/*! 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 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 woman with mass \(50 \mathrm{~kg}\) is standing on the rim of a large horizontal disk that is rotating at \(0.80 \mathrm{rev} / \mathrm{s}\) about an axis through its center. The disk has mass \(110 \mathrm{~kg}\) and radius \(4.0 \mathrm{~m} .\) Calculate the magnitude of the total angular momentum of the woman-disk system. (Assume that you can treat the woman as a point.)

A large uniform horizontal turntable rotates freely about a vertical axle at its center. You measure the radius of the turntable to be \(3.00 \mathrm{~m} .\) To determine the moment of inertia \(I\) of the turntable about the axle, you start the turntable rotating with angular speed \(\omega\), which you measure. You then drop a small object of mass \(m\) onto the rim of the turntable. After the object has come to rest relative to the turntable, you measure the angular speed \(\omega_{\mathrm{f}}\) of the rotating turntable. You plot the quantity \(\left(\omega-\omega_{\mathrm{f}}\right) / \omega_{\mathrm{f}}\) (with both \(\omega\) and \(\omega_{\mathrm{f}}\) in rad \(\left./ \mathrm{s}\right)\) as a function of \(m\) (in kg). You find that your data lie close to a straight line that has slope \(0.250 \mathrm{~kg}^{-1}\). What is the moment of inertia \(I\) of the turntable?

A thin-walled, hollow spherical shell of mass \(m\) and radius \(r\) starts from rest and rolls without slipping down a track (Fig. \(\mathbf{P 1 0 . 7 2}\) ). Points \(A\) and \(B\) are on a circular part of the track having radius \(R\). The diameter of the shell is very small compared to \(h_{0}\) and \(R,\) and the work done by rolling friction is negligible. (a) What is the minimum height \(h_{0}\) for which this shell will make a complete loop-the-loop on the circular part of the track? (b) How hard does the track push on the shell at point \(B,\) which is at the same level as the center of the circle? (c) Suppose that the track had no friction and the shell was released from the same height \(h_{0}\) you found in part (a). Would it make a complete loop-theloop? How do you know? (d) In part (c), how hard does the track push on the shell at point \(A,\) the top of the circle? How hard did it push on the shell in part (a)?

A uniform solid disk made of wood is horizontal and rotates freely about a vertical axle at its center. The disk has radius \(0.600 \mathrm{~m}\) and mass \(1.60 \mathrm{~kg}\) and is initially at rest. A bullet with mass \(0.0200 \mathrm{~kg}\) is fired horizontally at the disk, strikes the rim of the disk at a point perpendicular to the radius of the disk, and becomes embedded in its rim, a distance of \(0.600 \mathrm{~m}\) from the axle. After being struck by the bullet, the disk rotates at \(4.00 \mathrm{rad} / \mathrm{s}\). What is the horizontal velocity of the bullet just before it strikes the disk?

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