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A \(31.4-\mathrm{kg}\) wheel with radius \(1.21 \mathrm{~m}\) is rotating at 283 rev/min. It must be brought to a stop in \(14.8 \mathrm{~s}\). Find the required average power. Assume the wheel to be a thin hoop.

Short Answer

Expert verified
The required average power to stop the wheel is 1125 Watts.

Step by step solution

01

Calculate the initial angular velocity

Firstly, we have to convert the rotational speed from rev/min to rad/sec. 1 rev = \(2π\) radian and 1 minute = 60 seconds, so the initial angular velocity ω = \(283 rev/min \times \frac{2π rad}{1 rev} \times \frac{1 min}{60 sec} = 29.6 rad/sec\)
02

Calculate the Moment of Inertia

The moment of inertia I for a hoop of mass m and radius r is \(I = m * r^2\). So, \(I = 31.4 kg * (1.1 m)^2 = 38.4 kg.m^2).
03

Calculate the initial Rotational Kinetic Energy

A body with moment of inertia I and angular velocity ω has the rotational kinetic energy KE = \(0.5 * I * ω^2\). So the initial rotational kinetic energy KE = \(0.5 * 38.4 kg.m^2 * (29.6 rad/sec)^2 = 16625 J\)
04

Calculate the required Power

The power required P to do work W in a time t is \(P = \frac{W}{t}\). In stopping the wheel, we do work equal to its initial rotational kinetic energy. So the required power P = \( \frac{16625 J}{14.8 s} = 1125 W\).

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Key Concepts

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

Angular Velocity
Angular velocity is a measure of how fast something rotates or spins. It's a key part of understanding rotational motion. Unlike linear velocity, which deals with straight-line movement, angular velocity deals with circular paths.
To convert rotational speed from revolutions per minute (rev/min) to radians per second (rad/sec), a two-step conversion is necessary. One revolution is equivalent to the angle of a full circle, which is 2π radians. Also, one minute equals 60 seconds.
The formula to find angular velocity is:
  • revolutions per minute to rad/sec: \[ \omega = \text{rev/min} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ sec}} \]
This change is important in physics because radians are a consistent unit that simplifies calculations with angular measurements. Knowing angular velocity allows us to understand the wheel's speed in radian terms, aiding further computation and predictions of rotational motion and energy.
Moment of Inertia
The moment of inertia is a property of rotating bodies that determines how difficult it is to change their rotational motion. It’s analogous to mass in linear motion. For different shapes, the moment of inertia changes based on how mass is distributed relative to the axis of rotation.
For a thin hoop, as in our wheel example, the moment of inertia (I) depends on:
  • Mass (m) of the hoop
  • Radius (r) of the hoop
The formula is:
  • Moment of Inertia for a hoop: \[ I = m \cdot r^2 \]
This shows that both mass and radius play crucial roles. As radius increases, the moment of inertia increases quadratically, meaning it becomes much harder to rotate the hoop. Hence, for objects like wheels, the moment of inertia crucially affects how they behave under forces, impacting the planning of energy needs and power use to stop or start motion.
Average Power
Average power is the rate at which work is done or energy is transferred over a period of time. It’s essential in understanding how much energy is needed to perform a task within a certain timeframe.
In the context of stopping a wheel, the energy needed is equal to its initial rotational kinetic energy, which is calculated from its angular velocity and moment of inertia. The formula for average power (P) is:
  • Average Power: \[ P = \frac{W}{t} \]
Where:
  • W is the work done or energy used (equal to the initial kinetic energy in this exercise)
  • t is the time duration
In practical terms, understanding average power helps to determine how much energy needs to be applied or dissipated consistently over time. This concept not only applies to mechanics but also extends into various fields such as electrical engineering and thermodynamics, highlighting its wide-ranging relevance.

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Most popular questions from this chapter

A fully loaded freight elevator has a total mass of \(1220 \mathrm{~kg}\). It is required to travel downward \(54.5 \mathrm{~m}\) in \(43.0 \mathrm{~s}\). The counterweight has a mass of \(1380 \mathrm{~kg}\). Find the power output, in \(\mathrm{hp}\), of the elevator motor. Ignore the work required to start and stop the elevator; that is, assume that it travels at constant speed.

How much power, in horsepower, must be developed by the engine of a \(1600-\mathrm{kg}\) car moving at \(26 \mathrm{~m} / \mathrm{s}(=94 \mathrm{~km} / \mathrm{h})\) on a level road if the forces of resistance total \(720 \mathrm{~N}\) ?

Two vectors, \(\overrightarrow{\mathbf{r}}\) and \(\overrightarrow{\mathbf{s}}\), lie in the \(x y\) plane. Their magnitudes are \(4.5\) and \(7.3\) units, respectively, whereas their directions are \(320^{\circ}\) and \(85^{\circ}\) measured counterclockwise from the positive \(x\) axis. What is the value of \(\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{s}}\) ?

A spring has a force constant of \(15.0 \mathrm{~N} / \mathrm{cm} .(a)\) How much work is required to extend the spring \(7.60 \mathrm{~mm}\) from its relaxed position? ( \(b\) ) How much work is needed to extend the spring an additional \(7.60 \mathrm{~mm}\) ?

Suppose that your car averages \(30 \mathrm{mi} / \mathrm{gal}\) of gasoline. \((a)\) How far could you travel on \(1 \mathrm{~kW} \cdot \mathrm{h}\) of energy consumed? \((b)\) If you are driving at \(55 \mathrm{mi} / \mathrm{h}\), at what rate are you expending energy? The heat of combustion of gasoline is \(140 \mathrm{MJ} / \mathrm{gal}\).

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