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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

BIO Force During a Jump. When jumping straight up from a crouched position, an average person can reach a maximum height of about \(60 \mathrm{~cm} .\) During the jump, the person's body from the knees up typically rises a distance of around \(50 \mathrm{~cm} .\) To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump. (a) With what initial speed does the person leave the ground to reach a height of \(60 \mathrm{~cm} ?\) (b) Draw a free-body diagram of the person during the jump. (c) In terms of this jumper's weight \(w\), what force does the ground exert on him or her during the jump?

A horizontal wire holds a solid uniform ball of mass \(m\) in place on a tilted ramp that rises \(35.0^{\circ}\) above the horizontal. The surface of this ramp is perfectly smooth, and the wire is directed away from the center of the ball (Fig. P5.64). (a) Draw a free-body diagram of the ball. (b) How hard does the surface of the ramp push on the ball? (c) What is the tension in the wire?

A stockroom worker pushes a box with mass \(16.8 \mathrm{~kg}\) on a horizontal surface with a constant speed of \(3.50 \mathrm{~m} / \mathrm{s}\). The coefficient of kinetic friction between the box and the surface is \(0.20 .\) (a) What horizontal force must the worker apply to maintain the motion? (b) If the force calculated in part (a) is removed, how far does the box slide before coming to rest?

Two crates connected by a rope lie on a horizontal surface (Fig. E5.37). Crate \(A\) has mass \(m_{A}\), and crate \(B\) has mass \(m_{B}\). The coefficient of kinetic friction between each crate and the surface is \(\mu_{\mathrm{k}} .\) The crates are pulled to the right at constant velocity by a horizontal force \(\overrightarrow{\boldsymbol{F}}\). Draw one or more free-body diagrams to calculate the following in terms of \(m_{A}, m_{B},\) and \(\mu_{\mathrm{k}}:\) (a) the magnitude of \(\overrightarrow{\boldsymbol{F}}\) and \((\mathrm{b})\) the tension in the rope connecting the blocks.

The cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of \(100 \mathrm{~m}\). Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every \(60.0 \mathrm{~s}\) ). (a) Find the speed of the passengers when the Ferris wheel is rotating at this rate. (b) A passenger weighs \(882 \mathrm{~N}\) at the weight-guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel? (c) What would be the time for one revolution if the passenger's apparent weight at the highest point were zero? (d) What then would be the passenger's apparent weight at the lowest point?
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