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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 .


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.


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

define realitives motion
Zahid Reply
Relative motion is the calculation of the motion of an object with regard to some other moving object. Thus, the motion is not calculated with reference to the earth, but is the velocity of the object in reference to the other moving object as if it were in a static state.
I am unable to access the mcq can someone help me with it?
Harkamal Reply
What is free fall?
Barham Reply
V=E½-P-½ where v; velocity, P; density and E; constant. Find dimension and it's units of E (constant)
michael Reply
derivation of simple harmonic equation
Daud Reply
if an equation is dimensionally correct does this mean that equation must be true?
michael Reply
how do I calculate angular velocity
Priscilla Reply
w=vr where w, angular velocity. v; velocity and r; radius of a circle
sorry I meant Maximum positive angular velocity of
please can u tell me the formular for tension in angular velocity I kind of forget it please don't ignore this msg I need it nw
Does the mass of the object affect the rate at which it accelerates ?
Can any one give me the definition for Bending moment plz...
Prema Reply
I need a question for moment
paul Reply
what is charge
An attribution of particle that we have thought about to explain certain things like Electomagnetism
Does the mass of the object affect the rate at which it accelerates ?
please what is the formula instantaneous velocity in projectile motion
Isaiah Reply
A computer is reading from a CD-ROM that rotates at 780 revolutions per minute.What is the centripetal acceleration at a point that is 0.030m from the center of the disc?
Rapqueen Reply
change revolution per minute by multiplying from 2pie and devide by 60.and take r=.030 and use formula centripital acceleration =omega sqare r.
OK thank you
observation of body boulded
Anwer Reply
a gas is compressed to 1/10 0f its original volume.calculate the rise temperature if the original volume is 400k. gamma =1.4
Celine Reply
the specific heat of hydrogen at constant pressure and temperature is 14.16kj|k.if 0.8kg of hydrogen is heated from 55 degree Celsius to 80 degree Celsius of a constant pressure. find the external work done .
what is imaginary mass and how we express is
Yash Reply
what is imaginary mass how we express it
centre of mass is also called as imaginary mass
l'm from Algeria and fell these can help me
Khlil Reply
Practice Key Terms 1

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