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Chapter 9: Problem 29

The spin cycles of a washing machine have two angular speeds, 423 rev/min and 640 rev/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) 15.75 m/s and 107.32 g for radial acceleration.

Step by step solution

01

Understanding Maximum Radial Force Ratio

To find the ratio of maximum radial forces, use the formula for centripetal force \( F = m \omega^2 r \). Since we're asked for the ratio and mass and radius are constant, we need to compare \( \omega^2 \) for both speeds. First, convert angular speeds from \( \text{rev/min} \) to \( \text{rad/s} \): \( \omega = \frac{2 \pi \times \text{rev/min}}{60} \). For 423 rev/min: \( \omega_1 = \frac{2 \pi \times 423}{60} \approx 44.35 \text{ rad/s} \). For 640 rev/min: \( \omega_2 = \frac{2 \pi \times 640}{60} \approx 66.98 \text{ rad/s} \). The ratio of forces is \( \left(\frac{\omega_2}{\omega_1}\right)^2 \). Substitute calculated values: \( \left(\frac{66.98}{44.35}\right)^2 \approx 2.2756 \).
02

Understanding Maximum Tangential Speed Ratio

The maximum tangential speed \( v \) is given by \( v = \omega r \). To find the ratio of tangential speeds, compare \( v_2 \) and \( v_1 \) for the two angular speeds. Since the radius is constant, the ratio is simple: \( \frac{v_2}{v_1} = \frac{\omega_2}{\omega_1} \). Substituting \( \omega_2 \) and \( \omega_1 \): \( \frac{66.98}{44.35} \approx 1.51 \).
03

Calculating Maximum Tangential Speed and Radial Acceleration

First, calculate the maximum tangential speed for the higher speed using \( v = \omega_2 r \). With \( r = 0.235 \text{ m} \) (since diameter is 0.470 m), \( v_2 = 66.98 \times 0.235 \approx 15.75 \text{ m/s} \). For maximum radial acceleration, use \( a_r = \omega^2 r \). Maximum \( a_r = (66.98)^2 \times 0.235 \approx 1052.46 \text{ m/s}^2 \). Since \( g \approx 9.81 \text{ m/s}^2 \), express \( a_r \) in terms of \( g \): \( \frac{1052.46}{9.81} \approx 107.32 \ g \).

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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 the force required to keep an object moving in a circular path. It acts towards the center of the circle and can be calculated using the formula:
  • \( F = m \omega^2 r \).
Here,
  • \( F \) is the centripetal force,
  • \( m \) is mass,
  • \( \omega \) is angular speed in radians per second,
  • \( r \) is the radius of the circular path.
The centripetal force increases with both the square of the angular speed and the radius of the path. When comparing forces at different angular speeds, like in the washing machine example, the mass and radius remain constant. Thus, the force's ratio depends entirely on the square of the angular velocity ratio, making higher speeds result in dramatically increased forces.
Angular Speed
Angular Speed represents how fast an object spins around a circle, described in \( \text{revolutions per minute (rev/min)} \). In physics, it is often more useful to express this in radians per second, to match other rotational formulas.
  • To convert from \( \text{rev/min} \) to \( \text{rad/s} \), use the conversion \( \omega = \frac{2 \pi \times \text{rev/min}}{60} \).

Angular speed provides a measure of how quickly something like the drum in a washing machine is spinning. The washing machine example uses two speeds, 423 and 640 rev/min, equating to approximately 44.35 rad/s and 66.98 rad/s respectively. Angular speed is a vital determinant in finding the centripetal force required to maintain motion in a circular path.
Tangential Speed
Tangential Speed is the linear speed of something moving along a circular path and is directly tied to angular speed. Calculating tangential speed is straightforward:
  • \( v = \omega r \),
where
  • \( v \) is tangential speed,
  • \( \omega \) is angular speed,
  • \( r \) is the radius of the circle.

In the context of a washing machine, the tangential speed changes as the angular speed changes. For instance, with a drum radius of 0.235 m (half of the 0.470 m diameter), the higher angular speed results in a tangential speed of approximately 15.75 m/s. In a rotating system, the outer edges typically move the fastest owing to this linear relationship between tangential speed and angular speed.
Radial Acceleration
Radial Acceleration, also known as centripetal acceleration, is the acceleration directed towards the center of the circular path. It can be determined using the formula:
  • \( a_r = \omega^2 r \).
In this,
  • \( a_r \) is radial acceleration,
  • \( \omega \) is angular speed,
  • \( r \) is the radius.

This acceleration must be present for any object moving in a circular path, ensuring the object doesn't fly off but continues its path. For instance, at the higher angular speed, the radial acceleration in the washing machine reaches about 107.32 times the force of gravity \( (g) \). Understanding radial acceleration explains why laundry sticks to the drum鈥檚 wall during fast spins. It underscores the massive forces at play even over small radial distances as speed increases.

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Most popular questions from this chapter

What angle in radians is subtended by an arc 1.50 \(\mathrm{m}\) long on the circumference of a circle of radius 2.50 \(\mathrm{m} ?\) What is this angle in degrees? (b) An arc 14.0 \(\mathrm{cm}\) long on the circumference of a circle subtends an angle of \(128^{\circ} .\) What is the radius of the circle? (c) The angle between two radii of a circle with radius 1.50 \(\mathrm{m}\) is 0.700 rad. What length of arc is intercepted on the circumference of the circle by the two radii?

The rotating blade of a blender turns with constant angular acceleration 1.50 rad/s\(^{2}\) (a) How much time does it take to reach an angular velocity of 36.0 rad/s, starting from rest? (b) Through how many revolutions does the blade turn in this time interval?

Energy is to be stored in a 70.0 -kg flywheel in the shape of a uniform solid disk with radius \(R=1.20 \mathrm{m.}\) To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3500 \(\mathrm{m} / \mathrm{s}^{2} .\) What is the maximum kinetic energy that can be stored in the flywheel?

A uniform 2.00 -m ladder of mass 9.00 kg is leaning against a vertical wall while making an angle of \(53.0^{\circ}\) with the floor. A worker pushes the ladder up against the wall until it is vertical. What is the increase in the gravitational potential energy of the ladder?

At \(t=3.00\) s a point on the rim of a 0.200 -m-radius wheel has a tangential speed of 50.0 \(\mathrm{m} / \mathrm{s}\) as the wheel slows down with a tangential acceleration of constant magnitude 10.0 \(\mathrm{m} / \mathrm{s}^{2}\) . (a) Calculate the wheel's constant angular acceleration. (b) Calculate the angular velocities at \(t=3.00\) s and \(t=0 .\) (c) Through what angle did the wheel turn between \(t=0\) and \(t=3.00\) s? (d) At what time will the radial acceleration equal \(g ?\)
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