/*! 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 50 A 230 -kg beam \(2.7 \mathrm{~m}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 50

A 230 -kg beam \(2.7 \mathrm{~m}\) in length slides broadside down the ice with a speed of \(18 \mathrm{~m} / \mathrm{s}\) (Fig. 11-38). A 65-kg man at rest grabs one end as it goes past and hangs on as both he and the beam go spinning down the ice. Assume frictionless motion. ( \(a\) ) How fast does the center of mass of the system move after the collision? (b) With what angular velocity does the system rotate about its CM?

Short Answer

Expert verified
(a) The CM of the system moves at 14.03 m/s. (b) The angular velocity of the system is 21.55 rad/s.

Step by step solution

01

Identify the Given Variables

We are provided with the following:- Mass of the beam, \( m_b = 230 \text{ kg} \).- Length of the beam, \( L = 2.7 \text{ m} \).- Speed of the beam, \( v_b = 18 \text{ m/s} \).- Mass of the man, \( m_m = 65 \text{ kg} \).- Initial speed of the man, \( v_m = 0 \text{ m/s} \).The system consists of both the beam and the man after the collision.
02

Determine the Velocity of the Center of Mass

The velocity of the center of mass \( v_{\text{cm}} \) after the collision can be found using the conservation of linear momentum. The total initial momentum must equal the total final momentum. Therefore:\[ m_b v_b + m_m v_m = (m_b + m_m) v_{\text{cm}} \]Since the man is initially at rest, \( v_m = 0 \), so:\[ 230 \times 18 = (230 + 65) v_{\text{cm}} \]\[ 4140 = 295 v_{\text{cm}} \]\[ v_{\text{cm}} = \frac{4140}{295} \approx 14.03 \text{ m/s} \]
03

Calculate the Angular Momentum Before Collision

To determine the system's angular velocity, we first find the angular momentum before the collision. The angular momentum \( L \) around the center of mass can be calculated as:\[ L = m_b v_b \left(\frac{L}{2}\right) \]Where \( \frac{L}{2} \) is the distance from the center of mass of the beam to where the man grabs it:\[ L = 230 \times 18 \times \frac{2.7}{2} \]\[ L = 230 \times 18 \times 1.35 \]\[ L = 5589 \text{ kg m}^2/\text{s} \]
04

Calculate the Moment of Inertia

The moment of inertia \( I \) of the system around the center of mass can be calculated as:For the beam:\[ I_b = \frac{1}{12} m_b L^2 \]\[ I_b = \frac{1}{12} \times 230 \times 2.7^2 \]\[ I_b = 140.66 \text{ kg } \text{m}^2 \]For the man:\[ I_m = m_m \left(\frac{L}{2}\right)^2 \]\[ I_m = 65 \times 1.35^2 \]\[ I_m = 118.6125 \text{ kg } \text{m}^2 \]Total moment of inertia:\[ I = I_b + I_m = 140.66 + 118.6125 = 259.2725 \text{ kg } \text{m}^2 \]
05

Find the Angular Velocity

The angular velocity \( \omega \) of the combined system can be determined by using the conservation of angular momentum, where the total angular momentum before the collision equals the total angular momentum after:\[ L = I \omega \]\[ 5589 = 259.2725 \omega \]\[ \omega = \frac{5589}{259.2725} \approx 21.55 \text{ rad/s} \]

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 91影视!

Key Concepts

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

Center of Mass
The concept of the center of mass is crucial when analyzing the motion of a system involving multiple objects, such as a beam and a man.
The center of mass (CM) is the average position of all the mass in a system.
After the man grabs onto the beam, the combined system's center of mass changes in velocity based on the conservation of momentum.
To find the velocity of the center of mass (\( v_{\text{cm}} \)) after collision, we used the equation for conservation of linear momentum:
  • Before collision, only the beam has momentum because the man is at rest.
  • Total initial momentum: \( m_b v_b + m_m v_m \)
  • After the collision, the momentum is shared by both the man and beam, calculated by \( (m_b + m_m) v_{\text{cm}} \).
Recognizing the importance of the combined system's actions helps predict how entities interact post-collision.
This understanding of center of mass dynamics is key in physics problems involving collision and motion.
Moment of Inertia
The moment of inertia represents an object's resistance to changes in its rotational motion around a specific axis.
It's the rotational analog to mass in linear motion and depends on how an object's mass is distributed about the axis.
For calculating the moment of inertia of the combined system (beam and man):
  • The beam is considered to be a uniform rod rotating about its center, calculated using \( I_b = \frac{1}{12} m_b L^2 \).
  • The man, treated as a point mass at a distance \( \frac{L}{2} \), uses \( I_m = m_m \left(\frac{L}{2}\right)^2 \).
Adding these gives the total moment of inertia: \( I = I_b + I_m \).
This value is essential in calculating how the system will accelerate or decelerate in rotational terms, influencing the rotational speed after collision.
Angular Velocity
Angular velocity describes how fast an object rotates around a particular axis and is usually measured in radians per second.
For systems with rotational motion like the beam and man in this exercise, angular velocity (\( \omega \)) is vital for understanding post-collision dynamics.
We use the conservation of angular momentum to determine angular velocity:
  • Before the collision, only the beam has angular momentum, calculated by multiplying its linear momentum by the distance from the grab point \( \left(m_b v_b \frac{L}{2}\right) \).
  • After, we equate this to \( I \omega \) to solve for angular velocity.
This ensures the initial rotation energy has transitioned into the new system's dynamics, maintaining the conservation laws.
Understanding angular velocity aids in predicting the rotational behavior of objects in motion after sudden interactions like collisions.

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

(II) The origin of a coordinate system is at the center of a whecl which rotates in the \(x y\) plane about its axle which is the \(z\) axis. A force \(F=215 \mathrm{N}\) acts in the \(x y\) plane, at a \(+33.0^{\circ}\) angle to the \(x\) axis at the point \(x=28.0 \mathrm{cm}, y=33.5 \mathrm{cm}\) . Determine the magnitude and direction of the torque produced by this force about the axis.

The time-dependent position of a point object which moves counterclockwise along the circumference of a circle (radius \(R\) ) in the \(x y\) plane with constant speed \(v\) is given by $$ \overrightarrow{\mathbf{r}}=\hat{\mathbf{i}} R \cos \omega t+\hat{\mathbf{j}} R \sin \omega t $$ where the constant \(\omega=v / R .\) Determine the velocity \(\overrightarrow{\mathbf{v}}\) and angular velocity \(\overrightarrow{\boldsymbol{\omega}}\) of this object and then show that these three vectors obey the relation \(\overrightarrow{\mathbf{v}}=\overrightarrow{\boldsymbol{\omega}} \times \overrightarrow{\mathbf{r}}\).

(II) A potter's whecl is rotating around a vertical axis through its center at a frequency of 1.5 \(\mathrm{rev} / \mathrm{s}\) . The whecl can be considered a uniform disk of mass 5.0 \(\mathrm{kg}\) and diameter 0.40 \(\mathrm{m}\) . The potter then throws a 2.6 -kg chunk of clay, approximately shaped as a flat disk of radius \(8.0 \mathrm{cm},\) onto the center of the rotating whecl. What is the frequency of the whecl after the clay sticks to it?

Why might tall narrow SUVs and buses be prone to "rollover"? Consider a vehicle rounding a curve of radius \(R\) on a flat road. When just on the verge of rollover, its tires on the inside of the curve are about to leave the ground, so the friction and normal force on these two tires are zero. The total normal force on the outside tires is \(F_{\mathrm{N}}\) and the total friction force is \(F_{\mathrm{fr}}\). Assume that the vehicle is not skidding. (a) Analysts define a static stability factor SSF \(=w / 2 h\) where a vehicle's "track width" \(w\) is the distance between tires on the same axle, and \(h\) is the height of the CM above the ground. Show that the critical rollover speed is(b) Determine the ratio of highway curve radii (minimum possible) for a typical passenger car with \(\mathrm{SSF}=1.40\) and an SUV with \(\mathrm{SSF}=1.05\) at a speed of \(90 \mathrm{~km} / \mathrm{h}\). $$ v_{\mathrm{C}}=\sqrt{R g\left(\frac{w}{2 h}\right)} $$

A particle of mass \(1.00 \mathrm{~kg}\) is moving with velocity \(\overrightarrow{\mathbf{v}}=(7.0 \hat{\mathbf{i}}+6.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} . \quad(a)\) Find the angular momentum \(\overrightarrow{\mathbf{L}}\) relative to the origin when the particle is at \(\overrightarrow{\mathbf{r}}=(2.0 \hat{\mathbf{j}}+4.0 \hat{\mathbf{k}}) \mathrm{m} .(b)\) At position \(\overrightarrow{\mathbf{r}}\) a force of \(\overrightarrow{\mathbf{F}}=4.0 \mathrm{Ni} \hat{\mathbf{i}}\) is applied to the particle. Find the torque relative to the origin.
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Waves Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Radiation

Read Explanation

Translational Dynamics

Read Explanation

Fundamentals of Physics

Read Explanation
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.