In a department store toy display, a small disk (disk 1) of radius \(0.100 \mathrm{~m}\) is driven by a motor and turns a larger disk (disk 2) of radius \(0.500 \mathrm{~m}\). Disk 2 , in turn, drives disk 3 , whose radius is \(1.00 \mathrm{~m}\). The three disks are in contact and there is no slipping. Disk 3 is observed to sweep through one complete revolution every \(30.0 \mathrm{~s}\) a) What is the angular speed of disk \(3 ?\) b) What is the ratio of the tangential velocities of the rims of the three disks? c) What is the angular speed of disks 1 and \(2 ?\) d) If the motor malfunctions, resulting in an angular acceleration of \(0.100 \mathrm{rad} / \mathrm{s}^{2}\) for disk 1 , what are disks 2 and 3's angular accelerations?

Short Answer

Expert verified
Question: Determine the angular acceleration of disk 1 if the angular acceleration of disks 2 and 3 is given as shown above. Answer: The angular acceleration of disk 1 is \(0.100 \mathrm{rad/s^2}\).

Step by step solution

01

Calculate the angular speed of disk 3

Given that disk 3 makes one full revolution in 30 seconds, we can calculate its angular speed using the formula: \(\omega = \frac{2\pi}{T}\), where \(T = 30s\) for disk 3. So, \(\omega_3 = \frac{2\pi}{30} = \frac{\pi}{15} \mathrm{rad/s}\). b) Finding the ratio of the tangential velocities of the rims of the three disks (\(V_1 : V_2 : V_3\))
02

Relate the tangential velocity and angular speed

The tangential velocity (\(V\)) of a point on the edge of a rotating object is given by the formula: \(V = r\omega\), where \(r\) is the radius and \(\omega\) is the angular speed.
03

Find the ratio of tangential velocities

Since the disks are in contact without slipping, the tangential velocity at the point of contact for each pair of disks is the same. Thus, \(V_1 : V_2 : V_3 = (r_1 \omega_1) : (r_2 \omega_2) : (r_3 \omega_3)\). c) Finding the angular speed of disks 1 and 2 (\(\omega_1\) and \(\omega_2\))
04

Use the no-slip condition at the point of contact

Since there is no slipping, we can say that the product of the radius times the angular speed at the point of contact for each pair of disks is equal. That is, \(r_1\omega_1 = r_2\omega_2 = r_3\omega_3\).
05

Calculate the angular speeds

Using the no-slip condition and the fact that we've already found the angular speed of disk 3, we can now calculate the angular speeds for disks 1 and 2: \(\omega_1 =\frac{r_3\omega_3}{r_1}=\frac{1.00\cdot\frac{\pi}{15}}{0.100}=10\frac{\pi}{15} \mathrm{rad/s}\) \(\omega_2 =\frac{r_3\omega_3}{r_2}=\frac{1.00\cdot\frac{\pi}{15}}{0.500}=2\frac{\pi}{15} \mathrm{rad/s}\) d) Finding the angular accelerations of disks 2 and 3 (\(\alpha_2\) and \(\alpha_3\))
06

Use the no-slip condition for angular acceleration

Similar to the no-slip condition for angular speeds, we can also use a similar condition for angular accelerations: \(r_1\alpha_1 = r_2\alpha_2 = r_3\alpha_3\).
07

Calculate the angular accelerations

Using the no-slip condition for angular accelerations, we can now calculate the angular accelerations of disks 2 and 3: \(\alpha_2=\frac{r_1\alpha_1}{r_2}=\frac{0.100\cdot 0.100}{0.500}=0.020 \mathrm{rad/s^2}\) \(\alpha_3=\frac{r_1\alpha_1}{r_3}=\frac{0.100\cdot 0.100}{1.00}=0.010 \mathrm{rad/s^2}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Tangential Velocity
When an object moves in a circular path, tangential velocity refers to the linear velocity of any point on the object. It's the rate at which the point is covering distance along the circular path and is tangent to the rotation circle itself, pointing in the direction of motion.

Mathematically, tangential velocity, denoted as V, is found using the equation:
\[V = r\omega\]
where r is the radius of the circular path, and \(\omega\) is the angular speed of the object. In a system where multiple objects are connected, like the toy disks in the exercise, we can assess the relative motion between them through their tangential velocities. If there's no slipping, the tangential velocity at the edge of one disk will be equal to the tangential velocity at the edge of the other where they contact each other. This ensures a common speed at the point of contact, which is crucial for predicting the behavior of connected rotating systems.
Angular Acceleration
Angular acceleration is the rate at which angular speed changes with time. In scenarios where discs or wheels are being spun up or down by forces or torques, we use angular acceleration to describe how quickly they are picking up or losing speed in their rotation.

The equation for angular acceleration, denoted by \(\alpha\), is:
\[\alpha = \frac{\Delta\omega}{\Delta t}\]
where \(\Delta\omega\) signifies the change in angular speed and \(\Delta t\) is the change in time. This concept extends the understanding of rotational motion into a dynamic realm where speeds aren't constant, as illustrated when the problem discusses the malfunction of the motor leading to an angular acceleration on disk 1. The direct proportionality between the radii of the disks and their angular accelerations is a critical point, mirroring the no-slip condition applied to angular speeds.
Rotational Motion
Rotational motion is the movement of a body about an axis. Everything that spins, from wheels to planets, exhibits this type of motion. The central concepts in rotational motion are angular speed and angular acceleration, which together describe how quickly an object rotates (\(\omega\)) and how that speed changes over time (\(\alpha\)).

In the toy disk example, each disk exhibits rotational motion. By understanding the relationships between the radii of the disks and the no-slip condition, we're able to determine that the ratio of their rotational speeds is inverse to the ratio of their radii. This is fundamental in systems with gears or interconnected wheels, ensuring that the motion is transmitted effectively from one object to another without energy losses due to slippage.
No-Slip Condition
The no-slip condition in mechanics refers to the assumption that two rolling bodies in contact move without slipping against each other. This means that at the point of contact, both bodies have the same tangential velocity.

When this condition is applied to a set of rotating disks or gears, it asserts that the speed at which the edges of two contacting bodies move relative to their centers is the same. This can be expressed with the formula:
\[r_1\omega_1 = r_2\omega_2\]
where r represents the radius, and \(\omega\) the angular speed of the rotating objects. The condition ensures a fixed transmission ratio, which is vital in mechanisms where exact timing between the rotation of different parts is essential, such as in clocks or the toy display described in the exercise. When considering angular acceleration, the analog condition is applied to maintain the proportional relationship across different radii.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You are holding the axle of a bicycle wheel with radius \(35.0 \mathrm{~cm}\) and mass \(1.00 \mathrm{~kg}\). You get the wheel spinning at a rate of 75.0 rpm and then stop it by pressing the tire against the pavement. You notice that it takes \(1.20 \mathrm{~s}\) for the wheel to come to a complete stop. What is the angular acceleration of the wheel?

What is the centripetal acceleration of the Moon? The period of the Moon's orbit about the Earth is 27.3 days, measured with respect to the fixed stars. The radius of the Moon's orbit is \(R_{M}=3.85 \cdot 10^{8} \mathrm{~m}\).

Consider a \(53-\mathrm{cm}\) -long lawn mower blade rotating about its center at 3400 rpm. a) Calculate the linear speed of the tip of the blade. b) If safety regulations require that the blade be stoppable within \(3.0 \mathrm{~s}\), what minimum angular acceleration will accomplish this? Assume that the angular acceleration is constant.

Unlike a ship, an airplane does not use its rudder to turn. It turns by banking its wings: The lift force, perpendicular to the wings, has a horizontal component, which provides the centripetal acceleration for the turn, and a vertical component, which supports the plane's weight. (The rudder counteracts yaw and thus it keeps the plane pointed in the direction it is moving.) The famous spy plane, the SR-71 Blackbird, flying at \(4800 \mathrm{~km} / \mathrm{h}\), has a turning radius of \(290 . \mathrm{km} .\) Find its banking angle.

Two masses hang from two strings of equal length that are attached to the ceiling of a car. One mass is over the driver's seat; the other is over the passenger's seat. As the car makes a sharp turn, both masses swing away from the center of the turn. In their resulting positions, will they be farther apart, closer together, or the same distance apart as they were when the car wasn't turning?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free