Angular momentum in case of rotation about a fixed axis - class-XI

By pulling her arms and legs inward, the skater decreases her moment of inertia. Since external torque is zero, conservation of angular momentum requires the rotational speed to increase, which in turn increases the rotational kinetic energy due to work done by the skater.

Angular momentum is conserved when the net external torque on a system is zero. Torque, not force or pressure, is what changes angular momentum. The absence of net external torque means angular momentum remains constant.

The law of conservation of linear momentum is generally true for isolated systems, but in the presence of external forces (which are ubiquitous), it is not always conserved for a specific body. However, in the context of fundamental physics, all listed conservation laws are highly robust, but linear momentum is often cited as the one most easily 'broken' by external forces.

Energy E = -13.6 / n^2 eV. Given E = -0.544 eV, n^2 = 13.6 / 0.544 = 25, so n = 5. Bohr's quantization condition is L = n * h / (2 * pi). For n = 5, L = 5h / (2 * pi).

In a rotational motion, only the angular parameters ($\theta, w,\alpha$) are same but not the linear parameters.

When the arms are stretched the rotational inertia is higher. When the man folds his arm he is basically decreasing his rotational inertia thereby gaining more angular velocity using the principle of conservation of angular momentum. This implies more angular distance is covered in the same time giving us positive work.

Angular momentum of planet revolving around the sun is constant because the line of force of the gravitational force passes through the  sun . Therefore torque about the sun is zero.
$\therefore$  Angular momentum will remain constant.  And there is no other force acting on the planet, which results into zero tangential force.

Newton's 2nd law says that External Force is directly proportional to the rate of change of momentum. Similarly, external torque is directly proportional to the rate of change of angular momentum. So if angular momentum is conserved i.e., the rate of change of angular momentum is zero then external torque is zero.

$E=\dfrac{-13.6}{n^{2}}eV$

Under the action of a central force, there is a conservation ofAngular momentum and mechanical energy.

The angular momentum of a single particle is $mw^2r$. If there are 4 particles, we can write,
$L _{tot}=mw^2r+2mw^2(2r)+3mw^2(3r)+4mw^2(4r)=30mw^2r$

The correct option is (c)

For two particles at distance r from the z-axis, 180 degrees apart, the angular momentum about the origin (z-axis) is L = 2 * mvr = 2 * mr(rw) = 2mr^2w.

For uniform circular motion, the angular momentum L = r x p is constant only when the origin is at the center of the circle, as r and p are perpendicular and their magnitudes are constant.

Kepler's second law states that the radius vector sweeps out equal areas in equal times. February 2012 had 29 days. The area swept in 7 days is 7/29 of the total area.

Given,

L = m * v * d, where d is the perpendicular distance from the origin to the line y = x + 4 (or x - y + 4 = 0). d = |0 - 0 + 4| / sqrt(1^2 + (-1)^2) = 4 / sqrt(2) = 2 * sqrt(2). L = 5 * (3 * sqrt(2)) * (2 * sqrt(2)) = 5 * 3 * 2 * 2 = 60.

Radius of the nth orbit r_n ∝ n²/Z      ⇒    n ∝ (r n)½

Angular momentum L_n = nh/(2π) ∝ (r n)½

$=r^{1/2}$

According to Bohr modal of hydrogen like atoms, negatively charged electrons revolves around the positively charged nucleus. This uniform circular motion of electrons is equivalent to a current loop which possesses a magnetic dipole moment =IA

I:- current in the loop

A:-area

Consider an electron revolving anticlockwise around a nucleus in an orbit of radius r with speed v and time period T.

Equivalent current,

I= charge/time

=e/t = e/2πr/v

=ev/2πr

Area of current loop ,A=πr²

So , orbital magnetic moment of electron is=IA

=(ev/2πr). πr²

=evr/2

As J is angular momentum. J=mvr

m- mass of electron

So orbital magnetic moment = eJ/2m

In the Bohr model, E is proportional to 1/n^2 and J (angular momentum) is proportional to n. Therefore, E is proportional to 1/J^2.

According to conservation of Angular Momentum,
$I \omega$ = constant
Hence, when a person sitting on a rotating stool suddenly  lowers his hands, then his angular velocity will increase and moment of inertia will decrease. 

The moment of inertia can be increased or angular velocity can be decreased by stretching hands outside. Also, smaller the moment of inertia means less resistance to rotation (some rapid corrections have to made to achieve rotational equilibrium), hence change in moment of inertia leads the change in rotational kinetic energy. While angular momentum remains conserved.

The change in angular momentum is equal to the impulse of the torque, which is torque multiplied by time (delta L = tau * delta t).

Angular momentum is defined as the cross product of the position vector and linear momentum. For a mass rotating in a plane, the direction of this vector is determined by the right-hand rule, which is always perpendicular to the plane of rotation.

Angular momentum is conserved when the net external torque acting on a system is zero. This is the rotational analog to the conservation of linear momentum, which occurs when the net external force is zero.

By conservation of angular momentum, I1 * w1 = I2 * w2. Given I2 = 0.6 * I1 and w1 = 120 rpm, we have 120 * I1 = w2 * 0.6 * I1, which simplifies to w2 = 120 / 0.6 = 200 rpm.

When a ball strikes a rod on a smooth table, there is an impulsive reaction force at the point of contact. Angular momentum is conserved about any point where the net external torque is zero; for the end point, the impulsive reaction force at the pivot (if it were fixed) or the specific geometry makes the calculation valid.

For two masses M revolving about their center of mass at distance r/2, the orbital velocity v is found from GM^2 / r^2 = M * v^2 / (r/2). Thus v^2 = GM / 2, so v = sqrt(GM / 2). Angular momentum L = M * v * (r/2) = M * sqrt(GM / 2) * (r/2) = (1/2) * sqrt(GM^3 * r^2 / 2) = (1/2) * sqrt(GM^3 * r / 2).

If a rigid body is in translational equilibrium, the net force is zero, but the angular velocity can be constant or changing depending on the net torque. The statement that angular velocity must be constant is incorrect.

$K.E.$ for a given momentum is inversely proportional to the mass$.$

When the person contracts his hands he is decreasing his moment of inertia thereby increasing his angular velocity using the principle of conservation of angular momentum $I _1\omega _1=I _2\omega _2$
As the angular velocity increases, its kinetic energy also increases. Also energy would be required to draw the weights inwards from its position.

Moment = F * d_perpendicular. The force is vertical, so the perpendicular distance is the difference in x-coordinates: |2 - 1| = 1 m. 20 Nm = F * 1 m, so F = 20 N.

As the distance between the stone and the rotational axis is changing the moment of inertia thus changes. Momentum keeps on changing around the circular path. As the moment of inertia  changes we see that the kinetic energy changes too. The angular momentum is the only conserved quantity since there is no external torque acting on the stone.

by pulling his arms and legs in he decreases his moment of inertia, thereby increasing his angular velocity using the principle of conservation of angular momentum.

No force means no acceleration implying constant velocity. Also as velocity is constant kinetic energy remains constant. The angular momentum may change due to the change in moment of inertia. Also the total mechanical energy may change due the change in potential energy.

As the particle moves on a circular path, its position vector and linear momentum vector change directions, but if we consider angular momentum about the center of the circular path, its magnitude changes due to decreasing speed, but its direction (perpendicular to the plane of the circle) remains constant.

L = r x p = m * (r x v). r = i + j, v = 2i - 2j + 2k. r x v = (i + j) x (2i - 2j + 2k) = (i x 2i) - (i x 2j) + (i x 2k) + (j x 2i) - (j x 2j) + (j x 2k) = 0 - 2k - 2j - 2k - 0 + 2i = 2i - 2j - 4k. L = 2 * (2i - 2j - 4k) = 4i - 4j - 8k. The z-component is -8.

If we apply a torque on a body, then angular momentum of the body changes according to the relation
$\bar{\tau} = \displaystyle\frac{d\bar{L}}{dt} \Rightarrow if \space \bar{\tau} = 0$ then, $\overrightarrow{L} = constant$

Since no external torque act on gymnast, so angular momentum $(L =

I\omega)$ is conserved. After pulling her arms & legs, the angular velocity increases but moment of inertia of gymnast, decreases in, such a way that angular momentum remains constant.

The boy does not exert a torque to rotating table by jumping, so angular momentum is conserved i.e., 
$\displaystyle\frac{d\bar{L}}{dt} = 0 \Rightarrow \bar{L} = constant$

Angular momentum is conserved if net external torque is zero
(A) torque about sun is zero , $\because$ line of focus passer though it and no external force is present.
(B) Torque of friction is present 
(C) Force on the electron is internal , $\therefore$ torque is zero 
(D) No external force is present which may produce changes in angular momentum.

Moment of inertia $I$ inerease. Therefore, $\omega $ decreases (as $I \omega = $ constant). Hence time period increases
$\displaystyle \left( as: T=\frac{2 \pi}{\omega} \right) $

 Mass will move towards equator (away from axis) So, I will increase. Therefore $\omega$ will decrease
(as $I \omega =$ constant) Hence, time period will increase
$\displaystyle \left( as T=\frac{2 \pi}{\omega} \right) $
Duration of day-night increase

$acceleration\quad of\quad the\quad cylinder\quad is\quad given\quad by\ mgsin\theta -f=ma....(1)\ fr=I\alpha ....(2)\ and\quad a=\alpha r...(3)\ acceleration\quad is\quad not\quad always\quad zero\quad so\quad the\quad velocity\quad of\quad the\quad \ cyclinder\quad is\quad aloways\quad changing\quad and\quad the\quad angular\quad momentum\quad \ about\quad the\quad point\quad of\quad contact\quad is\quad mvr,\quad which\quad is\quad not\quad always\ constant.\ $

Moments of inertia (MOI) of the ring before attaching the masses $I$= $ MR^2$,

Moment of inertia of solid cylinder       $I _{cylinder}= \dfrac{1}{2}mr^2$

As the motion is circular, friction opposes the centripetal acceleration, thus its direction changes regularly.

The law of conservation of angular momentum states that , in an isolated system , total angular momentum remains constant . A rotating dancer is an isolated system therefore when she folds hands , her moment of inertia decreases but her angular speed increases such that angular momentum remains constant as L i.e.

Potential energy $U = mV$
$\Rightarrow U=(50x^2+100)10^{-2}$
     $F=-\dfrac{dU}{dx}=-(100x)10^{-2}$
$\Rightarrow m\omega ^2x=-(100\times 10^{-2})x$
      $10\times 10^{-3}\omega ^2x=100\times 10^{-2}x\Rightarrow \omega ^2=100,\omega =10$
$\Rightarrow f=\dfrac{\omega }{2\pi }=\dfrac{10}{2\pi}=\dfrac{5}{\pi }$

Initial angular momentum $={ I } _{ t }{ w } _{ 1 }4$

Now another disc is dropped coaxially on to the rotating disc

Final angular momentum

$=\left( { I } _{ t }+{ I } _{ b } \right) { w } _{ 2 }$

According to the law of conservation of momentum

${ I } _{ t }{ w } _{ 1 }=\left( { I } _{ t }+{ I } _{ b } \right) { w } _{ 2 }$

or$ { w } _{ 2 }=\cfrac{ { I } _{ t } }{ \left( { I } _{ t }+{ I } _{ b } \right)  } { w } _{ 1 }$

Loss in kinetic energy

$\Delta K=\cfrac{ 1 }{ 2 } { I } _{ t }{ w } _{ 1 }^{ 2 }-\cfrac{ 1 }{ 2 } \left( { I } _{ t }+{ I } _{ b } \right) { \left[ \cfrac{ { I } _{ t } }{ \left( { I } _{ t }+{ I } _{ b } \right)  } { w } _{ 1 } \right]  }^{ 2 }\\ \quad \quad =\cfrac{ 1 }{ 2 } { w } _{ 1 }^{ 2 }\cfrac{ { I } _{ t }{ I } _{ b } }{ \left( { I } _{ t }+{ I } _{ b } \right)  } $

Since both are moving in same line through origin total angular momentum is zero.

The shape of the orbit a planet depends on angular momentum.

For  conservation of angular momentum about origin, external torque acting about origin should be zero.

By conservation of energy

In this case, the angular momentum of  bit of putty about the axis of rotation = angular momentum of system of disc and bit of putty about the axis of rotation.

Let:

$M$ = Mass of puty

$m$ = Mass of disc

$ \therefore MvR=\left( \dfrac{m{{R}^{2}}}{2}+M{{R}^{2}} \right)\omega  $

 $ \omega =\dfrac{MvR}{\left( \dfrac{m{{R}^{2}}}{2}+M{{R}^{2}} \right)} $

 Putting all the values

 $ \omega =14.298\ m/s $

Initially body is stationary (zero velocity)

This is identical to question 478324. For line Y = X + 4, perpendicular distance from origin is 4/√2 = 2√2. L = mvr_perpendicular = 5 × 3√2 × 2√2 = 60 units.

As there is no external torque, thus $L$ is conserved.