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QIC S A solid uniform sphere of mass \(m\) and radius \(R\) rolls without slipping down an incline of height \(h\). (a) What forms of mechanical energy are associated with the sphere at any point along the incline when its angular speed is \(\omega\) ? Answer in words and symbolically in terms of the quantities \(m, g, y, I, \omega\), and \(v\). (b) What force acting on the sphere causes it to roll rather than slip down the incline? (c) Determine the ratio of the sphere's rotational kinetic energy to its total kinetic energy at any instant.

Short Answer

Expert verified
The sphere has three forms of mechanical energy: linear (Kinetic) \(\frac{1}{2}mv^2\), rotational (Kinetic) \(\frac{1}{2}I\omega^2\), and gravitational potential energy \(mgy\). Friction is the force allowing the sphere to roll. The ratio of the rotational kinetic energy (\(\frac{1}{5}mv^2\)) to the total kinetic energy (\(\frac{7}{10}mv^2\)) is \(2/7\) at any instant.

Step by step solution

01

Identifying mechanical energy forms

The sphere possesses two forms of mechanical energy - kinetic and potential. Kinetic energy itself is twofold: linear kinetic energy and rotational kinetic energy. The linear kinetic energy is given by \(\frac{1}{2}mv^2\) and rotational kinetic energy by \(\frac{1}{2}I\omega^2\), where \(I\) is the moment of inertia of the sphere and \(\omega\) its angular velocity. The potential energy is given by \(mgy\), where \(y\) is the height above the base of the incline.
02

Identifying the locomotion force

The force allowing the sphere to roll rather than slip is the static friction force. Static friction provides torque for the sphere to rotate while moving down the incline without slipping.
03

Computing the ratio of rotational to total kinetic energy

For the sphere, the moment of inertia \(I\) is given by \(\frac{2}{5}mR^2\). When the sphere rolls without slipping, the linear velocity \(v\) and angular velocity \(\omega\) relate as \(v = \omega R\). Therefore, the rotational kinetic energy is \(\frac{1}{2}I\omega^2 = \frac{1}{2} *\frac{2}{5}mR^2*\omega^2 = \frac{1}{5}m\omega^2R^2 = \frac{1}{5}mv^2\). The total kinetic energy is the sum of linear and rotational kinetic energy, which is \(\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2\). So, the ratio of rotational kinetic energy to total kinetic energy is \(\frac{1}{5}mv^2 / \frac{7}{10}mv^2 = 2/7\). This ratio is constant at any instant.

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Key Concepts

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

Rotational Kinetic Energy
When an object like a solid sphere rotates, it possesses a specific kind of kinetic energy due to this motion called rotational kinetic energy. This energy is analogous to the kinetic energy of an object that moves in a linear path but applies to rotation.

For an object with a moment of inertia, represented by the symbol 'I', and an angular speed 'ω' (omega), the expression for the rotational kinetic energy is given by the formula: \[\frac{1}{2}I\omega^2\].

In the context of the exercise involving a sphere descending an incline, we apply this concept to understand that as the sphere gains speed down the hill, the energy associated with its spinning - or rotational kinetic energy - increases. Since the sphere rolls without slipping, we also know that there is a direct relationship between its rotational speed and its linear speed. Understanding the distribution of kinetic energy between translation (linear motion) and rotation is fundamental in physics, particularly when analyzing rolling objects.
Potential Energy in Physics
Potential energy is the energy stored by an object due to its position relative to a force. In physics, when we consider a height 'h' in a gravitational field represented by the acceleration due to gravity 'g', the potential energy, often gravitational potential energy, is given by the equation: \[mgh\].

In our textbook exercise, the sphere starts at a certain height on an incline, indicating that it has potential energy at the beginning of its motion. As the sphere rolls down the incline, this potential energy is converted into kinetic energy — both linear and rotational. The potential energy directly relates to how far an object can fall under the influence of gravity, and in this case, 'y' represents the vertical distance above the base of the incline. This potential-to-kinetic energy transformation is a vivid illustration of the conservation of mechanical energy in action.
Static Friction Force
Static friction force is the force that resists the start of sliding motion between two surfaces that are in contact. It is the frictional force that acts when an object is at rest or when there is no relative motion between the surfaces.

In our sphere on an incline scenario, the static friction force is crucial because it prevents slipping; it allows the sphere to roll, matching linear and angular accelerations. It does this by applying a torque at the point of contact with the surface, which results in the sphere spinning rather than just sliding down. Essentially, static friction is what enables the rotational kinetic energy to manifest in rolling objects be it wheels, balls, or cylinders, making it intrinsic to understanding rolling motion in physics.

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Most popular questions from this chapter

W A potter's wheel having a radius of \(0.50 \mathrm{~m}\) and a moment of inertia of \(12 \mathrm{~kg} \cdot \mathrm{m}^{2}\) is rotating freely at \(50 \mathrm{rev} / \mathrm{min}\). The potter can stop the wheel in \(6.0 \mathrm{~s}\) by pressing a wet rag against the rim and exerting a radially inward force of \(70 \mathrm{~N}\). Find the effective coefficient of kinetic friction between the wheel and the wet rag.

A large grinding wheel in the shape of a solid cylinder of radius \(0.330 \mathrm{~m}\) is free to rotate on a frictionless, vertical axle. A constant tangential force of \(250 \mathrm{~N}\) applied to its edge causes the wheel to have an angular acceleration of \(0.940 \mathrm{rad} / \mathrm{s}^{2}\). (a) What is the moment of inertia of the wheel? (b) What is the mass of the wheel? (c) If the wheel starts from rest, what is its angular velocity after \(5.00 \mathrm{~s}\) have elapsed, assuming the force is acting during that time?

An airliner lands with a speed of \(50.0 \mathrm{~m} / \mathrm{s}\). Each wheel of the plane has a radius of \(1.25 \mathrm{~m}\) and a moment of inertia of \(110 \mathrm{~kg} \cdot \mathrm{m}^{2}\). At touchdown, the wheels begin to spin under the action of friction. Each wheel supports a weight of \(1.40 \times 10^{4} \mathrm{~N}\), and the wheels attain their angular speed in \(0.480 \mathrm{~s}\) while rolling without slipping. What is the coefficient of kinetic friction between the wheels and the runway? Assume that the speed of the plane is constant.

A car is designed to get its energy from a rotating flywheel with a radius of \(2.00 \mathrm{~m}\) and a mass of \(500 \mathrm{~kg}\). Before a trip, the flywheel is attached to an electric motor, which brings the flywheel's rotational speed up to \(5000 \mathrm{rev} / \mathrm{min}\). (a) Find the kinetic energy stored in the flywheel. (b) If the flywheel is to supply energy to the car as a \(10.0\)-hp motor would, find the length of time the car could run before the flywheel would have to be brought back up to speed.

A light rod of length \(\ell=1.00 \mathrm{~m}\) rotates about an axis perpendicular to its length and passing through its center as in Figure P8.45. Two particles of masses \(m_{1}=4.00 \mathrm{~kg}\) and \(m_{2}=3.00 \mathrm{~kg}\) are connected to the ends of the rod. (a) Neglecting the mass of the rod, what is the system's kinetic energy when its angular speed is \(2.50 \mathrm{rad} / \mathrm{s}\) ? (b) Repeat the problem, assuming the mass of the rod is taken to be \(2.00 \mathrm{~kg}\).

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