Here we have used the definition of
$\overrightarrow{p}$ and the fact that a vector crossed into itself is zero. From Newton’s second law,
$\frac{d\overrightarrow{p}}{dt}={\displaystyle \sum \overrightarrow{F}},$ the net force acting on the particle, and the definition of the net torque, we can write
Note the similarity with the linear result of Newton’s second law,
$\frac{d\overrightarrow{p}}{dt}={\displaystyle \sum \overrightarrow{F}}$ . 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
Choose a coordinate system about which the angular momentum is to be calculated.
Write down the radius vector to the point particle in unit vector notation.
Write the linear momentum vector of the particle in unit vector notation.
Take the cross product
$\overrightarrow{l}=\overrightarrow{r}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\overrightarrow{p}$ and use the right-hand rule to establish the direction of the angular momentum vector.
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
$\frac{d\overrightarrow{l}}{dt}={\displaystyle \sum \overrightarrow{\tau}}$ 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
$\overrightarrow{r}=25\phantom{\rule{0.2em}{0ex}}\text{km}\widehat{i}+25\phantom{\rule{0.2em}{0ex}}\text{km}\widehat{j}$ gives the position of the meteor with respect to the observer. At the instant the observer sees the meteor, it has linear momentum
$\overrightarrow{p}=15.0\phantom{\rule{0.2em}{0ex}}\text{kg}(\mathrm{-2.0}\text{km}\text{/}\text{s}\widehat{j})$ , and it is accelerating at a constant
$2.0\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}{\text{s}}^{2}(\text{\u2212}\widehat{j})$ 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?
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
$90.0\text{\xb0}$ below the horizontal, so the components of the acceleration in the
x - and
y -directions are
This is the instant that the observer sees the meteor.
To find the torque, we take the time derivative of the angular momentum. Taking the time derivative of
$\overrightarrow{l}$ as a function of time, which is the second equation immediately above, we have
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
$\overrightarrow{\text{r}}$ 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
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
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
Sadiku
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.
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;
Shaibu
thanks
Aarti
You are welcome
Shaibu
thnx
Junaid
what is drag force
Junaid
A backward acting force that tends to resist thrust
Ian
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
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)
Sharath
density or relative density
Shaibu
density
Ian
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
1046.8
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
Shaibu
Kg /cubic meters
how please
Shaibu
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
Ian
Upthrust =720-34.3=685.7
Upthrust 720-34.3
=685.7N
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
Shaibu
Yes
Okay, thanks
Shaibu
try finding the volume then
Ian
Vol of man = vol of water displayed
I've done that; I got 0.0687m^3
Shaibu
okay i got it thanks
Ian
u welcome
Shaibu
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
Same here, the function looks very ambiguous. please restate the question properly.
Sharath
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
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.)
Sharath
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!
Sharath
thank you so much Sharath Kumar
Liteboho
thank you, have also learned alot
Duncan
No issues at all. I love the subject and teaching it is fun. Cheers!
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
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
Sharath
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)
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.
Kumaga
if you want this to become intuitive to you then you should state it
Shii
changing the state of rest or uniform motion of a body
koffi
if a body is in rest or motion it is always rest or motion, upto external force appied on it. it explains inertia
Omsai
what is a vector
smith
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.
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.
An object continues to be in its state of rest or motion unless compelled by some external force
Alem
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.