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Chapter 10: Problem 27

A playground merry-go-round has radius 2.40 \(\mathrm{m}\) and moment of inertia 2100 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) about a vertical axle through its center, and it turns with negligible friction. (a) A child applies an 18.0 \(\mathrm{N}\) force tangentially to the edge of the merry-go-round for 15.0 \(\mathrm{s}\) . If the merry-go-round is initially at rest, what is its angular speed after this 15.0 -s interval? (b) How much work did the child do on the merry-go-round? (c) What is the average power supplied by the child?

Short Answer

Expert verified
(a) 0.3086 rad/s; (b) 100.05 J; (c) 6.67 W

Step by step solution

01

Calculate the torque applied

To find the torque \( \tau \) applied by the force, use the formula \( \tau = r \times F \), where \( r = 2.40 \, \mathrm{m} \) is the radius and \( F = 18.0 \, \mathrm{N} \) is the force applied. Substituting the values, we have:\[\tau = 2.40 \, \mathrm{m} \times 18.0 \, \mathrm{N} = 43.2 \, \mathrm{N} \cdot \mathrm{m}\]
02

Compute angular acceleration

The angular acceleration \( \alpha \) can be calculated using the relationship \( \tau = I \alpha \), where \( I = 2100 \, \mathrm{kg} \cdot \mathrm{m}^2 \) is the moment of inertia. Rearrange to find \( \alpha \):\[\alpha = \frac{\tau}{I} = \frac{43.2 \, \mathrm{N} \cdot \mathrm{m}}{2100 \, \mathrm{kg} \cdot \mathrm{m}^2} = 0.02057 \, \mathrm{rad/s^2}\]
03

Determine angular speed after 15 seconds

To find the angular speed \( \omega \) after 15 seconds, use the formula \( \omega = \omega_0 + \alpha t \). Since the merry-go-round starts from rest, \( \omega_0 = 0 \). Given \( \alpha = 0.02057 \, \mathrm{rad/s^2} \) and \( t = 15 \, \mathrm{s} \):\[\omega = 0 + 0.02057 \, \mathrm{rad/s^2} \times 15 \, \mathrm{s} = 0.3086 \, \mathrm{rad/s}\]
04

Calculate work done by the child

Work done \( W \) by the child can be found using \( W = \tau \theta \), where \( \theta = 0.5 \alpha t^2 \) is the angular displacement. Calculate \( \theta \):\[\theta = 0.5 \times 0.02057 \, \mathrm{rad/s^2} \times (15 \, \mathrm{s})^2 = 2.3156 \, \mathrm{rad}\]Then calculate \( W \):\[W = 43.2 \, \mathrm{N} \cdot \mathrm{m} \times 2.3156 \, \mathrm{rad} = 100.05 \, \mathrm{J}\]
05

Determine the average power

The average power \( P \) is the work done divided by the time interval, \( P = \frac{W}{t} = \frac{100.05 \, \mathrm{J}}{15 \, \mathrm{s}} \):\[P = 6.67 \, \mathrm{W}\]
06

Final Answer Summary

(a) Angular speed: \( 0.3086 \, \mathrm{rad/s} \)(b) Work done: \( 100.05 \, \mathrm{J} \)(c) Average power: \( 6.67 \, \mathrm{W} \)

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Key Concepts

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

Torque
Torque is the measure of the rotational force applied to an object causing it to rotate around an axis. This concept is similar to how force causes linear motion. When dealing with rotational motion, it's important to consider the distance from the axis of rotation to where the force is applied, also known as the lever arm. Torque is calculated by multiplying the force (F) by the radius (r).
  • Formula: \( \tau = r \times F \).
  • In our example, the child applies a force of 18.0 N at the edge of a merry-go-round with a radius of 2.40 m, calculating to \( \tau = 43.2 \, \mathrm{N} \cdot \mathrm{m} \).

This shows how applying a force at a greater distance increases the torque and, consequently, the rotational effect.
Applications of torque are everywhere, from opening a door to using a wrench; understanding torque can help optimize the effort needed in various tasks.
Moment of Inertia
Moment of inertia is a property of a body that determines its resistance to angular acceleration, similar to mass in linear motion. It's thought of as the "rotational mass" of an object, and its value depends on both the material distribution and the axis of rotation.
  • Formula: For a simple mass distribution, \( I = \sum m_i r_i^2 \) where \( m_i \) is a mass element and \( r_i \) is its distance from the axis.
  • In the merry-go-round problem, the moment of inertia given is 2100 kg鈰卪虏.

This value shows how much it resists the angular motion induced by the applied torque. Higher moments of inertia indicate greater resistance to changes in rotation.
Moments of inertia vary widely, with simple shapes having conventional formulas鈥攆or example, a thin hoop versus a solid disk.
Work and Power
Work and power are concepts in physics that describe how forces cause changes and energy is transferred over time. Work is the energy transferred by a force acting over a distance, and power is the rate at which this work is done.
  • Work (W) can be calculated in rotational systems using \( W = \tau \theta \), where 胃 represents the total angle turned in radians.
  • In our problem, after calculating 胃 as 2.3156 radians, the work done by the child is 100.05 J.

Power (P) is determined by dividing the work by the time interval, resulting in an average power in this case of 6.67 W over 15 seconds.
These calculations help us understand how effectively and efficiently a force performs its task, revealing how long and hard a person鈥攐r geared device鈥攏eeds to exert effort to maintain or change the rotational state of an object.
Angular Acceleration
Angular acceleration describes how the rotational speed of an object changes with time. It is the rate at which the angular velocity changes, and it can result from applying a torque to the object.
  • The formula to determine angular acceleration (\alpha) from torque (\tau) and moment of inertia (I) is \( \alpha = \frac{\tau}{I} \).
  • For the merry-go-round, inserting the calculated torque (43.2 \, \mathrm{N} \cdot \mathrm{m}) and given moment of inertia (2100 \, \mathrm{kg} \cdot \mathrm{m}^2), we find \( \alpha = 0.02057 \, \mathrm{rad/s}^2 \).

This shows how much the rotational velocity changes each second. Angular acceleration is key in determining the final rotational speed of an object when starting from rest or needing to reach a new speed.
Understanding angular acceleration is crucial in many fields, from engineering to everyday applications, whenever rotational motion is involved.

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Most popular questions from this chapter

The flywheel of an engine has moment of inertia 2.50 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 \(\mathrm{rev} / \mathrm{min}\) in 8.00 \(\mathrm{s}\) s, starting from rest?

A high-wheel antique bicycle has a large front wheel with the foot-powered crank mounted on its axle and a small rear wheel turning independently of the front wheel; there is no chain connecting the wheels. The radius of the front wheel is \(65.5 \mathrm{cm},\) and the radius of the rear wheel is 22.0 \(\mathrm{cm} .\) Your modern bike has awheel diameter of 66.0 \(\mathrm{cm}(26 \text { inches) and front and rear sprockets }\) with radii of 11.0 \(\mathrm{cm}\) and \(5.5 \mathrm{cm},\) respectively. The rear sprocket is rigidly attached to the axle of the rear wheel. You ride your modern bike and turn the front sprocket at 1.00 rev \(/ \mathrm{s}\) . The wheels of both bikes roll along the ground without slipping. (a) What is your linear speed when you ride your modern bike? (b) At what rate must you turn the crank of the antique bike in order to travel at the same speed as in part (a)? (c) What then is the angular speed (in rev/s) of the small rear wheel of the antique bike?

A demonstration gyroscope wheel is constructed by removing the tire from a bicycle wheel 0.650 \(\mathrm{m}\) in diameter, wrapping lead wire around the rim, and taping it in place. The shaft projects 0.200 \(\mathrm{m}\) at each side of the wheel, and a woman holds the ends of the shaft in her hands. The mass of the system is 8.00 \(\mathrm{kg}\) ; its entire mass may be assumed to be located at its rim. The shaft is horizontal, and the wheel is spinning about the shaft at 5.00 rev \(/ \mathrm{s}\) . Find the magnitude and direction of the force each hand exerts on the shaft (a) when the shaft is at rest; (b) when the shaft is rotating in a horizontal plane about its center at 0.050 rev \(/ \mathrm{s} ;\) (c) when the shaft is rotating in a horizontal plane about its center at 0.300 \(\mathrm{rev} / \mathrm{s} .\) (d) At what rate must the shaft rotate in order that it may be supported at one end only?

Neutron Star Glitches. Occasionally, a rotating neutron star (see Exercise 10.39 ) undergoes a sudden and unexpected speedup called a glitch. One explanation is that a ghitch occurs when the crust of the neutron star settles slightly, decreasing the moment of inertia about the rotation axis. A neutron star with angular speed \(\omega_{0}=70.4\) rad/s underwent such a glitch in October 1975 that increased its angular speed to \(\omega=\omega_{0}+\Delta \omega,\) where \(\Delta \omega / \omega_{0}=2.01 \times 10^{-6} .\) If the radius of the neutron star before the glitch was \(11 \mathrm{km},\) by how much did its radius decrease in the star-quake? Assume that the neutron star is a uniform sphere.

A \(2.00-\mathrm{kg}\) textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is \(0.150 \mathrm{m},\) to a hanging book with mass 3.00 \(\mathrm{kg}\) . The system is released from rest, and the books are observed to move 1.20 \(\mathrm{m}\) in 0.800 \(\mathrm{s}\) s. (a) What is the tension in each part of the cord? \((b)\) What is the moment of inertia of the pulley about its rotation axis?
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