/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 A wheel rotates without friction... [FREE SOLUTION] | 91Ó°ÊÓ

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A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to \(80.0 \mathrm{~N}\) is applied to the rim of the wheel. The wheel has radius \(0.120 \mathrm{~m}\) Starting from rest, the wheel has an angular speed of \(12.0 \mathrm{rev} / \mathrm{s}\) after \(2.00 \mathrm{~s}\). What is the moment of inertia of the wheel?

Short Answer

Expert verified
The moment of inertia of the wheel is \(0.255 \mathrm{~kg m²}\).

Step by step solution

01

Calculating the Torque

Torque (Ï„) can be calculated using the equation Ï„ = F*r, where F is the force and r is the radius. So, Ï„ = \(80.0 \mathrm{~N}\) * \(0.120 \mathrm{~m}\) = \(9.6 \mathrm{~Nm}\).
02

Angular speed conversion

The angular speed is given as 12.0 revolutions per second. We need to convert this into radian per second for our calculations. Since each revolution is \(2\pi\) radians, the angular speed is \(12.0 \mathrm{rev}/\mathrm{s} * 2\pi \mathrm{rad}/ \mathrm{rev} = 24\pi \mathrm{~rad/s}\).
03

Calculating Angular acceleration

Angular acceleration (α) can be calculated by using the equation α = Δω/t, where Δω is the change in angular speed and t is the time. The wheel starts from rest so initial angular speed is 0. So, α = \(24\pi \mathrm{~rad/s}\) / \(2.00 \mathrm{~s}\) = \(12\pi \mathrm{~rad/s²}\).
04

Calculating moment of inertia

Now using τ = Iα, where I is the moment of inertia and α is the angular acceleration. Re-arranging the equation, we get I = τ/α = \(9.6 \mathrm{~Nm}\) / \(12\pi \mathrm{~rad/s²}\).

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Key Concepts

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

Torque and Angular Momentum
Torque is a measure of the force that can cause an object to rotate about an axis. Just like force is what causes an object to accelerate in linear motion, torque is what causes an object to gain angular acceleration in rotational motion. The larger the torque, the greater the object’s angular acceleration will be, assuming the mass and distribution of mass (moment of inertia) remain constant.

Torque (\tau) is calculated by multiplying the force applied (F) by the radius (r) of the lever arm (distance from the axis of rotation). Mathematically, it is expressed as \tau = F * r. It's important to apply the force perpendicularly to the lever arm to achieve maximum torque.

In the context of the exercise, a constant tangential force is applied at the wheel's rim, where the force is perpendicular to the radius, resulting in maximal torque. As the wheel starts from rest and begins to rotate, it gains angular momentum. Angular momentum is the rotational equivalent of linear momentum and is dependent on the object's moment of inertia (I) and its angular velocity (\(\omega\)). The relationship between torque, angular momentum (L), and time (t) can be described by the equation: \( \tau = \frac{\text{d}L}{\text{d}t} \).

The concept can be summarized by knowing this relationship: as torque is applied to an object, it changes the object's angular momentum, causing it to rotate faster, and this change in rotation can be quantified by the moment of inertia.
Angular Speed
Angular speed, or angular velocity, is the rate at which an object rotates or revolves relative to another point, ie., how fast the angular position or orientation of an object changes with time. It is given in radians per second (\text{rad/s}) in the International System of Units (SI). For practical problems, including the wheel exercise, we often convert from revolutions per second (rev/s) to rad/s since calculations involving rotational motion are typically done in radians.

To proceed with the conversion from rev/s to rad/s, we use the fact that one full revolution equals \(2\pi\) radians: hence, \(12.0 \text{rev/s} * 2\pi \text{rad/rev} = 24\pi \text{rad/s}\). With the angular speed known, it’s possible to analyze other aspects of rotational motion such as the kinetic energy of the wheel or the centrifugal forces that may act upon it at a given speed.

For students, remembering that angular speed is the rotational analog of linear speed can help contextualize the concept. Just like how fast you're walking or running, angular speed tells you how fast an object is spinning.
Angular Acceleration
Angular acceleration is the rate of change of angular velocity. In other words, it’s how quickly an object’s rotating speed increases or decreases. It is measured in radians per second squared (\(\text{rad/s}^2\)).

The formula to compute angular acceleration (\(\alpha\)) is analogous to the formula for linear acceleration, \(\alpha = \frac{\Delta \omega}{t}\), where \Delta \omega is the change in angular velocity and t is the time over which the change occurs. From a standstill, the wheel in our exercise reaches an angular velocity of \(24\pi \text{rad/s}\), thus the angular acceleration can be computed as \(\alpha = \frac{24\pi \text{rad/s}}{2.00 \text{s}} = 12\pi \text{rad/s}^2\).

Understanding angular acceleration is crucial when analyzing how forces affect rotational motion. It is also especially relevant in the design of mechanical systems where controlling the speed of rotation is vital for functionality and safety. For instance, a car’s wheels must be able to accelerate and decelerate efficiently to allow for safe driving conditions.

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Most popular questions from this chapter

A teenager is standing at the rim of a large horizontal uniform wooden disk that can rotate freely about a vertical axis at its center. The mass of the disk (in \(\mathrm{kg}\) ) is \(M\) and its radius (in \(\mathrm{m}\) ) is \(R\). The mass of the teenager (in \(\mathrm{kg}\) ) is \(m .\) The disk and teenager are initially at rest. The teenager then throws a large rock that has a mass (in kg) of \(m_{\text {rock }}\). As it leaves the thrower's hands, the rock is traveling horizontally with speed \(v\) (in \(\mathrm{m} / \mathrm{s}\) ) relative to the earth in a direction tangent to the rim of the disk. The teenager remains at rest relative to the disk and so rotates with it after throwing the rock. In terms of \(M, R, m, m_{\text {rock }}\) and \(v,\) what is the angular speed of the disk? Treat the teenager as a point mass.

A \(15.0 \mathrm{~kg}\) bucket of water is suspended by a very light rope wrapped around a solid uniform cylinder \(0.300 \mathrm{~m}\) in diameter with mass \(12.0 \mathrm{~kg}\). The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls \(10.0 \mathrm{~m}\) to the water. (a) What is the tension in the rope while the bucket is falling? (b) With what speed does the bucket strike the water? (c) What is the time of fall? (d) While the bucket is falling, what is the force exerted on the cylinder by the axle?

A Gyroscope on the Moon. A certain gyroscope precesses at a rate of \(0.50 \mathrm{rad} / \mathrm{s}\) when used on earth. If it were taken to a lunar base, where the acceleration due to gravity is \(0.165 g,\) what would be its precession rate?

A grindstone in the shape of a solid disk with diameter \(0.520 \mathrm{~m}\) and a mass of \(50.0 \mathrm{~kg}\) is rotating at \(850 \mathrm{rev} / \mathrm{min} .\) You press an ax against the rim with a normal force of \(160 \mathrm{~N}\) (Fig. \(\mathrm{P} 10.58\) ), and the grindstone comes to rest in \(7.50 \mathrm{~s}\). Find the coefficient of friction between the ax and the grindstone. You can ignore friction in the bearings.

A metal bar is in the \(x y\) -plane with one end of the bar at the origin. A force \(\overrightarrow{\boldsymbol{F}}=(7.00 \mathrm{~N}) \hat{\imath}+(-3.00 \mathrm{~N}) \hat{\jmath}\) is applied to the bar at the point \(x=3.00 \mathrm{~m}, y=4.00 \mathrm{~m}\). (a) In terms of unit vectors \(\hat{\imath}\) and \(\hat{\jmath},\) what is the position vector \(\vec{r}\) for the point where the force is applied? (b) What are the magnitude and direction of the torque with respect to the origin produced by \(\overrightarrow{\boldsymbol{F}} ?\)

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