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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 .


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.


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

how can chip be made from sand
Eke Reply
is this allso about nanoscale material
are nano particles real
Missy Reply
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
Lale Reply
no can't
where is the latest information on a no technology how can I find it
where we get a research paper on Nano chemistry....?
Maira Reply
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
what are the products of Nano chemistry?
Maira Reply
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Preparation and Applications of Nanomaterial for Drug Delivery
Hafiz Reply
Application of nanotechnology in medicine
has a lot of application modern world
what is variations in raman spectra for nanomaterials
Jyoti Reply
ya I also want to know the raman spectra
I only see partial conversation and what's the question here!
Crow Reply
what about nanotechnology for water purification
RAW Reply
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
yes that's correct
I think
Nasa has use it in the 60's, copper as water purification in the moon travel.
nanocopper obvius
what is the stm
Brian Reply
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
scanning tunneling microscope
how nano science is used for hydrophobicity
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
what is differents between GO and RGO?
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
analytical skills graphene is prepared to kill any type viruses .
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
what is Nano technology ?
Bob Reply
write examples of Nano molecule?
The nanotechnology is as new science, to scale nanometric
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
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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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