




Check Your Understanding Which has greater angular momentum: a solid sphere of mass
m rotating at a constant angular frequency
${\omega}_{0}$ about the
z axis, or a solid cylinder of same mass and rotation rate about the
z axis?
${I}_{\text{sphere}}=\frac{2}{5}m{r}^{2},\phantom{\rule{0.5em}{0ex}}{I}_{\text{cylinder}}=\frac{1}{2}m{r}^{2}$ ; Taking the ratio of the angular momenta, we have:
$\frac{{L}_{\text{cylinder}}}{{L}_{\text{sphere}}}=\frac{{I}_{\text{cylinder}}{\omega}_{0}}{{I}_{\text{sphere}}{\omega}_{0}}=\frac{\frac{1}{2}m{r}^{2}}{\frac{2}{5}m{r}^{2}}=\frac{5}{4}$ . Thus, the cylinder has
$25\%$ more angular momentum. This is because the cylinder has more mass distributed farther from the axis of rotation.
Got questions? Get instant answers now!
Summary
 The angular momentum
$\overrightarrow{l}=\overrightarrow{r}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\overrightarrow{p}$ of a single particle about a designated origin is the vector product of the position vector in the given coordinate system and the particle’s linear momentum.
 The angular momentum
$\overrightarrow{l}={\displaystyle \sum _{i}{\overrightarrow{l}}_{i}}$ of a system of particles about a designated origin is the vector sum of the individual momenta of the particles that make up the system.
 The net torque on a system about a given origin is the time derivative of the angular momentum about that origin:
$\frac{d\overrightarrow{L}}{dt}={\displaystyle \sum \overrightarrow{\tau}}$ .
 A rigid rotating body has angular momentum
$L=I\omega $ directed along the axis of rotation. The time derivative of the angular momentum
$\frac{dL}{dt}={\displaystyle \sum \tau}$ gives the net torque on a rigid body and is directed along the axis of rotation.
Conceptual questions
For a particle traveling in a straight line, are there any points about which the angular momentum is zero? Assume the line intersects the origin.
All points on the straight line will give zero angular momentum, because a vector crossed into a parallel vector is zero.
Got questions? Get instant answers now!
If a particle is moving with respect to a chosen origin it has linear momentum. What conditions must exist for this particle’s angular momentum to be zero about the chosen origin?
The particle must be moving on a straight line that passes through the chosen origin.
Got questions? Get instant answers now!
Problems
A 0.2kg particle is travelling along the line
$y=2.0\phantom{\rule{0.2em}{0ex}}\text{m}$ with a velocity
$5.0\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}\text{s}$ . What is the angular momentum of the particle about the origin?
Got questions? Get instant answers now!
A bird flies overhead from where you stand at an altitude of 300.0 m and at a speed horizontal to the ground of 20.0 m/s. The bird has a mass of 2.0 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
xy plane. What is the bird’s angular momentum about the point where you are standing?
The magnitude of the cross product of the radius to the bird and its momentum vector yields
$rp\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta $ , which gives
$r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta $ as the altitude of the bird
h . The direction of the angular momentum is perpendicular to the radius and momentum vectors, which we choose arbitrarily as
$\widehat{k}$ , which is in the plane of the ground:
$\overrightarrow{L}=\overrightarrow{r}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\overrightarrow{p}=hmv\widehat{k}=(300.0\phantom{\rule{0.2em}{0ex}}\text{m})(2.0\phantom{\rule{0.2em}{0ex}}\text{kg})(20.0\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}\text{s})\widehat{k}=\mathrm{12,000.0}\phantom{\rule{0.2em}{0ex}}\text{kg}\xb7{\text{m}}^{2}\text{/}\text{s}\widehat{k}$
Got questions? Get instant answers now!
A Formula One race car with mass 750.0 kg is speeding through a course in Monaco and enters a circular turn at 220.0 km/h in the counterclockwise direction about the origin of the circle. At another part of the course, the car enters a second circular turn at 180 km/h also in the counterclockwise direction. If the radius of curvature of the first turn is 130.0 m and that of the second is 100.0 m, compare the angular momenta of the race car in each turn taken about the origin of the circular turn.
Got questions? Get instant answers now!
Questions & Answers
A spring with 50g mass suspended from it,has its length extended by 7.8cm
1.1 determine the spring constant?
1.2 it is observed that the length of the spring decreases by 4.7cm,from its original length, when a toy is place on top of it. what is the mass of the toy?
solution
mass = 50g= 0.05kg
force= 50 x 10= 500N
extension= 7.8cm = 0.078m
using the formula
Force= Ke
K = force/extension
500/.078
= 6410.25N/m
Sampson
1.2
Decrease in length= 4.7cm =0.047m
mass=?
acceleration due to gravity= 10
force = K x e
force= mass x acceleration
m x a = K x e
mass = K x e/acceleration
= 6410.25 x 0.047/10
= 30.13kg
Sampson
you should explain how yoy got the answer Anita
Grant
with the fomular F=mg I got the value for force because now the force acting on the spring is the weight of the object and also you have to convert from grams to kilograms and cm to meter
Anita
so the spring constant K=F/e
where
F is force and e is extension
Anita
In this first example why didn't we use P=P° + ¶hg
where ¶ is density
Density = force applied x area p=fA =p = mga, then a=h therefore substitute =p =mgh
Hlehle
sorry I had a little typo in my question
Anita
Density = m/v (mass/volume) simple as that
Augustine
Hlehle vilakazi how density is equal to force * area and you also wrote p= mgh which is machenical potential energy ? how ?
Manorama
who can state the third equation of motion
Alfred
wave is a distrubance that travelled in medium from one point to another with carry energy .
Manorama
wave is a periodic disturbance that carries energy from one medium to another..
Augustine
two particles rotate in a rigid body then acceleration will be ?
same acceleration for all particles because all prticles will be moving with same angular velocity.so at any time interval u find same acceleration of all the prticles
Zaheer
It is the study of the electromagnetic force, one of the four fundamental forces of nature. ... It includes the electric force, which pushes all charged particles, and the magnetic force, which only pushes moving charges.
Energy
What is th formular for force
State newton's second law of motion
can u tell me I cant remember
Indigo
force is equal to mass times acceleration
Santos
The acceleration of a system is directly proportional to the and in the same direction as the external force acting on the system and inversely proportional to its mass
that is f=ma
David
The uniform seesaw shown below is balanced on a fulcrum located 3.0 m from the left end. The smaller boy on the right has a mass of 40 kg and the bigger boy on the left has a mass 80 kg. What is the mass of the board?
Consider a wave produced on a stretched spring by holding one end and shaking it up and down. Does the wavelength depend on the distance you move your hand up and down?
no, only the frequency and the material of the spring
Chun
how to read physics ncert?
Tech
beat line read important. line under line
Rahul
how can one calculate the value of a given quantity
To determine the exact value of a percent of a given quantity we need to express the given percent as fraction and multiply it by the given number.
AMIT
briefly discuss rocket in physics
physics is the study of natural phenomena with concern with matter and energy and relationships between them
Ibrahim
a potential difference of 10.0v is connected across a 1.0AuF in an LC circuit. calculate the inductance of the inductor that should be connected to the capacitor for the circuit to oscillate at 1125Hza potential difference of 10.0v is connected across a 1.0AuF in an LC circuit. calculate the inducta
how did you get it?
Favour
is the magnetic field of earth changing
what is thought to be the energy density of multiverse and is the space between universes really space
tibebeab
Source:
OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
Google Play and the Google Play logo are trademarks of Google Inc.