# 11.2 Angular momentum  (Page 4/8)

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## Angular momentum of a rigid body

We have investigated the angular momentum of a single particle, which we generalized to a system of particles. Now we can use the principles discussed in the previous section to develop the concept of the angular momentum of a rigid body. Celestial objects such as planets have angular momentum due to their spin and orbits around stars. In engineering, anything that rotates about an axis carries angular momentum, such as flywheels, propellers, and rotating parts in engines. Knowledge of the angular momenta of these objects is crucial to the design of the system in which they are a part.

To develop the angular momentum of a rigid body, we model a rigid body as being made up of small mass segments, $\text{Δ}{m}_{i}.$ In [link] , a rigid body is constrained to rotate about the z -axis with angular velocity $\omega$ . All mass segments that make up the rigid body undergo circular motion about the z -axis with the same angular velocity. Part (a) of the figure shows mass segment $\text{Δ}{m}_{i}$ with position vector ${\stackrel{\to }{r}}_{i}$ from the origin and radius ${R}_{i}$ to the z -axis. The magnitude of its tangential velocity is ${v}_{i}={R}_{i}\omega$ . Because the vectors ${\stackrel{\to }{v}}_{i}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{\stackrel{\to }{r}}_{i}$ are perpendicular to each other, the magnitude of the angular momentum of this mass segment is

${l}_{i}={r}_{i}\left(\text{Δ}m{v}_{i}\right)\text{sin}\phantom{\rule{0.2em}{0ex}}90\text{°}.$

Using the right-hand rule, the angular momentum vector points in the direction shown in part (b). The sum of the angular momenta of all the mass segments contains components both along and perpendicular to the axis of rotation. Every mass segment has a perpendicular component of the angular momentum that will be cancelled by the perpendicular component of an identical mass segment on the opposite side of the rigid body. Thus, the component along the axis of rotation is the only component that gives a nonzero value when summed over all the mass segments. From part (b), the component of ${\stackrel{\to }{l}}_{i}$ along the axis of rotation is

$\begin{array}{cc}\hfill {\left({l}_{i}\right)}_{z}& ={l}_{i}\text{sin}\phantom{\rule{0.2em}{0ex}}{\theta }_{i}=\left({r}_{i}\text{Δ}{m}_{i}{v}_{i}\right)\text{sin}\phantom{\rule{0.2em}{0ex}}{\theta }_{i},\hfill \\ & =\left({r}_{i}\text{sin}\phantom{\rule{0.2em}{0ex}}{\theta }_{i}\right)\left(\text{Δ}{m}_{i}{v}_{i}\right)={R}_{i}\text{Δ}{m}_{i}{v}_{i}.\hfill \end{array}$

The net angular momentum of the rigid body along the axis of rotation is

$L=\sum _{i}{\left({\stackrel{\to }{l}}_{i}\right)}_{z}=\sum _{i}{R}_{i}\text{Δ}{m}_{i}{v}_{i}=\sum _{i}{R}_{i}\text{Δ}{m}_{i}\left({R}_{i}\omega \right)=\omega {\sum _{i}\text{Δ}{m}_{i}\left({R}_{i}\right)}^{2}.$

The summation ${\sum _{i}\text{Δ}{m}_{i}\left({R}_{i}\right)}^{2}$ is simply the moment of inertia I of the rigid body about the axis of rotation. For a thin hoop rotating about an axis perpendicular to the plane of the hoop, all of the ${R}_{i}$ ’s are equal to R so the summation reduces to ${R}^{2}\sum _{i}\text{Δ}{m}_{i}=m{R}^{2},$ which is the moment of inertia for a thin hoop found in [link] . Thus, the magnitude of the angular momentum along the axis of rotation of a rigid body rotating with angular velocity $\omega$ about the axis is

$L=I\omega .$

This equation is analogous to the magnitude of the linear momentum $p=mv$ . The direction of the angular momentum vector is directed along the axis of rotation given by the right-hand rule.

what is matar
The uniform boom shown below weighs 500 N, and the object hanging from its right end weighs 400 N. The boom is supported by a light cable and by a hinge at the wall. Calculate the tension in the cable and the force on the hinge on the boom. Does the force on the hinge act along the boom?
A 11.0-m boom, AB , of a crane lifting a 3000-kg load is shown below. The center of mass of the boom is at its geometric center, and the mass of the boom is 800 kg. For the position shown, calculate tension T in the cable and the force at the axle A .
Jave
what is the S.I unit of coefficient of viscosity
Derived the formula of Newton's law of universal gravitation Fg=G(M1M2)/R2
a non-uniform boom of a crane 15m long, weighs 2800nts, with its center of gravity at 40% of its lenght from the hingr support. the boom is attached to a hinge at the lower end. rhe boom, which mAKES A 60% ANGLE WITH THE HORIZONTAL IS SUPPORTED BY A HORIZONTAL GUY WIRE AT ITS UPPER END. IF A LOAD OF 5000Nts is hung at the upper end of the boom, find the tension in the guywire and the components of the reaction at the hinge.
what is the centripetal force
Of?
John
centripetal force of attraction that pulls a body that is traversing round the orbit of a circle toward the center of the circle. Fc = MV²/r
Sampson
centripetal force is the force of attraction that pulls a body that is traversing round the orbit of a circle toward the center of the circle. Fc = MV²/r
Sampson
I do believe the formula for centripetal force is F=MA or F=m(v^2/r)
John
I mean the formula is Fc= Mass multiplied by square of velocity all over the Radius of the circle
Sampson
Yes
John
The force is equal to the mass times the velocity squared divided by the radius
John
That's the current chapter I'm on in my engineering physics class
John
Centripetal force is a force of attraction which keeps an object round the orbit towards the center of a circle. Mathematically Fc=mv²/r
In Example, we calculated the final speed of a roller coaster that descended 20 m in height and had an initial speed of 5 m/s downhill. Suppose the roller coaster had had an initial speed of 5 m/s uphill instead, and it coasted uphill, stopped, and then rolled back down to a final point 20 m bel
A steel lift column in a service station is 4 meter long and .2 meter in diameter. Young's modulus for steel is 20 X 1010N/m2.  By how much does the column shrink when a 5000- kg truck is on it?
hi
Abdulrahman
mola mass
Abdulrahman
what exactly is a transverse wave
does newton's first law mean that we don't need gravity to be attracted
no, it just means that a brick isn't gonna move unless something makes it move. if in the air, moves down because of gravity. if on floor, doesn't move unless something has it move, like a hand pushing the brick. first law is that an object will stay at rest or motion unless another force acts upon
Grant
yeah but once gravity has already been exerted .. i am saying that it need not be constantly exerted now according to newtons first law
Dharmee
gravity is constantly being exerted. gravity is the force of attractiveness between two objects. you and another person exert a force on each other but the reason you two don't come together is because earth's effect on both of you is much greater
Grant
maybe the reason we dont come together is our inertia only and not gravity
Dharmee
this is the definition of inertia: a property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force.
Grant
the earth has a much higher affect on us force wise that me and you together on each other, that's why we don't attract, relatively speaking of course
Grant
quite clear explanation but i just want my mind to be open to any theory at all .. its possible that maybe gravity does not exist at all or even the opposite can be true .. i dont want a fixed state of mind thats all
Dharmee
why wouldn't gravity exist? gravity is just the attractive force between two objects, at least to my understanding.
Grant
earth moves in a circular motion so yes it does need a constant force for a circular motion but incase of objects on earth i feel maybe there is no force of attraction towards the centre and its our inertia forcing us to stay at a point as once gravity had acted on the object
Dharmee
why should it exist .. i mean its all an assumption and the evidences are empirical
Dharmee
We have equations to prove it and lies of evidence to support. we orbit because we have a velocity and the sun is pulling us. Gravity is a law, we know it exists.
Grant
yeah sure there are equations but they are based on observations and assumptions
Dharmee
g is obtained by a simple pendulum experiment ...
Dharmee
gravity is tested by dropping a rock...
Grant
and also there were so many newtonian laws proved wrong by einstein . jus saying that its a law doesnt mean it cant be wrong
Dharmee
pendulum is good for showing energy transfer, here is an article on the detection of gravitational waves: ***ligo.org/detections.php
Grant
yeah but g is calculated by pendulum oscillations ..
Dharmee
thats what .. einstein s fabric model explains that force of attraction by sun on earth but i am talking about force of attraction by earth on objects on earth
Dharmee
no... this is how gravity is calculated:F = G*((m sub 1*m sub 2)/r^2)
Grant
gravitational constant is obtained EXPERIMENTALLY
Dharmee
the G part
Dharmee
Calculate the time of one oscillation or the period (T) by dividing the total time by the number of oscillations you counted. Use your calculated (T) along with the exact length of the pendulum (L) in the above formula to find "g." This is your measured value for "g."
Dharmee
G is the universal gravitational constant. F is the gravity
Grant
search up the gravity equation
Grant
yeahh G is obtained experimentally
Dharmee
sure yes
Grant
thats what .. after all its EXPERIMENTALLY calculated so its empirical
Dharmee
yes... so where do we disagree?
Grant
its empirical whixh means it can be proved wrong
Dharmee
so cant just say why wouldnt gravity exists
Dharmee
the constant, sure but extremely unlikely it is wrong. gravity however exists, there are equations and loads of support surrounding the concept. unfortunately I don't have a high enough background in physics but have this discussion with a physicist
Grant
can u suggest a platform where i can?
Dharmee
stack overflow
Grant
stack exchange, physics section***
Grant
its an app?
Dharmee
there is! it is also a website as well
Grant
okayy
Dharmee
nice talking to you
Dharmee
***physics.stackexchange.com/
Grant
likewise :)
Grant
Gravity surely exist
muhammed
hi guys
Diwash
hi
muhammed
what is mathematics
What is the percentage by massof oxygen in Al2(so4)3
molecular mass of Al2(SO4)3 = (27×2)+3{(32×1)+(16×4)} =54+3(32+64) =54+3×96 =54+288 =342 g/mol molecular mass of Oxygen=12×16 =192 g/mol % of Oxygen= (molecular mass of Oxygen/ molecular mass of the compound)×100% =(192/342)×100% =19200/342% =56.14%
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
1.1 6.28Nm-¹
Anita
1.2 0.03kg or 30g
Anita
I used g=9.8ms-²
Anita
you should explain how yoy got the answer Anita
Grant
ok
Anita
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
mass=50g=50/1000 kg m=0.05kg extension=7.8 cm=7.8/100 e=0.078 m g=9.8 m/s² 1.1 F=ke k=F/e k=mg/e k=0.05×9.8/0.078 k=0.49/0.078 k=6.28 N/m 1.2 F=6.28e mg=6.28e m=6.28e/g e=4.7 cm =4.7/100 e=0.047 m=6.28×0.047/9.8 m=0.29516/9.8 m=0.0301 kg
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
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
what is wave
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
what exactly is a transverse wave then?
Dharmee
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