/*! 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 85 A girl of mass \(M\) stands on t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A girl of mass \(M\) stands on the rim of a frictionless merrygo-round of radius \(R\) and rotational inertia \(I\) that is not moving. She throws a rock of mass \(m\) horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is \(v\). Afterward, what are (a) the angular speed of the merry-go-round and (b) the linear speed of the girl?

Short Answer

Expert verified
(a) Angular speed \( \omega = \frac{m v R}{I + MR^2} \). (b) Linear speed \( v_g = \frac{m v R^2}{I + MR^2} \).

Step by step solution

01

Identify Concepts

This problem involves the conservation of angular momentum since no external torques act on the system of the girl, merry-go-round, and rock.
02

Write the Conservation Equation for Angular Momentum

Initially, the system has zero angular momentum as nothing is moving. After the rock is thrown, the angular momentum must still be zero, so the angular momentum of the rock and the angular momentum of the girl and the merry-go-round must cancel each other. Thus, \[ L_{ ext{rock}} + L_{ ext{merry-go-round and girl}} = 0 \]Given the rock's linear momentum, we can calculate its angular momentum as \[ L_{ ext{rock}} = m v R \].
03

Express Angular Speed of the Merry-Go-Round

Let the angular speed of the merry-go-round and girl together be \( \omega \). Since the merry-go-round and the girl rotate together, their combined rotational inertia is \( I + MR^2 \) due to the parallel axis theorem. The angular momentum for the merry-go-round and girl is \[ L_{ ext{merry-go-round and girl}} = (I + MR^2) \omega \].
04

Solve for Angular Speed of the Merry-Go-Round

From the conservation of angular momentum,\[ m v R = (I + MR^2) \omega \].Solving for \( \omega \), we have\[ \omega = \frac{m v R}{I + MR^2} \].
05

Determine Linear Speed of the Girl

The linear speed of the girl on the rim is related to the angular speed by\[ v_g = \omega R \].Substituting the expression for \( \omega \), we get\[ v_g = \frac{m v R^2}{I + MR^2} \].

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.

Rotational Inertia
Rotational inertia, also known as the moment of inertia, is a fundamental concept in physics that determines how difficult it is to change the rotational state of an object. It plays a similar role to mass in linear motion but applies to rotation. The rotational inertia of an object depends on both its mass and how that mass is distributed with respect to the axis of rotation.

For instance, in the problem involving the merry-go-round, the combined rotational inertia of the girl and the merry-go-round is calculated as \(I + MR^2\). Here, \(I\) refers to the merry-go-round's rotational inertia, while \(MR^2\) represents the additional inertia due to the girl's mass \(M\) at a distance \(R\) from the axis, applying the parallel axis theorem.
  • Greater mass or mass further from the axis increases rotational inertia.
  • An object with higher rotational inertia requires more torque to change its rotational speed.
Understanding rotational inertia helps in predicting how various systems will react to forces, particularly in rotational dynamics.
Angular Speed
Angular speed is the rate at which an object rotates or revolves around an axis. It is commonly denoted by \( \omega \) (omega) and is measured in radians per second. This concept is pivotal in understanding rotations and circular motion.

In the exercise, the angular speed of the merry-go-round \( \omega \) is calculated using the conservation of angular momentum, which states that the initial angular momentum of a system is equal to its final angular momentum, assuming no external torques. Initially, the system's angular momentum is zero. After the rock is thrown, the resulting equation \( m v R = (I + MR^2) \omega \) solves for the system's new angular speed as \( \omega = \frac{m v R}{I + MR^2} \).
  • Angular speed can be converted to linear speed by multiplying with the radius (for circular paths).
  • Angular speed remains constant if no external torques are applied.
This principle allows us to understand how the system reacts when internal components (like the rock in this scenario) are in motion.
Linear Speed
Linear speed refers to the distance an object travels per unit of time along a path. For a point on a rotating object, linear speed is directly related to its angular speed. It varies depending on the distance from the axis of rotation. Thus, linear speed is calculated by the formula: \( v = \omega R \), where \( v \) is the linear speed, \( \omega \) is the angular speed, and \( R \) is the radius from the rotation axis.

In the context of the girl on the merry-go-round, after the rock is thrown, the girl's linear speed \( v_g \) is determined by \( v_g = \omega R \). Plugging the value of \( \omega \) we found earlier: \( v_g = \frac{m v R^2}{I + MR^2} \). This relationship illustrates how even though the angular speed remains the same regardless of the distance for all points in the rigid body, the linear speed increases with the radius.
  • As the distance from the axis increases, linear speed also increases.
  • Linear speed tells us how fast a point on the outer edge moves through its path.
Such relationships are vital in understanding the dynamics of rotational systems, especially when translating rotational movement into linear speeds.

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 thin-walled pipe rolls along the floor. What is the ratio of its translational kinetic energy to its rotational kinetic energy about the central axis parallel to its length?

A \(2.50 \mathrm{~kg}\) particle that is moving horizontally over a floor with velocity \((-3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) undergoes a completely inelastic collision with a \(4.00 \mathrm{~kg}\) particle that is moving horizontally over the floor with velocity \((4.50 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}\). The collision occurs at \(x y\) coordinates \((-0.500 \mathrm{~m},-0.100 \mathrm{~m})\). After the collision and in unit- vector notation, what is the angular momentum of the stuck-together particles with respect to the origin?

A uniform disk of mass \(10 \mathrm{~m}\) and radius \(3.0 r\) can rotate freely about its fixed center like a merry-go-round. A smaller uniform disk of mass \(m\) and radius \(r\) lies on top of the larger disk, concentric with it. Initially the two disks rotate together with an angular velocity of \(20 \mathrm{rad} / \mathrm{s}\). Then a slight disturbance causes the smaller disk to slide outward across the larger disk, until the outer edge of the smaller disk catches on the outer edge of the larger disk. Afterward, the two disks again rotate together (without further sliding). (a) What then is their angular velocity about the center of the larger disk? (b) What is the ratio \(K / K_{0}\) of the new kinetic energy of the two-disk system to the system's initial kinetic energy?

At the instant the displacement of a \(2.00 \mathrm{~kg}\) object relative to the origin is \(\vec{d}=(2.00 \mathrm{~m}) \hat{\mathrm{i}}+(4.00 \mathrm{~m}) \hat{\mathrm{j}}-(3.00 \mathrm{~m}) \hat{\mathrm{k}}\), its velocity is \(\vec{v}=-(6.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}+(3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{k}}\) and it is subject to a force \(\vec{F}=(6.00 \mathrm{~N}) \hat{\mathrm{i}}-(8.00 \mathrm{~N}) \hat{\mathrm{j}}+(4.00 \mathrm{~N}) \hat{\mathrm{k}}\). Find (a) the acceleration of the object, (b) the angular momentum of the object about the origin, (c) the torque about the origin acting on the object, and (d) the angle between the velocity of the object and the force acting on the object.

A uniform solid sphere rolls down an incline. (a) What must be the incline angle if the linear acceleration of the center of the sphere is to have a magnitude of \(0.10 g ?\) (b) If a frictionless block were to slide down the incline at that angle, would its acceleration magnitude be more than, less than, or equal to \(0.10 \mathrm{~g}\) ? Why?

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.