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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

when I click on the links in the topics noting shows. what should I do.
Emmanuel Reply
can we regard torque as a force?
Emmanuel Reply
Torque is only referred a force to rotate objects.
I need lessons on Simple harmonic motion
what is the formulae for elastic modulus
Ark Reply
Given two vectors, vector C which is 3 units, and vector D which is 5 units. If the two vectors form an angle of 45o, determine C D and direction.
At time to = 0 the current to the DC motor is reverse, resulting in angular displacement of the motor shafts given by angle = (198rad/s)t - (24rad/s^2)t^2 - (2rad/s^3)t^3 At what time is the angular velocity of the motor shaft zero
Princston Reply
what is angular velocity
In three experiments, three different horizontal forces are ap- plied to the same block lying on the same countertop. The force magnitudes are F1 " 12 N, F2 " 8 N, and F3 " 4 N. In each experi- ment, the block remains stationary in spite of the applied force. Rank the forces according to (a) the
Given two vectors, vector C which is 3 units, and vector D which is 5 units. If the two vectors form an angle of 45o, determine C D and direction.
CD=5.83 n direction is NE
state Hooke's law of elasticity
Aarti Reply
Hooke's law states that the extension produced is directly proportional to the applied force provided that the elastic limit is not exceeded. F=ke;
You are welcome
what is drag force
A backward acting force that tends to resist thrust
solve:A person who weighs 720N in air is lowered in to tank of water to about chin level .He sits in a harness of negligible mass suspended from a scale that reads his apparent weight .He then dumps himself under water submerging his body .If his weight while submerged is 34.3N. find his density
Ian Reply
please help me solve this 👆👆👆
The weight inside the tank is lesser due to the buoyancy force by the water displaced. Weight of water displaced = His weight outside - his weight inside tank = 720 - 34.3 = 685.7N Now, the density of water = 997kg/m³ (this is a known value) Volume of water displaced = Mass/Density (next com)
density or relative density
Upthrust =720-34.3=685.7N mass of water displayed = 685.7/g vol of water displayed = 685.7/g/997 hence, density of man = 720/g / (685.7/g/997) =1046.6 kg/m3
R.d=weight in air/upthrust in water =720/34.3=20.99 R.d=density of substance/density of water 20.99=x/1 x=20.99g/cm^3
Kg /cubic meters
how please
Upthrust = 720-34.3=685.7N vol of water = 685.7/g/density of water = 685.7/g/997 so density of man = 720/g /(685.7/g/997) =1046.8 kg/m3
is there anyway i can see your calculations
Upthrust =720-34.3=685.7
Upthrust 720-34.3
Vol of water = 685.7/g/997
Hence density of man = 720/g / (685.7/g/997)
=1046.8 kg/m3
so the density of water is 997
Okay, thanks
try finding the volume then
Vol of man = vol of water displayed
I've done that; I got 0.0687m^3
okay i got it thanks
u welcome
HELLO kindly assist me on this...(MATHS) show that the function f(x)=[0 for xor=0]is continuous from the right of x->0 but not from the left of x->0
Duncan Reply
I do not get the question can you make it clearer
Same here, the function looks very ambiguous. please restate the question properly.
please help me solve this problem.a hiker begins a trip by first walking 25kmSE from her car.she stops and sets her tent for the night . on the second day, she walks 40km in a direction 60°NorthofEast,at which she discovers a forest ranger's tower.find components of hiker's displacement for each day
Liteboho Reply
Take a paper. put a point (name is A), now draw a line in the South east direction from A. Assume the line is 25 km long. that is the first stop (name the second point B) From B, turn 60 degrees to the north of East and draw another line, name that C. that line is 40 km long. (contd.)
Now, you know how to calculate displacements, I hope? the displacement between two points is the shortest distance between the two points. go ahead and do the calculations necessary. Good luck!
thank you so much Sharath Kumar
thank you, have also learned alot
No issues at all. I love the subject and teaching it is fun. Cheers!
cheers too
hii too
you mean
solution problems
what is the definition of model
matthew Reply
please is there any way that i can understand physics very well i know am not support to ask this kind of question....
prove using vector algebra that the diagonals of a rhombus perpendicular to each other.
Baijnath Reply
A projectile is thrown with a speed of v at an angle of theta has a range of R on the surface of the earth. For same v and theta,it's range on the surface of moon will be
Roshani Reply
what is soln..
Using some kinematics, time taken for the projectile to reach ground is (2*v*g*Sin (∆)) (here, g is gravity on Earth and ∆ is theta) therefore, on Earth, R = 2*v²*g*Sin(∆)*Cos(∆) on moon, the only difference is the gravity. Gravity on moon = 0.166*g substituting that value in R, we get the new R
Some corrections to my old post. Time taken to reach ground = 2*v*Sin (∆)/g R = (2*v²*Sin(∆)*Cos(∆))/g I put the g in the numerator by mistake in my old post. apologies for that. R on moon = (R on Earth)/(0.166)
state Newton's first law of motion
Awal Reply
Every body will continue in it's state of rest or of uniform motion in a straight line, unless it is compelled to change that state by an external force.
if you want this to become intuitive to you then you should state it
changing the state of rest or uniform motion of a body
if a body is in rest or motion it is always rest or motion, upto external force appied on it. it explains inertia
what is a vector
a ship move due north at 100kmhr----1 on a River flowing be due east on at 25kmperhr. cal the magnitude of the resultant velocity of the ship.
Emmanuel Reply
The result is a simple vector addition. The angle between the vectors is 90 degrees, so we can use Pythagoras theorem to get the result. V magnitude = sqrt(100*100 + 25*25) = 103.077 km/hr. the direction of the resultant vector can be found using trigonometry. Tan (theta) = 25/100.
state Newton's first law of motion
Kansiime Reply
An object continues to be in its state of rest or motion unless compelled by some external force
First law (law of inertia)- If a body is at rest, it would remain at rest and if the body is in the motion, it would be moving with the same velocity until or unless no external force is applied on it. If force F^=0 acceleration a^=0 or v^=0 or constant.
Practice Key Terms 1

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