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\(\bullet\) The spin cycles of a washing machine have two angular speeds, 423 \(\mathrm{rev} / \mathrm{min}\) and 640 \(\mathrm{rev} / \mathrm{min.}\) The internal diameter of the drum is 0.470 \(\mathrm{m} .\) (a) What is the ratio of the maximum radial force on the laundry for the higher angular speed to that for the lower speed? (b) What is the ratio of the maximum tangential speed of the laundry for the higher angular speed to that for the lower speed? (c) Find the laundry's maximum tangential speed and the maximum radial acceleration, in terms of \(g\) .

Short Answer

Expert verified
(a) 2.28 (b) 1.51 (c) Max tangential speed = 15.74 m/s; Max radial acceleration = 107.58g

Step by step solution

01

Understanding Angular Speed

The washing machine spins at two speeds, 423 revolutions per minute (rpm) and 640 rpm. We need to find the ratios of various forces and speeds using these given speeds.
02

Find the Angular Speeds in Radians per Second

Convert the angular speeds from revolutions per minute to radians per second. Use the conversion factor: 1 revolution = \(2\pi\) radians and 1 minute = 60 seconds. For 423 rpm: \(\text{Angular Speed 1} = \frac{423 \times 2\pi}{60} = 44.34 \, \text{rad/s} \)For 640 rpm: \(\text{Angular Speed 2} = \frac{640 \times 2\pi}{60} = 67.02 \, \text{rad/s} \)
03

Calculate Radial Force Ratio

The radial force for an object moving in a circle is given by \(F_r = m \cdot \omega^2 \cdot r\), where \(m\) is the mass, \(\omega\) is angular speed, and \(r\) is the radius.Since mass and radius are constant, the ratio of the radial forces is:\( \text{Force Ratio} = \left(\frac{\omega_2}{\omega_1}\right)^2 = \left(\frac{67.02}{44.34}\right)^2 \approx 2.28 \)
04

Calculate Tangential Speed Ratio

Tangential speed \(v\) is given by \(v = \omega \cdot r\). So the ratio of tangential speeds is:\( \text{Tangential Speed Ratio} = \frac{\omega_2}{\omega_1} = \frac{67.02}{44.34} \approx 1.51 \)
05

Calculate Maximum Tangential Speed

Using \(r = \frac{0.470}{2} = 0.235\) m (since the radius is half the diameter), For the higher speed:\(\text{Max Tangential Speed} = \omega_2 \cdot r = 67.02 \cdot 0.235 = 15.74 \, \text{m/s} \)
06

Calculate Maximum Radial Acceleration

Radial acceleration \(a_r\) is given by \(a_r = \omega^2 \cdot r\).For the higher speed:\(a_r = 67.02^2 \cdot 0.235 = 1055.37 \, \text{m/s}^2\)Express in terms of \(g\), where \(g\) is the acceleration due to gravity (\(9.81 \, \text{m/s}^2\)):\(a_r \approx \frac{1055.37}{9.81} \approx 107.58 \, g\)

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Key Concepts

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

Radial Force
Radial force plays a crucial role in understanding how objects behave when they move in a circle. When objects, like laundry in a washing machine drum, spin in a circular path, a force acts perpendicular to their velocity, pulling them towards the center of the circle. This force is known as radial or centripetal force.

The formula to calculate radial force is:
  • \( F_r = m \cdot \omega^2 \cdot r \)
where:
  • \( F_r \) is the radial force,
  • \( m \) is the mass of the object,
  • \( \omega \) is the angular speed in radians per second, and
  • \( r \) is the radius of the circular path.
For the washing machine problem, the radial force changes with speed. As we increase the angular speed from 423 to 640 revolutions per minute, the centrifugal force increases dramatically. Using the given formula, we found that the maximum radial force for the higher speed is approximately 2.28 times greater than at the lower speed.

This increase is because radial force is proportional to the square of the angular speed. When you double the speed, you quadruple the force exerted inwards.
Tangential Speed
Tangential speed is another important concept when dealing with circular motion. Unlike radial force which acts towards the center, tangential speed refers to how fast the object moves along the circular path. It is the linear speed of any point rotating at radius \( r \).

For an object in circular motion, tangential speed \( v \) is calculated as:
  • \( v = \omega \cdot r \)
where:
  • \( v \) is the tangential speed,
  • \( \omega \) is the angular speed, and
  • \( r \) is the radius of the circular path.
In the context of the washing machine exercise, we calculated the tangential speed ratios for speeds of 423 rpm and 640 rpm, resulting in a ratio of approximately 1.51. This means that the tangential speed at the higher speed is about 1.51 times that at the lower speed.

Understanding tangential speed helps in visualizing how quickly a point on a spinning drum is moving. It is especially critical in applications like washing machines, where high tangential speeds help clean clothes by forcing water out more efficiently.
Radial Acceleration
Radial acceleration, also known as centripetal acceleration, is the rate at which an object's velocity changes direction towards the center of its circular path. This happens not because the speed itself changes, but because the direction of the velocity changes continuously during circular motion.

We calculate radial acceleration using the formula:
  • \( a_r = \omega^2 \cdot r \)
where:
  • \( a_r \) is the radial acceleration,
  • \( \omega \) is the angular speed, and
  • \( r \) is the radius.
In the solution, for a drum with a diameter of 0.470 m, the maximum radial acceleration during the higher spin cycle (640 rpm) was very high, approximately 1055.37 \( \text{m/s}^2 \). When expressed in terms of gravity ('\( g \) where \( g = 9.81 \text{m/s}^2 \)'), this equated to about 107.58 \( g \).

This high value indicates how powerful circular motions are regarding force and acceleration. It explains why it's essential for machines and rotating systems to be engineered to endure these high forces without suffering damage.

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Most popular questions from this chapter

\(\bullet\) (a) A cylinder 0.150 \(\mathrm{m}\) in diameter rotates in a lathe at 620 \(\mathrm{rpm} .\) What is the tangential speed of the surface of the cylinder? (b) The proper tangential speed for machining cast iron is about 0.600 \(\mathrm{m} / \mathrm{s} .\) At how many revolutions per minute should a piece of stock 0.0800 \(\mathrm{m}\) in diameter be rotated in a lathe to produce this tangential speed?

A flywheel in a motor is spinning at 500.0 \(\mathrm{rpm}\) when a power failure suddenly occurs. The flywheel has mass 40.0 \(\mathrm{kg}\) and diameter 75.0 \(\mathrm{cm} .\) The power is off for 30.0 \(\mathrm{s}\) , and during this time the flywheel slown down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 200.0 complete revolutions. (a) At what rate is the flywheel spinning when the power comes back on? (b) How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

\(\bullet\) A 7300 \(\mathrm{N}\) elevator is to be given an acceleration of 0.150 \(\mathrm{g}\) by connecting it to a cable of negligible weight wrapped around a turning cylindrical shaft. If the shaft's diameter can be no larger than 16.0 \(\mathrm{cm}\) due to space limitations, what must be its minimum angular acceleration to provide the required acceleration of the elevator?

\(\bullet\) When a toy car is rapidly scooted across the floor, it stores energy in a flywheel. The car has mass \(0.180 \mathrm{kg},\) and its fly- wheel has moment of inertia \(4.00 \times 10^{-5} \mathrm{kg} \cdot \mathrm{m}^{2} .\) The car is 15.0 \(\mathrm{cm}\) long. An advertisement claims that the car can travel at a scale speed of up to 700 \(\mathrm{km} / \mathrm{h}(440 \mathrm{mi} / \mathrm{h}) .\) The scale speed is the speed of the toy car multiplied by the ratio of the length of an actual car to the length of the toy. Assume a length of 3.0 \(\mathrm{m}\) for a real car. (a) For a scale speed of \(700 \mathrm{km} / \mathrm{h},\) what is the actual translational speed of the car? (b) If all the kinetic energy that is initially in the flywheel is converted to the translational kinetic energy of the toy, how much energy is originally stored in the flywheel? (c) What ini- tial angular velocity of the flywheel was needed to store the amount of energy calculated in part (b)?

\(\bullet\) The kinetic energy of walking. If a person of mass \(M\) simply moved forward with speed \(V,\) his kinetic energy would be \(\frac{1}{2} M V^{2} .\) However, in addition to possessing a forward motion, various parts of his body (such as the arms and legs) undergo rotation. Therefore, his total kinetic energy is the sum of the energy from his forward motion plus the rotational kinetic energy of his arms and legs. The purpose of this problem is to see how much this rotational motion contributes to the person's kinetic energy. Biomedical measurements show that the arms and hands together typically make up 13\(\%\) of a person's mass, while the legs and feet together account for 37\(\% .\) For a rough (but reasonable) calculation, we can model the arms and legs as thin uniform bars pivoting about the shoulder and hip, respectively. In a brisk walk, the arms and legs each move through an angle of about \(\pm 30^{\circ}\) (a total of \(60^{\circ}\) ) from the vertical in approximately 1 second. We shall assume that they are held straight, rather than beingbent, which is not quite true. Let us consider a 75 kg person walking at 5.0 \(\mathrm{km} / \mathrm{h}\) , having arms 70 \(\mathrm{cm}\) long and legs 90 \(\mathrm{cm}\) long. (a) What is the avertage angular velocity of his arms and legs? (b) Using the average angular velocity from part (a), calculate the amount of rotational kinetic energy in this person's arms and legs as he walks. (c) What is the total kinetic energy due to both his forward motion and his rotation? (d) What percentage of his kinetic energy is due to the rotation of his legs and arms?

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