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Chapter 9: Problem 19

Spin cycles of washing machines remove water from clothes by producing a large radial acceleration at the rim of the cylindrical tub that holds the water and clothes. Estimate the diameter of the tub in a typical home washing machine. (a) What is the rotation rate, in rev/min, of the tub during the spin cycle if the radial acceleration of points on the tub wall is \(3 g ?\) (b) At this rotation rate, what is the tangential speed in \(\mathrm{m} / \mathrm{s}\) of a point on the tub wall?

Short Answer

Expert verified
Based on the steps and calculations provided, the tub's rotation rate during the spin cycle would be around 26.63 rev/min and the tangential speed of a point on the tub wall would be about 2.94 m/s.

Step by step solution

01

Calculate the Rotation Rate

To find the rotation rate, we can use the formula for radial acceleration \(a_r = r蠅^2\), where \(蠅\) is the angular velocity and \(r\) is the radius. Given that \(a_r = 3g\), we can solve for \(蠅\).
02

Formulate the Method to Calculate Rotation Rate in rev/min

The angular velocity is usually in radians per second. To convert it to rev/min which is the unit required in the problem, use the conversion \(1 rev = 2蟺 rad\) and \(1 min = 60 s\). So to change it from rad/sec to rev/min multiply the angular velocity by \( \frac{60}{2蟺}\).
03

Calculate the Tangential Speed

From circular motion physics, we know that the tangential speed \(v_t = r蠅\). Since we know the value of \(蠅\) from the first part of the problem and r, which is half the diameter, we can substitute these values to find \(v_t\).

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Key Concepts

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

Circular Motion
Circular motion refers to the movement of an object along the circumference of a circle or circular path. This motion is characterized by a constant change in the direction of the moving object, while maintaining a constant speed. The forces involved in maintaining this motion are primarily centripetal, pointing towards the center of the circular path.
  • For a washing machine, the tub rotates in a circular motion to effectively remove water from the clothes during the spin cycle.
  • The radial acceleration faced by any point on the tub's wall is directed towards the center, helping keep the clothes pressed against the wall of the tub as it spins.
  • Radial acceleration acts perpendicular to the tangential speed of points on the circular path.
Understanding circular motion is crucial for figuring out how machines like washing machines, amusement park rides, and even satellites function, as they all rely on the principles of circular motion to operate effectively.
Angular Velocity
Angular velocity describes how fast something spins around an axis in a circular path. It is defined as the rate of change of the angular position of a rotating object and is usually expressed in radians per second (rad/s). Expressing it in revolutions per minute (rev/min) is common in practical scenarios.
  • In this exercise, the angular velocity helps determine how fast the tub of the washing machine spins, enabling effective water extraction.
  • To calculate angular velocity, formulas such as the radial acceleration formula, \(a_r = r蠅^2\), are used. Here, \(r\) is the radius of the tub, and \(蠅\) is the angular velocity.
  • The conversion from rad/s to rev/min requires the knowledge that \(1 \text{ rev} = 2蟺 \text{ rad}\) and \(1 \text{ min} = 60 \text{ s}\).
This conversion is essential for practical understanding as most people relate better to concepts like revolutions per minute rather than equations in physics terminology.
Tangential Speed
Tangential speed is essentially the linear speed of something moving along a circular path. It tells us how fast a point on the edge of a rotating object is moving. The formula for tangential speed \(v_t\) is \(v_t = r蠅\), where \(r\) is the radius, and \(蠅\) is the angular velocity.
  • In washing machines, tangential speed informs us about how quickly points on the tub wall are moving, which is vital for understanding the washing machine鈥檚 efficiency in drying fabrics.
  • Since the radius is half the diameter, knowing the radius and angular velocity allows us to calculate the tangential speed quite quickly.
  • Higher tangential speeds typically result in more effective water removal, as clothes are squeezed tighter against the tub wall.
Comprehending tangential speed is not just limited to understanding washing machines; it applies to any rotating system where the speed at the edge impacts the system鈥檚 functionality, from engines to wheel dynamics.

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Most popular questions from this chapter

A uniform, solid disk with mass \(m\) and radius \(R\) is pivoted about a horizontal axis through its center. A small object of the same mass \(m\) is glued to the rim of the disk. If the disk is released from rest with the small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.

Two metal disks, one with radius \(R_{1}=2.50 \mathrm{~cm}\) and mass \(M_{1}=0.80 \mathrm{~kg}\) and the other with radius \(R_{2}=5.00 \mathrm{~cm}\) and mass \(M_{2}=1.60 \mathrm{~kg},\) are welded together and mounted on a frictionless axis through their common center (Fig. \(\mathbf{P 9 . 7 9}\) ). (a) What is the total moment of inertia of the two disks? (b) A light string is wrapped around the edge of the smaller disk, and a \(1.50 \mathrm{~kg}\) block is suspended from the free end of the string. If the block is released from rest at a distance of \(2.00 \mathrm{~m}\) above the floor, what is its speed just before it strikes the floor? (c) Repeat part (b), this time with the string wrapped around the edge of the larger disk. In which case is the final speed of the block greater? Explain.

A child is pushing a merry-go-round. The angle through which the merry-go- round has turned varies with time according to \(\theta(t)=\gamma t+\beta t^{3}, \quad\) where \(\quad \gamma=0.400 \mathrm{rad} / \mathrm{s} \quad\) and \(\quad \beta=0.0120 \mathrm{rad} / \mathrm{s}^{3}\) (a) Calculate the angular velocity of the merry-go-round as a function of time. (b) What is the initial value of the angular velocity? (c) Calculate the instantaneous value of the angular velocity \(\omega_{z}\) at \(t=5.00 \mathrm{~s}\) and the average angular velocity \(\omega_{\mathrm{av}-z}\) for the time interval \(t=0\) to \(t=5.00 \mathrm{~s}\) Show that \(\omega_{\mathrm{av}-z}\) is \(n o t\) equal to the average of the instantaneous angular velocities at \(t=0\) and \(t=5.00 \mathrm{~s},\) and explain.

Three small blocks, each with mass \(m\), are clamped at the ends and at the center of a rod of length \(L\) and negligible mass. Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through (a) the center of the rod and (b) a point onefourth of the length from one end.

The motor of a table saw is rotating at 3450 rev \(/\) min. A pulley attached to the motor shaft drives a second pulley of half the diameter by means of a V-belt. A circular saw blade of diameter \(0.208 \mathrm{~m}\) is mounted on the same rotating shaft as the second pulley. (a) The operator is careless and the blade catches and throws back a small piece of wood. This piece of wood moves with linear speed equal to the tangential speed of the rim of the blade. What is this speed? (b) Calculate the radial acceleration of points on the outer edge of the blade to see why sawdust doesn't stick to its teeth.
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