/*! 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 33 A potter is shaping a bowl on a ... [FREE SOLUTION] | 91Ó°ÊÓ

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A potter is shaping a bowl on a potter's wheel rotating at constant angular speed (Fig. \(10-51\) ). The friction force between her hands and the clay is \(1.5 \mathrm{~N}\) total. \((a)\) How large is her torque on the wheel, if the diameter of the bowl is \(12 \mathrm{~cm} ?\) (b) How long would it take for the potter's wheel to stop if the only torque acting on it is due to the potter's hand? The initial angular velocity of the wheel is \(1.6 \mathrm{rev} / \mathrm{s},\) and the moment of inertia of the wheel and the bowl is \(0.11 \mathrm{~kg} \cdot \mathrm{m}^{2}\)

Short Answer

Expert verified
The torque is 0.09 Nâ‹…m, and it takes approximately 12.3 seconds for the wheel to stop.

Step by step solution

01

Convert Diameter to Radius

The diameter of the bowl is given as 12 cm. To find the radius, divide the diameter by 2. Thus, the radius is \( r = \frac{12}{2} = 6 \) cm, or \( 0.06 \) m.
02

Calculate Torque

Torque \( \tau \) can be calculated using the formula \( \tau = F \times r \), where \( F \) is the force and \( r \) is the radius. Here, \( F = 1.5 \) N and \( r = 0.06 \) m. Hence, \( \tau = 1.5 \times 0.06 = 0.09 \) Nâ‹…m.
03

Convert Angular Velocity to Radians Per Second

The initial angular velocity is given in revolutions per second, \( 1.6 \) rev/s. Convert this to radians per second using the conversion factor \( 2\pi \) radians for one revolution: \( \omega_0 = 1.6 \times 2\pi = 3.2\pi \) rad/s.
04

Calculate Angular Deceleration

The angular deceleration \( \alpha \) can be found using the formula \( \alpha = \frac{\tau}{I} \), where \( I \) is the moment of inertia. Here, \( \tau = 0.09 \) N⋅m and \( I = 0.11 \) kg⋅m². Thus, \( \alpha = \frac{0.09}{0.11} = \frac{90}{110} = 0.818 \) rad/s².
05

Calculate Time to Stop

The time \( t \) it takes for the wheel to stop can be determined using the equation \( \omega = \omega_0 - \alpha t \), where the final angular velocity \( \omega \) is 0. Solving for \( t \), we get \( t = \frac{\omega_0}{\alpha} = \frac{3.2\pi}{0.818} \). Calculate \( t \) to find the time it takes for the wheel to stop.

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Key Concepts

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

Torque Calculation
Torque is an essential concept in rotational dynamics. It describes how much a force acting on an object causes that object to rotate. The principle behind torque can be visualized by thinking of opening a door; the push or pull exerted far from the hinge causes the door to rotate more easily.
In mathematical terms, torque (\( \tau \)) is calculated using the formula:
  • \[ \tau = F \times r \]
where:
  • \( F \) is the force applied, measured in newtons (N), and
  • \( r \) is the lever arm or the perpendicular distance from the axis of rotation to the line of action of the force, measured in meters (m).

In the potter's wheel example, the force of friction exerted by the potter's hands is 1.5 N, and the lever arm is the radius of the bowl, 0.06 m. Substituting these values into the torque formula gives us \( \tau = 1.5 \times 0.06 = 0.09 \) Nâ‹…m. This means the potter applies a torque of 0.09 Nâ‹…m on the wheel.
Angular Velocity
Angular velocity refers to how fast an object rotates or revolves relative to another point. It is usually measured in radians per second (rad/s) but can also be expressed in revolutions per second (rev/s).
To convert angular velocity from revolutions per second to radians per second:
  • Use the conversion factor that one revolution equals \(2\pi\) radians.

For example, if the potter's wheel starts spinning with an angular velocity of 1.6 rev/s, we convert this to rad/s by multiplying by \(2\pi\):
\[ \omega_0 = 1.6 \times 2\pi = 3.2\pi \] rad/s.
This conversion is crucial because most mathematical physics equations use radians for angular measurements.
Moment of Inertia
Moment of inertia is an object's resistance to changes in its rotation. Imagine trying to spin a bicycle wheel versus a heavier tire. The heavier tire has a larger moment of inertia, making it harder to start or stop its rotation.
In physics, moment of inertia (\( I \)) depends on the mass distribution relating to the axis of rotation. Gathering a detailed understanding involves the formula:
  • \[ I \] is often provided for standard shapes and can be calculated for custom shapes
  • It is measured in kgâ‹…m²

In the potter's simulation, the wheel and the clay have a combined moment of inertia of 0.11 kg⋅m². This value helps us find how much torque is needed to alter the wheel's movement to increase or decrease its rotating speed.
Angular Deceleration
Angular deceleration is the rate at which an object slows down its rotational speed. It’s fundamentally crucial when considering stopping scenarios, like the potter's wheel reaching rest.
To calculate angular deceleration (\( \alpha \)), use the formula:
  • \[ \alpha = \frac{\tau}{I} \]
where:
  • \( \tau \) is the torque applied, and
  • \( I \) is the moment of inertia.

For the potter's wheel, use the torque of 0.09 N⋅m and moment of inertia of 0.11 kg⋅m² to find:
\[ \alpha = \frac{0.09}{0.11} \approx 0.818 \] rad/s².
Knowing this, we can also explore how long it takes to stop the wheel by setting the final angular velocity to zero and solving for time.

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Most popular questions from this chapter

On a 12.0-cm-diameter audio compact disc (CD), digital bits of information are encoded sequentially along an outward spiraling path. The spiral starts at radius \(R_{1}=2.5 \mathrm{~cm}\) and winds its way out to radius \(R_{2}=5.8 \mathrm{~cm} .\) To read the digital information, a CD player rotates the CD so that the player's readout laser scans along the spiral's sequence of bits at a constant linear speed of \(1.25 \mathrm{~m} / \mathrm{s}\). Thus the player must accurately adjust the rotational frequency \(f\) of the \(\mathrm{CD}\) as the laser moves outward. Determine the values for \(f\) (in units of \(\mathrm{rpm}\) ) when the laser is located at \(R_{1}\) and when it is at \(R_{2}\).

A marble of mass \(m\) and radius \(r\) rolls along the looped rough track of Fig. \(10-67 .\) What is the minimum value of the vertical height \(h\) that the marble must drop if it is to reach the highest point of the loop without leaving the track? (a) Assume \(r \ll R ;\) (b) do not make this assumption. Ignore frictional losses.

The \(1100-\mathrm{kg}\) mass of a car includes four tires, each of mass (including wheels) \(35 \mathrm{~kg}\) and diameter \(0.80 \mathrm{~m}\). Assume each tire and wheel combination acts as a solid cylinder. Determine \((a)\) the total kinetic energy of the car when traveling \(95 \mathrm{~km} / \mathrm{h}\) and \((b)\) the fraction of the kinetic energy in the tires and wheels. ( \(c\) ) If the car is initially at rest and is then pulled by a tow truck with a force of \(1500 \mathrm{~N}\), what is the acceleration of the car? Ignore frictional losses. \((d)\) What percent error would you make in part \((c)\) if you ignored the rotational inertia of the tires and wheels?

A cyclist accelerates from rest at a rate of \(1.00 \mathrm{~m} / \mathrm{s}^{2} .\) How fast will a point at the top of the rim of the tire (diameter \(=68 \mathrm{~cm})\) be moving after \(2.5 \mathrm{~s}\) ? [Hint: At any moment, the lowest point on the tire is in contact with the ground and is at rest-see Fig. \(10-\)

A bowling ball of mass \(7.3 \mathrm{~kg}\) and radius \(9.0 \mathrm{~cm}\) rolls without slipping down a lane at \(3.7 \mathrm{~m} / \mathrm{s} .\) Calculate its total kinetic energy.

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