/*! 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 56 A \(\mathrm{A} 392\) -N wheel co... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(\mathrm{A} 392\) -N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.0 rad/s. The radius of the wheel is \(0.600 \mathrm{m},\) and its moment of inertia about its rotation axis is 0.800\(M R^{2} .\) Friction does work on the wheel as it rolls up the hill to a stop, a height \(h\) above the bottom of the hill; this work has absolute value 3500 J. Calculate \(h .\)

Short Answer

Expert verified
The height \( h \) is \( 0.255 \) m.

Step by step solution

01

Identify Known Values

Let's list the known values given in the problem:- Weight of the wheel, \(W = 392\, \text{N}\)- Rotational speed at the bottom, \(\omega = 25.0\, \text{rad/s}\)- Radius of the wheel, \(R = 0.600\, \text{m}\)- Moment of inertia, \(I = 0.800MR^2\)- Work done by friction, \(W_f = 3500\, \text{J}\).
02

Calculate the Mass of the Wheel

The mass of the wheel can be calculated using the weight and gravitational acceleration: \[ M = \frac{W}{g} = \frac{392}{9.8} = 40\, \text{kg} \].
03

Determine Rotational Kinetic Energy

Calculate the rotational kinetic energy at the bottom of the hill using the formula:\[ KE_{rot} = \frac{1}{2}I\omega^2 \].
04

Calculate the Moment of Inertia

Find the moment of inertia using the formula provided:\[ I = 0.800MR^2 = 0.800 \times 40 \times (0.600)^2 = 11.52\, \text{kg}\cdot\text{m}^2 \].
05

Compute Initial Rotational Kinetic Energy

Plug the values to find the initial rotational kinetic energy:\[ KE_{rot} = \frac{1}{2} \times 11.52 \times (25.0)^2 = 3600\, \text{J} \].
06

Use Energy Conservation Principle

Utilize conservation of energy, accounting for work done by friction:\[ PE_{top} = KE_{rot} - W_f \], where \(PE_{top}\) is the potential energy at the top of the hill.
07

Calculate Potential Energy at the Top

Find \(PE_{top}\) using:\[ PE_{top} = 3600 - 3500 = 100\, \text{J} \].
08

Calculate Height h

Use the potential energy formula \(PE = Mgh\) to find \(h\): \[ h = \frac{PE_{top}}{Mg} = \frac{100}{40 \times 9.8} = 0.255\, \text{m} \].

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Key Concepts

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

Rotational Kinetic Energy
Rotational kinetic energy is one form of energy in motion, typically observed in objects that spin or rotate. It is the rotational equivalent of linear kinetic energy. In our exercise, it's the energy the wheel possesses due to its rotation around its axis. Here's the formula to understand it better: \[ KE_{rot} = \frac{1}{2} I \omega^2 \]Where:
  • \( KE_{rot} \) is the rotational kinetic energy,
  • \( I \) is the moment of inertia, and
  • \( \omega \) is the angular velocity in radians per second.
The wheel at the bottom of the hill carries this energy, contributing to its ability to ascend until friction and gravity slow it down. Understanding how these variables interact provides insights into rotational dynamics.
Moment of Inertia
The moment of inertia is akin to the "rotational mass" of an object and it tells us how difficult it is to change its rotational motion. It depends on the mass of the object and how that mass is distributed relative to the axis of rotation. For our wheel, the formula provided is:\[ I = 0.800 M R^2 \]This means:
  • \( M \) is the mass of the wheel,
  • \( R \) is the radius, and
  • the 0.800 factor accounts for distribution specifics.
By calculating the moment of inertia, we determine how the wheel's mass and geometry affect its rotational motion. Having larger inertia means it will spin slower for the same amount of energy input, or it will require more energy to achieve the same rotational speed.
Work by Friction
Friction is the resistive force that occurs when two surfaces in contact move past each other. When it comes to the rolling wheel, friction does negative work by removing energy from the system. In our scenario, friction plays a key role as it dissipates some of the wheel's initial kinetic energy, converting it to other forms such as heat. The given work done by friction is:\[ W_f = 3500 \, \text{J} \]Why does this matter? Because it impacts how much energy is left for the wheel to climb uphill. By losing energy to friction, this loss must be considered when calculating potential energy gained by the wheel at the top of the hill.Understanding work by friction allows us to appreciate why rolling objects eventually stop — it's because energy isn't fully conserved in kinetic terms due to these resistive forces.

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Most popular questions from this chapter

\(\cdot\) A turntable that spins at a constant 78.0 rpm takes 3.50 s to reach this angular speed after it is turned on. Find (a) its angular acceleration (in \(\mathrm{rad} / \mathrm{s}^{2} ),\) assuming it to be constant, and (b) the number of degrees it turns through while speeding up.

\(\bullet\) (a) What angle in radians is subtended by an arc 1.50 \(\mathrm{m}\) in length on the circumference of a circle of radius 2.50 \(\mathrm{m} ?\) What is this angle in degrees? (b) An arc 14.0 \(\mathrm{cm}\) in length on the circumference of a circle subtends an angle of \(128^{\circ} .\) What is the radius of the circle? (c) The angle between two radii of a circle with radius 1.50 \(\mathrm{m}\) is 0.700 rad. What length of arc is intercepted on the circumference of the circle by the two radii?

\(\bullet\) (a) A cylinder 0.150 \(\mathrm{m}\) in diameter rotates in a lathe at 620 \(\mathrm{rpm} .\) What is the tangential speed of the surface of the cylinder? (b) The proper tangential speed for machining cast iron is about 0.600 \(\mathrm{m} / \mathrm{s} .\) At how many revolutions per minute should a piece of stock 0.0800 \(\mathrm{m}\) in diameter be rotated in a lathe to produce this tangential speed?

\(\bullet\) A uniform marble rolls down a symmetric bowl, starting from rest at the top of the left side. The top of each side is a distance \(h\) above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil. (a) How far up the smooth side will the marble go, measured vertically from the bottom? (b) How high would the marble go if both sides were as rough as the left side? (c) How do you account for the fact that the marble goes higher with friction on the right side than without friction?

Storing energy in flywheels. It has been suggested that we should use our power plants to generate energy in the off-hours (such as late at night) and store it for use during the day. One idea put forward is to store the energy in large fly wheels. Suppose we want to build such a flywheel in the shape of a hollow cylinder of inner radius 0.500 \(\mathrm{m}\) and outer radius \(1.50 \mathrm{m},\) using concrete of density \(2.20 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3} .\) (a) If, for stability, such a heavy flywheel is limited to 1.75 second for each revolution and has negligible friction at its axle, what must be its length to store 2.5 \(\mathrm{MJ}\) of energy in its rotational motion? (b) Suppose that by strengthening the frame you could safely double the flywheel's rate of spin. What length of flywheel would you need in that case? (Solve this part without reworking the entire problem!)

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