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Chapter 6: Problem 44

A small lead ball, attached to a 0.75-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled at a constant rate of three revolutions per second and is released on the upward part of the circular motion when it is \(1.5 \mathrm{m}\) above the ground. The ball travels straight upward. In the absence of air resistance, to what maximum height above the ground does the ball rise?

Short Answer

Expert verified
The ball rises to approximately 35.97 meters above the ground.

Step by step solution

01

Calculate Linear Speed

First, we convert the angular velocity into linear speed. The linear speed \( v \) of the ball is given by the formula \( v = 2\pi r f \), where \( r \) is the radius (0.75 m) and \( f \) is the frequency (3 revolutions per second). Thus, \( v = 2\pi \times 0.75 \times 3 = 4.5\pi \) m/s.
02

Use Kinetic Energy at Release

The kinetic energy at the point of release is \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass of the ball and \( v \) is the speed calculated in Step 1. This energy will be converted into gravitational potential energy as the ball rises.
03

Calculate Potential Energy Increase

The potential energy at the maximum height \( h_{max} \) is given by \( PE = mgh_{max} \), where \( g = 9.8 \) m/s虏. Set the kinetic energy equal to the potential energy, so \( \frac{1}{2} mv^2 = mg(h_{max} - 1.5) \).
04

Solve for Maximum Height

Cancel mass \( m \) from both sides and rearrange the equation to find \( h_{max} \):\[\frac{1}{2} v^2 = g(h_{max} - 1.5)\]Substitute \( v = 4.5\pi \) and \( g = 9.8 \):\[\frac{1}{2} (4.5\pi)^2 = 9.8(h_{max} - 1.5)\]Solve for \( h_{max} \):\[h_{max} = \frac{\frac{1}{2} (4.5\pi)^2}{9.8} + 1.5\]\[h_{max} \approx 35.97 \text{ m}\]
05

Calculate Total Height

Add the initial height from which the ball is released (1.5 m) to the calculated rise \( h_{rise} \) to get \( h_{max} = h_{rise} + 1.5 \). The maximum height above the ground reached by the ball is \( 35.97 \text{ m} \).

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Key Concepts

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

Kinetic Energy
Kinetic energy is a form of energy that an object possesses due to its motion. It is one of the primary concepts in physics that helps to describe how objects move. In the context of projectile motion, such as a ball being whirled in a circle, kinetic energy plays a crucial role in determining how high the ball will go once released.
The kinetic energy (\( KE \)) can be calculated using the formula:
  • \( KE = \frac{1}{2} mv^2 \)
where \( m \) is the mass of the object and \( v \) is its velocity. As the ball is released, its kinetic energy starts transforming into potential energy.
In our given problem, the ball had a calculated speed of \( 4.5\pi \) meters per second. With this speed, it carries enough energy to convert in the upward movement, following the law of conservation of energy. This is why understanding kinetic energy is essential here, as it helps explain the conversion of movement into height. Once the ball's movement is stopped at the top of its arc, all kinetic energy is converted into potential energy.
Potential Energy
Potential energy is the energy stored in an object due to its position or condition. In projectile motion, potential energy is often described in terms of height above the ground, which means it involves gravitational potential energy.
The potential energy (\( PE \)) in this context can be measured by the formula:
  • \( PE = mgh \)
where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (9.8 m/s虏 in this case), and \( h \) is the height above the ground.
As the whirled ball is released, its kinetic energy starts converting into potential energy as it climbs higher. The exercise beautifully illustrates how these two energies balance each other. In the problem, when the ball reaches its highest point, all the initial kinetic energy is converted into potential energy. This maximum distance from the ground can be calculated by equating the kinetic energy at release to the potential energy at the peak. By incorporating both types of energy, we observe the logical interplay of energy conversion that drives the motion of the ball.
Angular Velocity
Angular velocity refers to how fast an object rotates or revolves relative to another point, typically the center of a circle. This concept is crucial in understanding circular motion, such as that of a ball being spun on a rope.
Angular velocity is denoted by \( \omega \) and measured in radians per second. However, it can also be expressed using the frequency of revolutions, which was three revolutions per second in this case. In linear terms:
  • \( \omega = 2\pi f \)
where \( f \) represents the frequency of rotation.
To convert from angular velocity to linear velocity, we employ the relationship \( v = \omega r \), where \( r \) is the radius of the circle in which the ball is rotating. In the original problem, this conversion was crucial for applying the kinetic energy formula later on. Understanding angular velocity allows us to bridge rotational motion with linear attributes, making it a cornerstone in tackling physics problems involving circular paths.

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Most popular questions from this chapter

A 47.0-g golf ball is driven from the tee with an initial speed of \(52.0 \mathrm{m} / \mathrm{s}\) and rises to a height of \(24.6 \mathrm{m}\). (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is \(8.0 \mathrm{m}\) below its highest point?

The (nonconservative) force propelling a \(1.50 \times 10^{3}-\mathrm{kg}\) car up a mountain road does \(4.70 \times 10^{6} \mathrm{J}\) of work on the car. The car starts from rest at sea level and has a speed of \(27.0 \mathrm{m} / \mathrm{s}\) at an altitude of \(2.00 \times 10^{2} \mathrm{m}\) above sea level. Obtain the work done on the car by the combined forces of friction and air resistance, both of which are nonconservative forces.

A Sledding Contest. You are in a sledding contest where you start at a height of \(40.0 \mathrm{m}\) above the bottom of a valley and slide down a hill that makes an angle of \(25.0^{\circ}\) with respect to the horizontal. When you reach the valley, you immediately climb a second hill that makes an angle of \(15.0^{\circ}\) with respect to the horizontal. The winner of the contest will be the contestant who travels the greatest distance up the second hill. You must now choose between using your flat-bottomed plastic sled, or your "Blade Runner," which glides on two steel rails. The hill you will ride down is covered with loose snow. However, the hill you will climb on the other side is a popular sledding hill, and is packed hard and is slick. The two sleds perform very differently on the two surfaces, the plastic one performing better on loose snow, and the Blade Runner doing better on hard-packed snow or ice. The performances of each sled can be quantified in terms of their respective coefficients of kinetic friction on the two surfaces. For the plastic sled: \(\mu=0.17\) on loose snow, and \(\mu=0.15\) on packed snow or ice. For the Blade Runner, \(\mu=0.19\) on loose snow, and \(\mu=0.07\) on packed snow or ice. Assuming the two hills are shaped like inclined planes, and neglecting air resistance, (a) how far does each sled make it up the second hill before stopping? (b) Assuming the total mass of the sled plus rider is \(55.0 \mathrm{kg}\) in both cases, how much work is done by nonconservative forces (over the total trip) in each case?

The hammer throw is a track-and-field event in which a \(7.3 \mathrm{kg}\) ball (the "hammer"), starting from rest, is whirled around in a circle several times and released. It then moves upward on the familiar curving path of projectile motion. In one throw, the hammer is given a speed of \(29 \mathrm{m} / \mathrm{s} .\) For comparison, a .22 caliber bullet has a mass of \(2.6 \mathrm{g}\) and, starting from rest, exits the barrel of a gun at a speed of \(410 \mathrm{m} / \mathrm{s}\). Determine the work done to launch the motion of (a) the hammer and (b) the bullet.

Go A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates. A 70.3 -kg water- skier has an initial speed of \(6.10 \mathrm{m} / \mathrm{s}\). Later, the speed increases to \(11.3 \mathrm{m} / \mathrm{s} .\) Determine the work done by the net external force acting on the skier.
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