/*! 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 24 A hemispherical bowl of radius \... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 24

A hemispherical bowl of radius \(R\) is rotated about its axis of symmetry which is kept vertical. A small block is kept in the bowl at a position where the radius makes an angle \(\theta\) with the vertical. The block rotates with the bowl without any slipping. The friction coefficient between the block and the bowl surface is \(\mu .\) Find the range of the angular speed for which the block will not slip.

Short Answer

Expert verified
The block will not slip if the angular speed is \( 0 \leq \omega \leq \sqrt{\frac{\mu g}{R}} \) and \( \tan \theta \leq \mu \).

Step by step solution

01

Identify Forces on the Block

First, we need to identify the forces acting on the block. The block experiences a gravitational force downward, a normal force perpendicular to the surface of the bowl, and a frictional force that prevents slipping. Additionally, since the bowl is rotating, there is a centripetal force that acts radially inward.
02

Set Up Equations of Motion

Consider the balance of forces. The gravitational force has a component parallel to the surface as \( mg \sin \theta \) and a component perpendicular to it as \( mg \cos \theta \). The centripetal force required to keep the block moving in a circle is \( mR \omega^2 \cos \theta \). The frictional force \( f \) can provide the necessary centripetal force.
03

Use Frictional Force Equation

The maximum frictional force is given by \( f_{\text{max}} = \mu N \), where \( N = mg \cos \theta \) is the normal force. So, \( f_{\text{max}} = \mu mg \cos \theta \). This frictional force must be equal to or greater than the required centripetal force \( mR \omega^2 \cos \theta \), which gives us: \( \mu mg \cos \theta \geq mR \omega^2 \cos \theta \).
04

Simplify the Inequality

Divide both sides of the inequality by \( m \cos \theta \), assuming \( \cos \theta eq 0 \). This results in \( \mu g \geq R \omega^2 \). Therefore, the maximum angular speed is \( \omega = \sqrt{\frac{\mu g}{R}} \).
05

Consider Minimum Angular Speed

For the block not to slip up the bowl, the component of gravitational force down the slope \( mg \sin \theta \) must be less than the friction force. Using a similar approach from earlier: \( mg \sin \theta \leq \mu mg \cos \theta \). Simplifying: \( \tan \theta \leq \mu \). This condition must also be satisfied for the entire range of \( \omega \).
06

Conclusion on Range of Angular Speed

The block will not slip if the angular speed satisfies \( 0 \leq \omega \leq \sqrt{\frac{\mu g}{R}} \), given that \( \tan \theta \leq \mu \). The lower bound \( 0 \) is the speed at which the block is at rest, and the upper bound is determined by the frictional force condition.

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Key Concepts

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

Centripetal Force
Centripetal force is a crucial aspect when dealing with rotational motion, especially in systems where objects move in a circular path. This force is responsible for keeping an object moving along a curved trajectory instead of flying off in a straight line. It is always directed towards the center of the circular path.

In the exercise context, the small block inside the hemispherical bowl experiences centripetal force due to the bowl's rotation. The formula to determine the amount of centripetal force required to maintain the block's circular motion is given by \( F_c = mR \omega^2 \cos \theta \). Here, \( m \) is the block's mass, \( R \) is the radius of the bowl, and \( \omega \) represents the angular speed.

  • Direction: Always points towards the center of rotation.
  • Origin: It results from the requirement to change the direction of the block's velocity as it moves in a circle.
  • Relation to motion: Centrally important for calculating the block鈥檚 required speed to stay in motion without slipping.
Frictional Force
Frictional force plays an essential role in preventing the block from slipping inside the rotating bowl. Without adequate friction, the block would slide out due to the centrifugal effect, which is an apparent force felt in a rotating system.

The frictional force provides the necessary grip between the block and the bowl's surface. The maximum possible frictional force can be calculated using the formula \( f_{\text{max}} = \mu N \), where \( \mu \) is the coefficient of friction and \( N = mg \cos \theta \) is the normal force.

  • Grip: Friction allows the block to move with the bowl without slipping off.
  • Maximum value: Determined by the coefficient of friction and the normal force.
  • Crucial balance: The frictional force must exceed the gravitational component trying to move the block down the slope, \( mg \sin \theta \).
Without sufficient friction, anything rotating would slip and potentially fall off the path of rotation.
Angular Speed
Angular speed is a measure of how fast something is rotating. It is a key factor in rotational systems, influencing every aspect of circular motion. In the given exercise, finding the range of angular speeds at which the block does not slip is critical.

The angular speed is represented by the symbol \( \omega \) and impacts the centripetal force required to keep the block in circular motion. The solution presents that the angular speed must satisfy an inequality to prevent slipping: \( 0 \leq \omega \leq \sqrt{\frac{\mu g}{R}} \), given \( \tan \theta \leq \mu \).

  • Relation to force: Faster rotations require more centripetal force.
  • Maximum limit: Defined to ensure the frictional force can keep the block in place.
  • Listens to friction: Its limits consider the material and surface properties expressed by the coefficient of friction \( \mu \).
Angular speed directly dictates the motion of the object and must be tuned correctly to maintain stable and stay within the slip-preventive limits.

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Most popular questions from this chapter

A steel wire fixed at both ends has a fundamental frequency of \(200 \mathrm{~Hz}\). A person can hear sound of maximum frequency \(14 \mathrm{kHz}\). What is the highest harmonic that can be played on this string which is audible to the person?

A \(2 \mathrm{~m}\) -long string fixed at both ends is set into vibrations in its first overtone. The wave speed on the string is \(200 \mathrm{~m} \mathrm{~s}^{-1}\) and the amplitude is \(0 \cdot 5 \mathrm{~cm}\). (a) Find the wavelength and the frequency. (b) Write the equation giving the displacement of different points as a function of time. Choose the \(X\) -axis along the string with the origin at one end and \(t=0\) at the instant when the point \(x=50 \mathrm{~cm}\) has reached its maximum displacement.

Two waves, each having a frequency of \(100 \mathrm{~Hz}\) and a wavelength of \(2 \cdot 0 \mathrm{~cm}\), are travelling in the same direction on a string. What is the phase difference between the waves (a) if the second wave was produced \(0-015\) s later than the first one at the same place, \((b)\) if the two waves were produced at the same instant but the first one was produced a distance \(4 \cdot 0 \mathrm{~cm}\) behind the second one? (c) If each of the waves has an amplitude of \(2 \cdot 0 \mathrm{~mm}\), what would be the amplitudes of the resultant waves in part (a) and (b)?

In a children's park a heavy rod is pivoted at the centre and is made to rotate about the pivot so that the rod always remains horizontal. Two kids hold the rod near the ends and thus rotate with the rod (figure 7-E2). Let the mass of each kid be \(15 \mathrm{~kg}\), the distance between the points of the rod where the two kids hold it be \(3 \cdot 0 \mathrm{~m}\) and suppose that the rod rotates at the rate of 20 revolutions per minute. Find the force of friction exerted by the rod on one of the kids.

A circular road of radius \(50 \mathrm{~m}\) has the angle of banking equal to \(30^{\circ}\). At what speed should a vehicle go on this road so that the friction is not used?
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