/*! 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 23 A solid sphere of radius \(R\) a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A solid sphere of radius \(R\) and mass \(M\) is placed at a height \(h_{0}\) on an inclined plane of slope \(\theta\). When released, it rolls without slipping to the bottom of the incline. Next, a cylinder of same mass and radius is released on the same incline. From what height \(h\) should it be released in order to have the same speed as the sphere at the bottom?

Short Answer

Expert verified
Answer: The cylinder should be released from a height of \(\frac{15}{14}h_0\), where \(h_0\) is the initial height of the solid sphere.

Step by step solution

01

Write the Conservation of Mechanical Energy for the Sphere

The total mechanical energy is the sum of gravitational potential energy and kinetic energy. For the sphere, the initial mechanical energy at height \(h_0\) is all gravitational potential energy. At the bottom of the incline, this potential energy has been converted into kinetic energy (both translational and rotational). The equation for the conservation of mechanical energy is: Initial Energy = Final Energy $$Mgh_0 = \frac{1}{2}Mv_{s}^2 + \frac{1}{2}I_{sphere}\omega^2$$
02

Find the Moment of Inertia and Angular Velocity for the Sphere and the Cylinder

Moment of inertia for a solid sphere is \(I_{sphere} = \frac{2}{5}MR^2\), and for a cylinder it is \(I_{cylinder} = \frac{1}{2}MR^2\). Since both objects are rolling without slipping, we can relate their linear and angular velocities by: \(v = R\omega\). We can substitute the moment of inertia and convert angular velocity to linear velocity: $$Mgh_0 = \frac{1}{2}Mv_{s}^2 + \frac{1}{2}\left(\frac{2}{5}MR^2\right)\left(\frac{v_{s}}{R}\right)^2$$
03

Simplify the Energy Conservation Equation for the Sphere

Now, we can simplify the equation for the sphere by cancelling out terms and collecting like terms: $$gh_0 = \frac{1}{2}v_{s}^2 + \frac{1}{5}v_{s}^2$$ Combine the fractions: $$gh_0 = \frac{7}{10}v_{s}^2$$ Solving for the final velocity of the sphere: $$v_{s} = \sqrt{\frac{10}{7}gh_0}$$
04

Write the Conservation of Mechanical Energy for the Cylinder

Now we need to repeat the process for the cylinder. The initial mechanical energy at height \(h\) is all gravitational potential energy. At the bottom of the incline, this potential energy has been converted into kinetic energy (both translational and rotational). The equation for the conservation of mechanical energy is: Initial Energy = Final Energy $$Mgh = \frac{1}{2}Mv_{c}^2 + \frac{1}{2}I_{cylinder}\omega^2$$ Substitute the moment of inertia and convert angular velocity to linear velocity: $$Mgh = \frac{1}{2}Mv_{c}^2 + \frac{1}{2}\left(\frac{1}{2}MR^2\right)\left(\frac{v_{c}}{R}\right)^2$$
05

Simplify the Energy Conservation Equation for the Cylinder

Now, we can simplify the equation for the cylinder by cancelling out terms and collecting like terms: $$gh = \frac{1}{2}v_{c}^2 + \frac{1}{4}v_{c}^2$$ Combine the fractions: $$gh = \frac{3}{4}v_{c}^2$$
06

Set the Sphere's and Cylinder's Final Velocities Equal

Since we want the cylinder to have the same speed as the sphere at the bottom, we can set their final velocities equal: $$v_{s} = v_{c}$$ $$\sqrt{\frac{10}{7}gh_0} = \sqrt{\frac{4}{3}gh}$$
07

Solve for the Cylinder's Release Height

Now, we can simplify the equation and solve for the cylinder's release height: $$\frac{10}{7}gh_0 = \frac{4}{3}gh$$ Dividing both sides by \(g\) and multiplying by \(\frac{3}{4}\), we find the value for \(h\): $$h = \frac{3}{4}\left(\frac{10}{7}h_0\right)$$ $$h = \frac{15}{14}h_0$$ So the cylinder should be released from a height of \(\frac{15}{14}h_0\) to have the same speed as the sphere at the bottom of the incline.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Rolling Motion
Rolling motion involves objects such as spheres and cylinders rolling down an incline without slipping. When something rolls without slipping, it means the point of contact on the object is momentarily at rest while the object rotates around it. This is crucial because it connects linear and rotational motion through the relationship:
  • Linear Velocity (\(v\)) relates to Angular Velocity (\(\omega\)) by: \(v = R\omega\). Here \(R\) is the radius of the object.
This relationship allows us to break down the motion into translational (straight-line) and rotational (spinning) kinetic energies. For instance, as a sphere rolls down an incline, its increased speed comprises both the speed at which its center moves and how fast it's spinning:
  • Translational Kinetic Energy: \(\frac{1}{2}Mv^2\)
  • Rotational Kinetic Energy: \(\frac{1}{2}I\omega^2\)
These are both components of the total kinetic energy we consider when applying the conservation of mechanical energy.
Moment of Inertia
Moment of inertia is a measure of an object's resistance to changes in its rotation. Think of it as rotational mass for simplicity. It's affected by the distribution of mass within an object and is calculated differently for different shapes:
  • For a solid sphere, \(I_{sphere} = \frac{2}{5}MR^{2}\)
  • For a solid cylinder, \(I_{cylinder} = \frac{1}{2}MR^{2}\)
These formulas reflect how mass distribution matters. A sphere has a smaller moment of inertia compared to a cylinder of the same mass and radius, meaning it's easier to start spinning and faster for a given incline height. Understanding moment of inertia is essential when calculating how fast an object will move or how it reacts to forces. It combines with angular velocity (\(\omega\)) to find rotational kinetic energy which changes how we apply the conservation of mechanical energy.
Inclined Plane
An inclined plane is a flat, sloped surface, and it's a classic setting in physics problems to study motion. As objects like spheres and cylinders roll down an incline, gravity does work on them, converting potential energy to kinetic energy. Here's how it works:
  • The slope angle (\(\theta\)) affects the component of gravitational force acting along the incline. Steeper slopes result in faster accelerations.
  • Potential Energy at height \(h\): \(Mg h\), where \(M\) is mass, and \(g\) is gravitational acceleration. This energy is what gets converted to kinetic energy at the bottom.
  • Kinetic Energy forms (both translational and rotational) increase as potential energy decreases, conserving total mechanical energy:
When tackling these problems, note that the height from which an object is released dictates its speed at the bottom. Different objects with different mass distributions may need to start from different heights to end at the same speed due to their moments of inertia. This is why the cylinder must start from a higher height (\(\frac{15}{14}h_0\)) to match the sphere's speed at the incline's bottom.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In another race, a solid sphere and a thin ring roll without slipping from rest down a ramp that makes angle \(\theta\) with the horizontal. Find the ratio of their accelerations, \(a_{\text {ring }} / a_{\text {sphere }}\)

A ball attached to the end of a string is swung in a vertical circle. The angular momentum of the ball at the top of the circular path is a) greater than the angular momentum at the bottom of the circular path. b) less than the angular momentum at the bottom of the circular path. c) the same as the angular momentum at the bottom of the circular path.

Consider a cylinder and a hollow cylinder, rotating about an axis going through their centers of mass. If both objects have the same mass and the same radius, which object will have the larger moment of inertia? a) The moment of inertia will be the same for both objects. b) The solid cylinder will have the larger moment of inertia because its mass is uniformly distributed. c) The hollow cylinder will have the larger moment of inertia because its mass is located away from the axis of rotation.

An oxygen molecule \(\left(\mathrm{O}_{2}\right)\) rotates in the \(x y\) -plane about the \(z\) -axis. The axis of rotation passes through the center of the molecule, perpendicular to its length. The mass of each oxygen atom is \(2.66 \cdot 10^{-26} \mathrm{~kg},\) and the average separation between the two atoms is \(d=1.21 \cdot 10^{-10} \mathrm{~m}\) a) Calculate the moment of inertia of the molecule about the \(z\) -axis. b) If the angular speed of the molecule about the \(z\) -axis is \(4.60 \cdot 10^{12} \mathrm{rad} / \mathrm{s},\) what is its rotational kinetic energy?

An ice skater rotating on frictionless ice brings her hands into her body so that she rotates faster. Which, if any, of the conservation laws hold? a) conservation of mechanical energy and conservation of angular momentum b) conservation of mechanical energy only c) conservation of angular momentum only d) neither conservation of mechanical energy nor conservation of angular momentum

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.