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Chapter 9: Problem 18

In a charming 19 th century hotel, an old-style elevator is connected to a counterweight by a cable that passes over a rotating disk 2.50 \(\mathrm{m}\) in diameter (Fig. E9.18). The elevator is raised and lowered by turning the disk, and the cable does not slip on the rim of the disk but turns with it. (a) At how many rpm must the disk turn to raise the elevator at 25.0 \(\mathrm{cm} / \mathrm{s} ?\) (b) To start the elevator moving, it must be accelerated at \(\frac{1}{8} g\) . (c) Through what angle (in radians and degrees) has the disk turned when it has raised the elevator 3.25 \(\mathrm{m}\) between floors? What must be the angular acceleration of the disk, in rad/s's?

Short Answer

Expert verified
(a) 1.91 rpm; (b) Angular acceleration: 0.981 rad/s虏; (c) Angle: 2.6 rad or 148.49掳.

Step by step solution

01

Determine Circumference of the Disk

First, calculate the circumference of the disk. Given the diameter of the disk is 2.50 m, the radius is half of that. So \( r = \frac{2.50}{2} = 1.25 \) m. The circumference \( C \) of the disk is given by the formula \( C = 2\pi r \). Thus, \( C = 2\pi \times 1.25 = 2.5\pi \) m.
02

Calculate RPM to Achieve Given Speed

To find the rpm (revolutions per minute) needed for the disk to raise the elevator at 25.0 cm/s (\(0.25\) m/s), determine how many revolutions per second are required. Since one full revolution moves the elevator the distance of the circumference (2.5\(\pi\) m), set 2.5\(\pi n = 0.25\) where \(n\) is the revolutions per second. Solving for \(n\) gives \( n = \frac{0.25}{2.5\pi} \) revolutions per second. Convert to rpm by multiplying by 60: \( n_{rpm} = 60 \times \frac{0.25}{2.5\pi} \approx 1.91 \text{ rpm} \).
03

Calculate Required Angular Acceleration

The elevator's acceleration is \( \frac{1}{8} g \), where \( g = 9.81 \text{ m/s}^2 \). Thus, the linear acceleration \( a = \frac{9.81}{8} = 1.22625 \text{ m/s}^2 \). This linear acceleration corresponds to angular acceleration \( \alpha \) via \( a = r\alpha \). With \( r = 1.25 \) m, rearrange to find \( \alpha = \frac{1.22625}{1.25} \approx 0.981 \text{ rad/s}^2 \).
04

Calculate the Angle for the Distance Raised

When the elevator is raised 3.25 m, determine how many revolutions this corresponds to. Since one revolution raises it by 2.5\(\pi\), set \(2.5\pi n = 3.25\) and solve for \(n\). Divide both sides to find \(n = \frac{3.25}{2.5\pi}\). Each revolution equals \(2\pi\) radians, thus the angle in radians is \(\theta = 2\pi n = 2\pi \times \frac{3.25}{2.5\pi} = \frac{6.5}{2.5} \approx 2.6 \) radians. Convert this angle to degrees: \(\theta_{degrees} = 2.6 \times \frac{180}{\pi} \approx 148.49\) degrees.

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Key Concepts

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

Angular Acceleration
Angular acceleration refers to how quickly an object speeds up its rotation. For the elevator in the problem, it starts from rest and is brought up to a moving speed. The rotational disk does this by gradually increasing its rate of spin. It's like pushing a merry-go-round harder and harder until it spins faster.

To calculate angular acceleration, we used the concept that linear acceleration is connected to angular motion. The formula linking them is:
  • Linear acceleration (\(a\) ) = Radius of the disk (\(r\) ) 脳 Angular acceleration (\(\alpha\) )
Since the elevator needs to accelerate at a rate of \(\frac{1}{8} g\), or 1.22625 m/s虏, and the radius of our disk is 1.25 m, we can solve for angular acceleration \(\alpha = \frac{1.22625}{1.25} \). This becomes approximately 0.981 rad/s虏.

Angular acceleration is crucial because it tells us how forcefully the elevator's motor has to work to get the elevator moving at just the right speed safely.
Revolutions Per Minute (RPM)
Revolutions per minute (RPM) is a measurement of how fast an object spins. Imagine watching a clock, where the second hand circles around the clock face 鈥� that's counting revolutions. If the disk attached to the elevator makes one full spin every minute, it has an RPM of 1.

In our problem, we're asked to find the RPM necessary for the elevator to rise at a speed of 25 cm/s. We first figured out how many revolutions occur per second based on how far a single revolution of the disk moves the elevator (which is the circumference of the disk).
  • The equation we used was: \(2.5\pi n = 0.25\), where \(n\) is revolutions per second.
  • Solving for \(n\), we converted revolutions per second to RPM by multiplying by 60.
  • The result was approximately 1.91 RPM.
This value tells us that for the elevator to ascend smoothly at the desired speed, the disk should spin approximately 1.91 times every minute.
Linear Acceleration
Linear acceleration is all about increasing speed in a straight line. In this problem, it's the rate at which the elevator speeds up as it heads from one floor to the next. In physics, it's often represented as "a" and is measured in meters per second squared (m/s虏).

Since the elevator has to start moving up from a standstill, we need to consider what rate or acceleration is required to do this. The problem specifies that this acceleration must equal \(\frac{1}{8} g\), where \(g\) is gravity (9.81 m/s虏). This simplifies to a linear acceleration of 1.22625 m/s虏.

Why is this important?
  • It determines the comfort and safety of passengers during the ride.
  • It affects how the motor of the elevator operates.
Just like how cars need to accelerate to merge onto a highway, elevators need to accelerate to start moving, often in a controlled manner that considers passenger comfort. This is why calculating and ensuring the correct linear acceleration is a major part of elevator design and function.

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Most popular questions from this chapter

A wheel changes its angular velocity with a constant angular acceleration while rotating about a fixed axis through its center. (a) Show that the change in the magnitude of the radial acceleration during any time interval of a point on the wheel is twice the product of the angular acceleration, the angular displacement, and the perpendicular distance of the point from the axis. (b) The radial acceleration of a point on the wheel that is 0.250 \(\mathrm{m}\) from the axis changes from 25.0 \(\mathrm{m} / \mathrm{s}^{2}\) to 85.0 \(\mathrm{m} / \mathrm{s}^{2}\) as the wheel rotates through 20.0 rad. Calculate the tangential acceleration of this point. (c) Show that the change in the wheel's kinetic energy during any time interval is the product of the moment of inertia about the axis, the angular acceleration, and the angular displacement. (d) During the 20.0 -rad angular displacement of part (b), the kinetic energy of the wheel increases from 20.0 \(\mathrm{J}\) to 45.0 \(\mathrm{J} .\) What is the moment of inertia of the wheel about the rotation axis?

9.75\(\cdots .\) It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball bearings. Consider a flywheel made of iron (density 7800 \(\mathrm{kg} / \mathrm{m}^{3}\) ) in the shape of a 10.0 -cm-thick uniform disk. (a) What would the diameter of such a disk need to be if it is to store 10.0 megajoules of kinetic energy when spinning at 90.0 rpm about an axis perpendicular to the disk at its center? (b) What would be the centripetal acceleration of a point on its rim when spinning at this rate?

A uniform disk with radius \(R=0.400 \mathrm{m}\) and mass 30.0 kg rotates in a horizontal plane on a frictionless vertical axle that passes through the center of the disk. The angle through which the disk has turned varies with time according to \(\theta(t)=\) \((1.10 \mathrm{rad} / \mathrm{s}) t+\left(8.60 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2} .\) What is the resultant linear acceleration of a point on the rim of the disk at the instant when the disk has turned through 0.100 \(\mathrm{rev}\) ?

A thin, light wire is wrapped around the rim of a wheel, as shown in Fig. E9.49. The wheel rotates about a stationary horizontal axle that passes through the center of the wheel. The wheel has radius 0.180 \(\mathrm{m}\) and moment of inertia for rotation about the axle of \(I=0.480 \mathrm{kg} \cdot \mathrm{m}^{2}\) . A small block with mass 0.340 \(\mathrm{kg}\) is suspended from the free end of the wire. When the system is released from rest, the block descends with constant acceleration. The bearings in the wheel at the axle are rusty, so friction there does \(-6.00 \mathrm{J}\) of work as the block descends 3.00 \(\mathrm{m}\) . What is the magnitude of the angular velocity of the wheel after the block has descended 3.00 \(\mathrm{m} ?\)

You need to design an industrial turntable that is 60.0 \(\mathrm{cm}\) in diameter and has a kinetic energy of 0.250 \(\mathrm{J}\) when turning at 45.0 \(\mathrm{rpm}(\mathrm{rev} / \mathrm{min}) .\) (a) What must be the moment of inertia of the turntable about the rotation axis? (b) If your workshop makes this turntable in the shape of a uniform solid disk, what must be its mass?
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