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Chapter 5: Problem 43

A stone with mass \(0.80 \mathrm{~kg}\) is attached to one end of a string \(0.90 \mathrm{~m}\) long. The string will break if its tension exceeds \(60.0 \mathrm{~N}\). The stone is whirled in a horizontal circle on a frictionless tabletop; the other end of the string remains fixed. (a) Draw a free-body diagram of the stone. (b) Find the maximum speed the stone can attain without the string breaking.

Short Answer

Expert verified
The maximum speed the stone can attain without the string breaking is approximately \( v = \sqrt{67.5} \) or 8.2 m/s.

Step by step solution

01

Drawing Free Body Diagram

To draw the diagram: First, draw a circle representing the stone. With the center of the circle as the origin, draw a horizontal arrow pointing towards the origin, which will represent the tension in the string. Now label this force as 'T', there are no other forces acting in the horizontal direction.
02

Calculation of Maximum Speed

To calculate speed, the tension should not exceed 60.0N. Therefore, we'll use the formula \( T = \frac{mv^2}{r} \). The mass of the stone \( m \) is 0.80 kg, the tension \( T \) is 60.0 N and the radius \( r \) is 0.90 m. Now insert these values in the formula and solve for the speed \( v \). Therefore it can be written as \( v = \sqrt{\frac{T*r}{m}} \).
03

Calculation

Using \( v = \sqrt{\frac{T*r}{m}} \), substitute the provided values to get \( v = \sqrt{\frac{60.0N * 0.90m}{0.80kg}} \), which simplifies to \( v = \sqrt{67.5} \).

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Key Concepts

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

Tension
Tension is a force that is transmitted through a string, rope, or any object that can be stretched. It is a pulling force that acts along the length of the object, trying to return it to its original shape.
For example, when a stone tied to a string is whirled in a circle, the tension in the string acts towards the center of the circle. This force is what keeps the stone moving along the circular path.
  • Tension is always directed along the length of a string or rope and is a force of pull.
  • In circular motion, tension contributes to the centripetal force that keeps the object moving in a circle.
  • The tension force increases with speed and decreases with a longer radius of motion if the mass stays constant.

In this scenario, tension must not exceed the string's tensile strength鈥攈ere, up to 60 N鈥攐r the string will break.
Free-body diagram
A free-body diagram (FBD) is a simple sketch that shows all the forces acting on an object. This tool helps us understand and solve problems involving mechanics and dynamics.
To illustrate: draw the stone and represent all related forces.
In this exercise with the stone:
  • The diagram contains a circle (the stone) with an arrow pointing towards the center (the tension force).
  • The tension ( T ) is the central force acting on the stone.
  • The diagram is simple, showing a top view with no vertical forces, as the stone is on a horizontal plane.

This diagram assists in visualizing the net force at play and simplifies the calculation of elements like speed and tension.
Maximum speed
The maximum speed in circular motion refers to the fastest an object can travel along a circular path without breaking any constraints, such as the tensile strength of a string.
The formula for calculating speed takes into account the balance of forces: T = \( \frac{mv^2}{r} \).Rearranging for speed, it becomes:v = \( \sqrt{\frac{T \cdot r}{m}} \).
  • Here, \( T \) is the maximum tension the string can handle without breaking.
  • \( m \) is the mass of the stone.
  • \( r \) is the radius of the circle, which is the length of the string.

Inserting the given values from the problem:v = \( \sqrt{\frac{60.0 \times 0.90}{0.80}} \),which simplifies tov = \( \sqrt{67.5} \).This gives the stone's maximum speed of approximately 8.21 m/s. Always ensure you do not exceed this calculated speed to prevent the string from snapping.

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Most popular questions from this chapter

A box with mass \(m\) sits at the bottom of a long ramp that is sloped upward at an angle \(\alpha\) above the horizontal. You give the box a quick shove, and after it leaves your hands it is moving up the ramp with an initial speed \(v_{0}\). The box travels a distance \(d\) up the ramp and then slides back down. When it returns to its starting point, the speed of the box is half the speed it started with; it has speed \(v_{0} / 2 .\) What is the coefficient of kinetic friction between the box and the ramp? (Your answer should depend on only \(\alpha\).)

A \(2540 \mathrm{~kg}\) test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force such that its vertical velocity as a function of time is given by \(v(t)=A t+B t^{2}\), where \(A\) and \(B\) are constants and time is measured from the instant the fuel is ignited. The rocket has an upward acceleration of \(1.50 \mathrm{~m} / \mathrm{s}^{2}\) at the instant of ignition and, 1.00 s later, an upward velocity of \(2.00 \mathrm{~m} / \mathrm{s}\). (a) Determine \(A\) and \(B\), including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket's weight. (d) What was the initial thrust due to the fuel?

An \(8.00 \mathrm{~kg}\) block of ice, released from rest at the top of a 1.50-m-long friction less ramp, slides downhill, reaching a speed of \(2.50 \mathrm{~m} / \mathrm{s}\) at the bottom. (a) What is the angle between the ramp and the horizontal? (b) What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of \(10.0 \mathrm{~N}\) parallel to the surface of the ramp?

On the ride "Spindletop" at the amusement park Six Flags Over Texas, people stood against the inner wall of a hollow vertical cylinder with radius \(2.5 \mathrm{~m} .\) The cylinder started to rotate, and when it reached a constant rotation rate of \(0.60 \mathrm{rev} / \mathrm{s},\) the floor dropped about \(0.5 \mathrm{~m}\). The people remained pinned against the wall without touching the floor. (a) Draw a force diagram for a person on this ride after the floor has dropped. (b) What minimum coefficient of static friction was required for the person not to slide downward to the new position of the floor? (c) Does your answer in part (b) depend on the person's mass? (Note: When such a ride is over, the cylinder is slowly brought to rest. As it slows down, people slide down the wall to the floor.

Some sliding rocks approach the base of a hill with a speed of \(12 \mathrm{~m} / \mathrm{s} .\) The hill rises at \(36^{\circ}\) above the horizontal and has coefficients of kinetic friction and static friction of 0.45 and \(0.65,\) respectively, with these rocks. (a) Find the acceleration of the rocks as they slide up the hill. (b) Once a rock reaches its highest point, will it stay there or slide down the hill? If it stays, show why. If it slides, find its acceleration on the way down.
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