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I = mr 2 + Mr 2 3 = m + M 3 r 2 . size 12{I'= ital "mr" rSup { size 8{2} } + { { ital "Mr" rSup { size 8{2} } } over {3} } = left (m+ { {M} over {3} } right )r rSup { size 8{2} } } {}

Entering known values in this equation yields,

I = 0 . 0500 kg + 0 . 667 kg 1 . 20 m 2 = 1 . 032 kg m 2 . size 12{I'= left (0 "." "0500"" kg"+0 "." "667 kg" right ) left (1 "." "20"" m" right ) rSup { size 8{2} } =1 "." "032"" kg" cdot m rSup { size 8{2} } } {}

The value of I is now entered into the expression for ω , which yields

ω = mvr I = 0 . 0500 kg 30 . 0 m/s 1 . 20 m 1 . 032 kg m 2 = 1 . 744 rad/s 1 . 74 rad/s .

Strategy for (b)

The kinetic energy before the collision is the incoming disk’s translational kinetic energy, and after the collision, it is the rotational kinetic energy of the two stuck together.

Solution for (b)

First, we calculate the translational kinetic energy by entering given values for the mass and speed of the incoming disk.

KE = 1 2 mv 2 = ( 0 . 500 ) 0 . 0500 kg 30 . 0 m/s 2 = 22.5 J size 12{"KE"= { {1} over {2} } ital "mv" rSup { size 8{2} } =0 "." "500" left (0 "." "0500"" kg" right ) left ("30" "." 0" m/s" right ) rSup { size 8{2} } ="22" "." 5" J"} {}

After the collision, the rotational kinetic energy can be found because we now know the final angular velocity and the final moment of inertia. Thus, entering the values into the rotational kinetic energy equation gives

KE′ = 1 2 I ω 2 = ( 0.5 ) 1.032 kg m 2 1 . 744 rad s 2 = 1.57 J.

Strategy for (c)

The linear momentum before the collision is that of the disk. After the collision, it is the sum of the disk’s momentum and that of the center of mass of the stick.

Solution of (c)

Before the collision, then, linear momentum is

p = mv = 0 . 0500 kg 30 . 0 m/s = 1 . 50 kg m/s . size 12{p= ital "mv"= left (0 "." "0500"" kg" right ) left ("30" "." 0" m/s" right )=1 "." "50"" kg" cdot "m/s"} {}

After the collision, the disk and the stick’s center of mass move in the same direction. The total linear momentum is that of the disk moving at a new velocity v = size 12{v'=rω'} {} plus that of the stick’s center of mass,

which moves at half this speed because v CM = r 2 ω = v 2 size 12{v rSub { size 8{"CM"} } = left ( { {r} over {2} } right )ω'= { {v'} over {2} } } {} . Thus,

p = mv + Mv CM = mv + Mv 2 . size 12{p'= ital "mv"'+ ital "Mv" rSub { size 8{"CM"} } '= ital "mv"'+ { { ital "Mv"'} over {2} } } {}

Gathering similar terms in the equation yields,

p = m + M 2 v size 12{p'= left (m+ { {M} over {2} } right )v'} {}

so that

p = m + M 2 . size 12{p'= left (m+ { {M} over {2} } right )rω'} {}

Substituting known values into the equation,

p = 1 . 050 kg 1 . 20 m 1 . 744 rad/s = 2.20 kg m/s .

Discussion

First note that the kinetic energy is less after the collision, as predicted, because the collision is inelastic. More surprising is that the momentum after the collision is actually greater than before the collision. This result can be understood if you consider how the nail affects the stick and vice versa. Apparently, the stick pushes backward on the nail when first struck by the disk. The nail’s reaction (consistent with Newton’s third law) is to push forward on the stick, imparting momentum to it in the same direction in which the disk was initially moving, thereby increasing the momentum of the system.

The above example has other implications. For example, what would happen if the disk hit very close to the nail? Obviously, a force would be exerted on the nail in the forward direction. So, when the stick is struck at the end farthest from the nail, a backward force is exerted on the nail, and when it is hit at the end nearest the nail, a forward force is exerted on the nail. Thus, striking it at a certain point in between produces no force on the nail. This intermediate point is known as the percussion point .

An analogous situation occurs in tennis as seen in [link] . If you hit a ball with the end of your racquet, the handle is pulled away from your hand. If you hit a ball much farther down, for example, on the shaft of the racquet, the handle is pushed into your palm. And if you hit the ball at the racquet’s percussion point (what some people call the “sweet spot”), then little or no force is exerted on your hand, and there is less vibration, reducing chances of a tennis elbow. The same effect occurs for a baseball bat.

In figure a, a disk hitting a stick is compared to a tennis ball being hit by a racquet. When the ball strikes the racquet near the end with a force denoted by f ball as shown by the direction of the arrow, a backward force, f hand is exerted on the hand, In figure b, when the racquet is struck much farther down by a force F ball, a forward force, f hand is exerted on the hand as shown by the arrows. In figure (c), when the racquet is struck by the ball with a force f ball at the percussion point, no force is delivered to the hand. This implies that f hand is equal to zero.
A disk hitting a stick is compared to a tennis ball being hit by a racquet. (a) When the ball strikes the racquet near the end, a backward force is exerted on the hand. (b) When the racquet is struck much farther down, a forward force is exerted on the hand. (c) When the racquet is struck at the percussion point, no force is delivered to the hand.

Is rotational kinetic energy a vector? Justify your answer.

No, energy is always scalar whether motion is involved or not. No form of energy has a direction in space and you can see that rotational kinetic energy does not depend on the direction of motion just as linear kinetic energy is independent of the direction of motion.

Section summary

  • Angular momentum L is analogous to linear momentum and is given by L = size 12{L=Iω} {} .
  • Angular momentum is changed by torque, following the relationship net τ = Δ L Δ t .
  • Angular momentum is conserved if the net torque is zero L = constant net τ = 0 or L = L net τ = 0 . This equation is known as the law of conservation of angular momentum, which may be conserved in collisions.

Problems&Exercises

Twin skaters approach one another as shown in [link] and lock hands. (a) Calculate their final angular velocity, given each had an initial speed of 2.50 m/s relative to the ice. Each has a mass of 70.0 kg, and each has a center of mass located 0.800 m from their locked hands. You may approximate their moments of inertia to be that of point masses at this radius. (b) Compare the initial kinetic energy and final kinetic energy.

Figure a shows two skaters from the top view approaching each other from opposite directions with velocity v. In figure b two skaters then lock their right hands and start to spin in the clockwise direction with angular velocity omega.
Twin skaters approach each other with identical speeds. Then, the skaters lock hands and spin.

(a) 3.13 rad/s

(b) Initial KE = 438 J, final KE = 438 J

Suppose a 0.250-kg ball is thrown at 15.0 m/s to a motionless person standing on ice who catches it with an outstretched arm as shown in [link] .

(a) Calculate the final linear velocity of the person, given his mass is 70.0 kg.

(b) What is his angular velocity if each arm is 5.00 kg? You may treat the ball as a point mass and treat the person's arms as uniform rods (each has a length of 0.900 m) and the rest of his body as a uniform cylinder of radius 0.180 m. Neglect the effect of the ball on his center of mass so that his center of mass remains in his geometrical center.

(c) Compare the initial and final total kinetic energies.

Figure a shows a skater through an overhead view with both his hands outstretched. A ball is seen approaching toward him in air with velocity v. Figure b shows that skater catching two balls in his left hand, and then, recoiling toward the left, in clockwise direction, with angular velocity omega and finally, the balls have velocity v prime.
The figure shows the overhead view of a person standing motionless on ice about to catch a ball. Both arms are outstretched. After catching the ball, the skater recoils and rotates.

Questions & Answers

do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
Akash Reply
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
Do somebody tell me a best nano engineering book for beginners?
s. Reply
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
SUYASH Reply
for screen printed electrodes ?
SUYASH
What is lattice structure?
s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
Sanket Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Unit 8 - rotational motion. OpenStax CNX. Feb 22, 2016 Download for free at https://legacy.cnx.org/content/col11970/1.1
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