80 \mathrm{~cm}\). (c) A full-length recording lasts for \(74 \mathrm{~min}, 33 \mathrm{~s}\). Find the average angular acceleration of the disc. (d) Assuming the acceleration is constant, find the total angular displacement of the disc as it plays. (e) Find the total length of the track.

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