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
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.
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?
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?
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
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.
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