# 10.5 Angular momentum and its conservation  (Page 3/7)

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What we have here is, in fact, another conservation law. If the net torque is zero , then angular momentum is constant or conserved . We can see this rigorously by considering $\text{net}\phantom{\rule{0.25em}{0ex}}\tau =\frac{\text{Δ}L}{\text{Δ}t}$ for the situation in which the net torque is zero. In that case,

$\text{net}\tau =0$

implying that

$\frac{\text{Δ}L}{\text{Δ}t}=0.$

If the change in angular momentum $\text{Δ}L$ is zero, then the angular momentum is constant; thus,

$L=\text{constant}\phantom{\rule{0.25em}{0ex}}\left(\text{net}\phantom{\rule{0.25em}{0ex}}\tau =0\right)$

or

$L=L\prime \text{}\left(\text{net}\tau =0\right).$

These expressions are the law of conservation of angular momentum    . Conservation laws are as scarce as they are important.

An example of conservation of angular momentum is seen in [link] , in which an ice skater is executing a spin. The net torque on her is very close to zero, because there is relatively little friction between her skates and the ice and because the friction is exerted very close to the pivot point. (Both $F$ and $r$ are small, and so $\tau$ is negligibly small.) Consequently, she can spin for quite some time. She can do something else, too. She can increase her rate of spin by pulling her arms and legs in. Why does pulling her arms and legs in increase her rate of spin? The answer is that her angular momentum is constant, so that

$L=L\prime .$

Expressing this equation in terms of the moment of inertia,

$\mathrm{I\omega }=I\prime \omega \prime ,$

where the primed quantities refer to conditions after she has pulled in her arms and reduced her moment of inertia. Because $I\prime$ is smaller, the angular velocity $\omega \prime$ must increase to keep the angular momentum constant. The change can be dramatic, as the following example shows.

## Calculating the angular momentum of a spinning skater

Suppose an ice skater, such as the one in [link] , is spinning at 0.800 rev/ s with her arms extended. She has a moment of inertia of $2\text{.}\text{34}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot {\text{m}}^{2}$ with her arms extended and of $0\text{.}\text{363}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot {\text{m}}^{2}$ with her arms close to her body. (These moments of inertia are based on reasonable assumptions about a 60.0-kg skater.) (a) What is her angular velocity in revolutions per second after she pulls in her arms? (b) What is her rotational kinetic energy before and after she does this?

Strategy

In the first part of the problem, we are looking for the skater’s angular velocity $\omega \prime$ after she has pulled in her arms. To find this quantity, we use the conservation of angular momentum and note that the moments of inertia and initial angular velocity are given. To find the initial and final kinetic energies, we use the definition of rotational kinetic energy given by

${\text{KE}}_{\text{rot}}=\frac{1}{2}{\mathrm{I\omega }}^{2}.$

Solution for (a)

Because torque is negligible (as discussed above), the conservation of angular momentum given in $\mathrm{I\omega }=I\prime \omega \prime$ is applicable. Thus,

$L=L\prime$

or

$\mathrm{I\omega }=I\prime \omega \prime$

Solving for $\omega \prime$ and substituting known values into the resulting equation gives

$\begin{array}{lll}\omega \prime & =& \frac{I}{I\prime }\omega =\left(\frac{\text{2.34 kg}\cdot {m}^{2}}{0\text{.363 kg}\cdot {m}^{2}}\right)\left(\text{0.800 rev/s}\right)\\ & =& \text{}\text{5.16 rev/s.}\end{array}$

can some one tell me how v=RW is dimensionally correct?
ms-1 = m X Hz
babar
What is displacement
shortest distance b/w two points
bilal
distance+direction
A.d
explain distanace+direction
bilal
the change of postion from one point to another with direction
A.d
if we change thrle direction then displacement is destroy?
bilal
change the direction then?
bilal
what do u mean by i didnt understand bro
A.d
displacement is one dimension...?
bilal
displacement is the total length an object cover from initial to the final with respect to direction as Well as time.
mohammed
thanks
bilal
what are the differences between vector and scalar quantity
vector is assigned to those physical quantity that has both direction and magnitude! example velocity ,scalar just has magnitude example Mass of an object. hope it helps
Mudang
velocity is produce in fan...?
how many electrons are there in 5 microcouloumb
can a given total amount of mechanical energy be totally converted into heat energy..if so give example
human running
Emmanuel
what is the fumula for calculating specific heat capacity, fusion,fission and vaporization?
Q=cm(∆t)
Emmanuel
Q=cm∆T
what is difference b/w vaporization and evaporation
evaporation is the process of extracting moisture while vaporization is process of becoming a vapor or gas
Emmanuel
From a molecular standpoint they are both cooling processes. Also, you may want to explore states of matter😊 #myTwoCents ~Shi~
Shii
cooling is a similarlity in both process I am confused in difference
1- Evaporation is a process where a liquid change to gas without reaching its boiling point. 2- Vaporization is a process where a liquid change to gas after reaching its boiling point. 3- Sublimation is a process where a solid changes into vapour without passing through a liquid state
Victor
I see. Evaporation is a type of vaporization, that occurs on the surface of a liquid as it changes into the gaseous phase before reaching its boiling point. hope that aids
Shii
vaporisation is cooling process while vaporization is heating process
Emmanuel
I mean to write evaporation is an heating process while vaporization is cooling process
Emmanuel
Yea here are two applications. 1- your wet washed clothes dry under the sun, the water EVAPORATES 2- when u are cooking, it reaches a point where u need to add more water because the water you added previously is getting dried. this is VAPORIZATION. Am not sure which is a cooling or heating process
Victor
vaporization occur only when the evaporation get to level where the above cloud is been (saturated) so cooling take place and started to change to liquid (eg rain fall)
Emmanuel
They are both properties of the same process so they're both cooling
Shii
what about sublimation? cooling or heating process?
Victor
exact
evaporation is the increase in kinetic energy of the liquid which can be gone by adding heat
Emmanuel
so its an heating process
Emmanuel
sublimation is when a solid change to gas
Emmanuel
evaporation is very definitely a cooling process. respectfully@Emmanuel when liquid turns to gas it requires more energy from its surroundings, this energy is in the form of heat, and when heat energy leaves the evaporating liquid it leaves it cooler. Thus, cooling process.
Shii
.
Shii
evaporation is very definitely a cooling process. respectfully@Emmanuel
Shii
kk
Emmanuel
You're right @Shi. I get your point
Victor
eascape velocity on the surface of Earth is 11.2 kms-1 the escape velocity on the surface of another planet of same mass as that of Earth but of 1/4 times of radius of Earth is a5.6kms-1 b11.2 kms-1 c22.4kms-1 d5.6ms-1
Emm.. is that a question? or..
Victor
it is McQ
a)5.6km/s
Alvis
c= Q/cm◇T
A.d
units...
Shii
vital
Shii
the time period of the artificial satellite is given by ?
raza
Why is there no 2nd harmonic in the classical electron orbit?
how to reform magnet after been demagneted
A petrol engine has a output of 20 kilowatts and uses 4.5 kg of fuel for each hour of running. The energy given out when 1 kg of petrol is burnt is 4.8 × 10 to the power of 7 Joules. a) What is the energy output of the engine every hour? b) What is the energy input of the engine every hour?
Issac Newton devised a genius way to calculate changing quantities...
Shii
what is the error during taking work done of a body..
what kind of error do you think? and work is held by which force?
Daniela
I am now in this group
smart
theory,laws,principles and what-a-view are not defined. why? you
A simple pendulum is used in a physics laboratory experiment to obtain an experimental value for the gravitational acceleration, g . A student measures the length of the pendulum to be 0.510 meters, displaces it 10 o from the equilibrium position, and releases it. Using a s
so what question are you passing across... sir?
Olalekan
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed?
54 joule
babar
how?
rakesh
Reduce that two body problem into one body problem. Apply potential and k. E formula to get total energy of the system
rakesh
i dont think dere is any potential energy... by d virtue of no height present
Olalekan
there is compressed energy,dats only potential energy na?
rakesh
yes.. but... how will u approach that question without The Height in the question?
Olalekan
Can you explain how you get 54J?
Emmanuel
Because mine is 36J
Emmanuel
got 36J too
Douglas
OK the answer is 54J Babar is correct
Emmanuel
Conservation of Momentum
Emmanuel
woow i see.. can you give the formula for this
joshua
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed? Asume there is no external force.
Emmanuel
Inuwa
By using the Quotient Rule dy/dx = 3y/(x +y)²
Emmanuel
3y/(x+y)²
Emmanuel
may be by using MC^2=MC^2 and Total energy=kinetic energy +potential energy so 1st find kinetic energy and den find potential energy which is stored energy
rakesh
i think i m correct
rakesh
But how?
Emmanuel
3y/(x+y)²
Douglas