/*! 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 4 Which rolls down an inclined pla... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Which rolls down an inclined plane faster, a hollow cylinder or a solid sphere? Both have the same mass and radius.

Short Answer

Expert verified
The solid sphere rolls down the inclined plane faster than the hollow cylinder.

Step by step solution

01

Determine the Moment of Inertia

The moment of Inertia for a hollow cylinder and a solid sphere are \(I_{cylinder} = MR^2\) and \(I_{sphere} = \frac{2}{5}MR^2\), respectively. M is mass, and R is the radius, which are the same for both objects.
02

Calculate the acceleration

We need to compute the acceleration of each object down the plane. The acceleration for an object rolling down a slope without slipping is given by \(a = \frac{g \sin(θ)}{1+I/(MR^2)}\), where θ is the angle of the slope, g is the acceleration due to gravity, and I is the moment of inertia. Substituting the moments of inertia from Step 1 will give us the accelerations \(a_{cylinder}\) and \(a_{sphere}\) for the cylinder and the sphere respectively.
03

Compare the accelerations

After obtaining the two accelerations, comparison can be made. Since the denominator in the acceleration formula for the sphere is greater than that for the cylinder (due to the moment of inertia being lower for the sphere), the acceleration \(a_{sphere}\) is greater than \(a_{cylinder}\). Consequently, the solid sphere rolls down faster than the hollow cylinder.

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.

Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotational motion. It plays a vital role in rotational dynamics, akin to how mass is related to linear motion. Understanding the concept of the moment of inertia is crucial for predicting how an object will behave when it starts to rotate, accelerates, or slows down.

For instance, in the exercise provided, we deal with a hollow cylinder and a solid sphere. The moment of inertia for any given object depends on its mass distribution relative to the rotation axis. For the hollow cylinder, the formula is \(I_{cylinder} = MR^2\), and for the solid sphere, it is \(I_{sphere} = \frac{2}{5}MR^2\). Here, 'M' stands for mass and 'R' for radius. Since the mass and radius are identical for both objects, it's the difference in the distribution of mass that results in different moments of inertia.

In simple terms, because the mass of a solid sphere is distributed closer to its axis of rotation in comparison to a hollow cylinder, its moment of inertia is less, meaning it will be easier to get rolling and change its rate of spin.
Rolling Without Slipping
Rolling without slipping is a term used to describe a common type of motion where an object, like a wheel or ball, rolls on a surface without any sliding. It is a condition where the rotational motion of the object is perfectly matched with its translational motion. In other words, at any instant, a point on the object in contact with the ground is momentarily at rest with respect to the ground.

The concept of rolling without slipping is essential when analyzing problems involving rotational motion. This is because it allows us to relate the translational speed of the object's center to its angular speed. For a rolling object, the condition is mathematically expressed as \(v = \omega R\), where 'v' is the linear speed of the center of the object, 'ω' (omega) is the angular speed, and 'R' is the radius of the object.

When applying this to our exercise, it's implied that both the hollow cylinder and the solid sphere are rolling without slipping as they descend the inclined plane. Utilizing this understanding, we can compute the acceleration of the objects using the formula given.
Acceleration Due to Gravity
Acceleration due to gravity, often denoted as 'g', is the acceleration that is imparted to objects due to the gravitational force of the Earth. This acceleration is what causes freely falling objects to speed up as they descend towards the Earth's surface. At sea level, it has an average value of approximately 9.81 m/s^2, depending on location.

This acceleration is always directed downward, towards the center of the Earth. In the case of our exercise, the acceleration due to gravity affects not just the rate at which the objects fall but also influences their rolling motion down the incline. By applying the formula \(a = \frac{g \sin(\theta)}{1+I/(MR^2)}\), where 'θ' represents the incline's angle, we can calculate how gravity, combined with the object's moment of inertia, determines the acceleration of the sphere and the cylinder as they roll down the incline.

Because 'g' is constant for both objects, it's the moment of inertia portion of the formula that results in the solid sphere, with its lower moment of inertia, accelerating faster than the hollow cylinder on the same incline.

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

A bird flies overhead from where you stand at an altitude of \(300.0 \mathrm{m}\) and at a speed horizontal to the ground of \(20.0 \mathrm{m} / \mathrm{s} .\) The bird has a mass of \(2.0 \mathrm{kg}\). The radius vector to the bird makes an angle \(\theta\) with respect to the ground. The radius vector to the bird and its momentum vector lie in the \(x y\) -plane. What is the bird's angular momentum about the point where you are standing?

Gyroscopes used in guidance systems to indicate directions in space must have an angular momentum that does not change in direction. When placed in the vehicle, they are put in a compartment that is separated from the main fuselage, such that changes in the orientation of the fuselage does not affect the orientation of the gyroscope. If the space vehicle is subjected to large forces and accelerations how can the direction of the gyroscopes angular momentum be constant at all times?

(a) What is the angular momentum of the Moon in its orbit around Earth? (b) How does this angular momentum compare with the angular momentum of the Moon on its axis? Remember that the Moon keeps one side toward Earth at all times.

The centrifuge at NASA Ames Research Center has a radius of \(8.8 \mathrm{m}\) and can produce forces on its payload of 20 \(g\) s or 20 times the force of gravity on Earth. (a) What is the angular momentum of a \(20-\mathrm{kg}\) payload that experiences 10 \(g\) s in the centrifuge? (b) If the driver motor was turned off in (a) and the payload lost \(10 \mathrm{kg}\), what would be its new spin rate, taking into account there are no frictional forces present?

A roller coaster has mass \(3000.0 \mathrm{kg}\) and needs to make it safely through a vertical circular loop of radius \(50.0 \mathrm{m}\). What is the minimum angular momentum of the coaster at the bottom of the loop to make it safely through? Neglect friction on the track. Take the coaster to be a point particle.

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.