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A cylindrical hoop, a cylinder, and a sphere of equal radius and mass are released at the same time from the top of an inclined plane. Using the conservation of mechanical energy, show that the sphere always gets to the bottom of the incline first with the fastest speed and that the hoop always arrives last with the slowest speed.

Short Answer

Expert verified
The sphere reaches the bottom first, followed by the cylinder, and finally the hoop because their velocities are in the order: sphere > cylinder > hoop.

Step by step solution

01

Understand the Problem

We have three objects: a cylindrical hoop, a cylinder, and a sphere. They have equal radius and mass, and are released from the same height on an inclined plane. We need to show that due to conservation of energy, the sphere reaches the bottom first, and the hoop reaches last.
02

Conservation of Mechanical Energy

For each object, apply the conservation of mechanical energy. The total mechanical energy at the top (potential energy) is converted to kinetic energy (translational + rotational) at the bottom:\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]where \( m \) is the mass, \( g \) is the gravitational acceleration, \( h \) is the height of the incline, \( v \) is the linear velocity, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity (\( \omega = \frac{v}{r} \), \( r \) is the radius).
03

Calculate Moment of Inertia for Each Object

Moment of inertia is different for each object:- Hoop: \( I = mr^2 \)- Cylinder: \( I = \frac{1}{2}mr^2 \)- Sphere: \( I = \frac{2}{5}mr^2 \)
04

Substitute Moment of Inertia into Energy Equation

Substitute the moment of inertia values into the energy equations for each object:- Hoop: \[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} (mr^2) \left(\frac{v}{r}\right)^2 = mv^2 \]- Cylinder: \[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{2}mr^2\right) \left(\frac{v}{r}\right)^2 = \frac{3}{4}mv^2 \]- Sphere: \[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5}mr^2\right) \left(\frac{v}{r}\right)^2 = \frac{7}{10}mv^2 \]
05

Solve for Velocity v

Solve each equation for \( v \):- Hoop: \[ v = \sqrt{gh} \]- Cylinder: \[ v = \sqrt{\frac{4gh}{3}} \]- Sphere: \[ v = \sqrt{\frac{10gh}{7}} \]
06

Compare Velocities

Compare the magnitudes:- For the hoop: \( v = \sqrt{gh} \)- For the cylinder: \( v = \sqrt{\frac{4gh}{3}} \)- For the sphere: \( v = \sqrt{\frac{10gh}{7}} \)Since \( \sqrt{\frac{10}{7}} > \sqrt{\frac{4}{3}} > 1 \), the sphere reaches fastest.

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Key Concepts

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

Moment of Inertia
The moment of inertia is a crucial concept in understanding rotational motion. It is a measure of an object's resistance to changes in its rotational motion. The formula for moment of inertia varies depending on the shape and mass distribution of the object.
For instance:
  • **Cylindrical Hoop:** Its moment of inertia is calculated as \( I = mr^2 \). This means the mass is concentrated farther from the axis of rotation, providing higher inertia.
  • **Cylinder:** The formula here is \( I = \frac{1}{2}mr^2 \), indicating that the mass is distributed more evenly around the axis.
  • **Sphere:** It has \( I = \frac{2}{5}mr^2 \), which signifies that it has the smallest moment of inertia among the three, allowing it to accelerate faster in rotational motion.
The smaller the moment of inertia, the easier it is for the object to rotate. This is why the sphere reaches the bottom of the incline first; its inertia is smaller, making it roll faster.
Rotational Motion
Rotational motion describes an object rotating around an axis. When objects roll down an incline, they simultaneously move linearly and rotate. The speed of this motion depends on both linear velocity and angular velocity.
The relationship between the two is given by \( \omega = \frac{v}{r} \), where \( \omega \) is the angular velocity, \( v \) is the linear velocity, and \( r \) is the radius of the object.
The kinetic energy of rotating objects is the sum of translational kinetic energy \( \left( \frac{1}{2}mv^2 \right) \) and rotational kinetic energy \( \left( \frac{1}{2}I\omega^2 \right) \).
  • Hoop: More energy is used in rotational motion due to higher inertia, resulting in a slower descent.
  • Cylinder: Balanced energy distribution between linear and rotational motion.
  • Sphere: Least rotational inertia leads to faster linear acceleration.
Understanding rotational motion clarifies why the sphere, with its efficient energy conversion, races ahead of the cylinder and hoop.
Conservation of Energy
The principle of conservation of energy states that energy in a closed system remains constant. For our inclined plane scenario, potential energy is converted to kinetic energy as objects roll down.
The energy equation is \( mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \), indicating that all potential energy \( mgh \) at the start transforms into kinetic energy at the bottom of the incline.
Different forms of kinetic energy are affected by inertia:
  • **Hoop:** Its higher inertia means more energy goes into rotation, reducing its speed.
  • **Cylinder:** A fair split between rotational and linear energy, making it moderately fast.
  • **Sphere:** The smallest moment of inertia allows more energy for linear motion, increasing speed.
This principle explains why objects, despite starting with the same energy, reach different speeds based on their shape and mass distribution.
Inclined Plane
An inclined plane is a simple physics concept used to analyze forces and motion. In this context, it's a flat surface tilted at an angle, causing objects to experience gravitational force components along the plane.
As objects descend the plane under gravity, their potential energy \( mgh \) is converted to kinetic energy.
The angle and friction do not change among the objects here, making inertia the main differentiator. However:
  • Smoother planes, or less friction, allow faster speeds.
  • A steeper incline increases the component of gravitational force along the plane, speeding up descent for all objects.
Understanding how an inclined plane works, combined with kinetics, provides insights into how different objects roll and accelerate. This helps clarify why, depending on their shape, objects like a hoop, cylinder, and sphere behave differently when let loose on such a surface.

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Most popular questions from this chapter

A planetary space probe is in the shape of a cylinder. To protect it from heat on one side (from the Sun's rays), operators on the Earth put it into a "barbecue mode," that is, they set it rotating about its long axis. To do this, they fire four small rockets mounted tangentially as shown in \(v\) Fig. 8.51 (the probe is shown coming toward you). The object is to get the probe to rotate completely once every \(30 \mathrm{~s}\), starting from no rotation at all. They wish to do this by firing all four rockets for a certain length of time. Each rocket can exert a thrust of \(50.0 \mathrm{~N}\). Assume the probe is a uniform solid cylinder with a radius of \(2.50 \mathrm{~m}\) and a mass of \(1000 \mathrm{~kg}\) and neglect the mass of each rocket engine. Determine the amount of time the rockets need to be fired.

A flat, solid cylindrical grinding wheel with a diameter of \(20.2 \mathrm{~cm}\) is spinning at 3000 rpm when its power is suddenly turned off. A workman continues to press his tool bit toward the wheel's center at the wheel's circumference so as to continue to grind as the wheel coasts to a stop. If the wheel has a moment of inertia of \(4.73 \mathrm{~kg} \cdot \mathrm{m}^{2}\), (a) determine the necessary torque that must be exerted by the workman to bring it to rest in \(10.5 \mathrm{~s}\). Ignore any friction at the axle. (b) If the coefficient of kinetic friction between the tool bit and the wheel surface is \(0.85,\) how hard must the workman push on the bit?

A fixed 0.15-kg solid-disk pulley with a radius of \(0.075 \mathrm{~m}\) is acted on by a net torque of \(6.4 \mathrm{~m} \cdot \mathrm{N}\). What is the angular acceleration of the pulley?

A stationary ice skater with a mass of \(80.0 \mathrm{~kg}\) and a moment of inertia (about her central vertical axis) of \(3.00 \mathrm{~kg} \cdot \mathrm{m}^{2}\) catches a baseball with her outstretched arm. The catch is made at a distance of \(1.00 \mathrm{~m}\) from the central axis. The ball has a mass of \(145 \mathrm{~g}\) and is traveling at \(20.0 \mathrm{~m} / \mathrm{s}\) before the catch. (a) What linear speed does the system (skater \(+\) ball) have after the catch? (b) What is the angular speed of the system (skater \(+\) ball) after the catch? (c) What percentage of the ball's initial kinetic energy is lost during the catch? Neglect friction with the ice.

A flat cylindrical grinding wheel is spinning at 2000 rpm (clockwise when viewed head-on) when its power is suddenly turned off. Normally, if left alone, it takes 45.0 s to coast to rest. Assume the grinder has a moment of inertia of \(2.43 \mathrm{~kg} \cdot \mathrm{m}^{2}\). (a) Determine its angular acceleration during this process. (b) Determine the tangential acceleration of a point on the grinding wheel if the wheel is \(7.5 \mathrm{~cm}\) in diameter. (c) The slowing down is caused by a frictional torque on the axle of the wheel. The axle is \(1.00 \mathrm{~cm}\) in diameter. Determine the frictional force on the axle. (d) How much work was done by friction on the system?

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