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Here we have used the definition of p and the fact that a vector crossed into itself is zero. From Newton’s second law, d p d t = F , the net force acting on the particle, and the definition of the net torque, we can write

d l d t = τ .

Note the similarity with the linear result of Newton’s second law, d p d t = F . The following problem-solving strategy can serve as a guideline for calculating the angular momentum of a particle.

Problem-solving strategy: angular momentum of a particle

  1. Choose a coordinate system about which the angular momentum is to be calculated.
  2. Write down the radius vector to the point particle in unit vector notation.
  3. Write the linear momentum vector of the particle in unit vector notation.
  4. Take the cross product l = r × p and use the right-hand rule to establish the direction of the angular momentum vector.
  5. See if there is a time dependence in the expression of the angular momentum vector. If there is, then a torque exists about the origin, and use d l d t = τ to calculate the torque. If there is no time dependence in the expression for the angular momentum, then the net torque is zero.

Angular momentum and torque on a meteor

A meteor enters Earth’s atmosphere ( [link] ) and is observed by someone on the ground before it burns up in the atmosphere. The vector r = 25 km i ^ + 25 km j ^ gives the position of the meteor with respect to the observer. At the instant the observer sees the meteor, it has linear momentum p = 15.0 kg ( −2.0 km / s j ^ ) , and it is accelerating at a constant 2.0 m / s 2 ( j ^ ) along its path, which for our purposes can be taken as a straight line. (a) What is the angular momentum of the meteor about the origin, which is at the location of the observer? (b) What is the torque on the meteor about the origin?

An x y coordinate system is shown, with positive x to the right, along the ground, and positive y vertically upward. An observer is shown near the origin. A vector r is shown from the origin to a meteor at some large positive x and positive y coordinates. The vector p at the location of the meteor points down.
An observer on the ground sees a meteor at position r with linear momentum p .

Strategy

We resolve the acceleration into x - and y -components and use the kinematic equations to express the velocity as a function of acceleration and time. We insert these expressions into the linear momentum and then calculate the angular momentum using the cross-product. Since the position and momentum vectors are in the xy -plane, we expect the angular momentum vector to be along the z -axis. To find the torque, we take the time derivative of the angular momentum.

Solution

The meteor is entering Earth’s atmosphere at an angle of 90.0 ° below the horizontal, so the components of the acceleration in the x - and y -directions are

a x = 0 , a y = −2.0 m / s 2 .

We write the velocities using the kinematic equations.

v x = 0 , v y = −2.0 × 10 3 m / s ( 2.0 m / s 2 ) t .
  1. The angular momentum is
    l = r × p = ( 25.0 km i ^ + 25.0 km j ^ ) × 15.0 kg ( 0 i ^ + v y j ^ ) = 15.0 kg [ 25.0 km ( v y ) k ^ ] = 15.0 kg[ 2.50 × 10 4 m ( −2.0 × 10 3 m / s ( 2.0 m / s 2 ) t ) k ^ ] .

    At t = 0 , the angular momentum of the meteor about the origin is
    l 0 = 15.0 kg [ 2.50 × 10 4 m ( −2.0 × 10 3 m / s ) k ^ ] = 7.50 × 10 8 kg · m 2 / s ( k ^ ) .

    This is the instant that the observer sees the meteor.
  2. To find the torque, we take the time derivative of the angular momentum. Taking the time derivative of l as a function of time, which is the second equation immediately above, we have
    d l d t = −15.0 kg ( 2.50 × 10 4 m ) ( 2.0 m / s 2 ) k ^ .

    Then, since d l d t = τ , we have
    τ = −7. 5 × 10 5 N · m k ^ .

    The units of torque are given as newton-meters, not to be confused with joules. As a check, we note that the lever arm is the x -component of the vector r in [link] since it is perpendicular to the force acting on the meteor, which is along its path. By Newton’s second law, this force is
    F = m a ( j ^ ) = 15.0 kg ( 2.0 m / s 2 ) ( j ^ ) = 30.0 kg · m / s 2 ( j ^ ) .

    The lever arm is
    r = 2.5 × 10 4 m i ^ .

    Thus, the torque is
    τ = r × F = ( 2.5 × 10 4 m i ^ ) × ( −30.0 kg · m / s 2 j ^ ) , = 7.5 × 10 5 N · m ( k ^ ) .

Questions & Answers

what is electromagnetism
David Reply
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 units?
Subhajit Reply
units as in how
praise
What is th formular for force
Joseph Reply
F = m x a
Santos
State newton's second law of motion
Seth Reply
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?
Asad Reply
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?
Sohail Reply
how can one calculate the value of a given quantity
Helen Reply
means?
Manorama
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
meaning
Winford
briefly discuss rocket in physics
Ibrahim Reply
ok let's discuss
Jay
What is physics
Nura Reply
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
Royalty Reply
L= 0.002H
NNAEMEKA
how did you get it?
Favour
is the magnetic field of earth changing
tibebeab Reply
what is thought to be the energy density of multiverse and is the space between universes really space
tibebeab
can you explain it
Guhan
Energy can not either created nor destroyed .therefore who created? and how did it come to existence?
Suzana Reply
this greatly depend on the kind of energy. for gravitational energy, it is result of the shattering effect violent collision of two black holes on the space-time which caused space time to be disturbed. this is according to recent study on gravitons and gravitational ripple. and many other studies
tibebeab
and not every thing have to pop into existence. and it could have always been there . and some scientists think that energy might have been the only entity in the euclidean(imaginary time T=it) which is time undergone wick rotation.
tibebeab
What is projectile?
Nana Reply
An object that is launched from a device
Grant
2 dimensional motion under constant acceleration due to gravity
Awais
Not always 2D Awais
Grant
no comments
Awais
why not? a bullet is a projectile, so is a rock I throw
Grant
bullet travel in x and y comment same as rock which is 2 dimensional
Awais
components
Awais
no all pf you are wrong. projectile is any object propelled through space by excretion of a force which cease after launch
tibebeab
for awais, there is no such thing as constant acceleration due to gravity, because gravity change from place to place and from different height
tibebeab
it is the object not the motion or its components
tibebeab
where are body center of mass on present.
Balwant Reply
on the mid point
Suzana
is the magnetic field of the earth changing?
tibebeab
does shock waves come to effect when in earth's inner atmosphere or can it have an effect on the thermosphere or ionosphere?
tibebeab
and for the question from bal want do you mean human body or just any object in space
tibebeab
A stone is dropped into a well of 19.6m deep and the impact of sound heared after 2.056 second ,find the velocity of sound in air.
Sisco Reply
9.53 m/s ?
Kyla
In this case, the velocity of sound is 350 m/s.
Zahangir
why?
Kyla
some calculations is need. then you will get exact result.
Zahangir
i mean how? isn't it just a d over t?
Kyla
calculate the time it takes the stone to hit the ground then minus the stone's time to the total time... then divide the total distance by the difference of the time
Snuggly
awit lenard. Hahahah ari ga to!
Kyla
Practice Key Terms 1

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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