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This is one of our many quizzes on Angular momentum. Click on this to access more such competitive quizzes on Dynamics.

Angular momentum Quiz 2

physics

 To loosen a pipe, you apply some force on a jolt. If the jolt is $1.5\,m$ long and you apply $230\,N$ at an angle of ${{74}^{\circ }}$ with respect to the jolt. Then, what amount of torque are you applying?

 The force applied by jolts on a pipe is an example of rotational analog of force in physics, which is referred to as torque. Thus, the amount of twists required to loosen the pipe can be considered by taking into consideration the equation of torque.

  $\text{Using the equation}, \\ \Rightarrow \tau =F.r\sin \theta  \\ \text{now, putting the values in the equation:} \\ \text{where, }F=230N,\,r=1.5m\text{ and }\theta ={{74}^{\circ }},\text{ we get:} \\ \Rightarrow \tau =230\times 1.5\times \sin {{74}^{\circ }}=230\times 1.5\times 0.985\text{ (}\because \sin {{74}^{\circ }}=0.985) \\ \text{Thus, we get:} \\ \therefore \tau =339.82\,Nm$

 Which of the following relations of angular momentum will hold true, if angular momentum is proportional to the moment of inertia and the angular speed?

 The relation of angular momentum proportional to the moment of inertia and the angular speed can be found by considering the cross product of the position vector and linear momentum of the particle.

  We know that angular momentum $L$ is proportional to the moment of inertia $I$ and angular speed $\omega$ $ \\ $ $\omega$ is measured in radians per second. Thus, we can say that, $ \\ \Rightarrow L=I\omega \text{    }...\text{(i)} \\ $  Moment of inertia $I$, also depends on the position of the axis of rotation and shape of the matter. $ \\ $  For a single particle, $I={{r}^{2}}m$ and for circular motion, $\omega =\dfrac{\upsilon }{r}$ $ \\ $ Now, putting the values of $I$ and $\omega$ in equation $(i)$, we get, $ \\ \Rightarrow L={{r}^{2}}m.\dfrac{\upsilon }{r}=rm\upsilon  \\ $  where, $r$ is the position vector of the particle, $ \\ $ $\upsilon$  is the linear velocity of the particle and $m$ is the mass of the particle.

 Suppose a light ray carries an orbital angular momentum in the paraxial limit. Which one of the following measures would you consider while generating OAM $($orbital angular momentum$)$

 OAM generation takes place along a collinear path when a ray of light is passed through it. Recall the methods of generating OAM to figure out the measures that must be considered to generate OAM.

 Orbital angular momentum $(OAM)$ is a polarized state of light that is capable of carrying the huge quantities of data through optical fibres. It arises from the amplitude cross-section of the ray and is independent of the spin angular momentum. To generate OAM, following things must be considered to attain $100\%$ efficiency:- $ \\ $ 1) Phase singularities: This can be measured by computer-generated holograms by coupling holograms to a single-mode fiber set.  $ \\ $ 2) Rotational Doppler effect: It can be measured using an interferometer that studies the spin of trapped particles and the optical transformations. $ \\ $ 3) Fringe pattern: This pattern can be observed by interfering with a single ray of light along a collinear path. $ \\ $ 4) Spatial light modulators: This measure is done to generate OAM in coherent superposition that will provide for quantum information protocols. $ \\ $ Thus, using all these measures you can generate OAM in a paraxial field.

 Consider the Atwood’s machine shown in the figure. The masses are $m$ and $2m$, and the pulley is a uniform disk having mass $m$ and radius $r$. The string is massless and does not slip with respect to the pulley. Find the rate of change of velocity of the masses? $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/09204ba6-90d1-45e7-93a4-22e54c6e01025467881778841528275.png"/>

 To find the rate of change of velocities, first obtain the change in potential energy and then the total kinetic energy. After this, obtain the non-slipping condition that will provide us with our usual kinematic result.  

 Let $v$ be their common speed after they both have moved a distance $d$. $ \\ $  If $2m$ falls at distance $d$, then the change in potential energy is $ \\ \Rightarrow -2mgd+mgd=-mgd \\ $ The total kinetic energy is $ \\ \Rightarrow \dfrac{1}{2}m{{v}^{2}}\text{+}\dfrac{1}{2}\left( 2m \right){{v}^{2}}\text{+}\dfrac{1}{2}\left( \dfrac{1}{2}m{{r}^{2}} \right){{\left( \dfrac{v}{r} \right)}^{2}}\text{=}\dfrac{7}{4}m{{v}^{2}} \\ $ where, we have used the non-slipping condition, $ \\ \Rightarrow v=r\omega  \\ \therefore \dfrac{1}{2}m{{v}^{2}}\text{+}\dfrac{1}{2}\left( 2m \right){{v}^{2}}+\dfrac{1}{2}I{{\omega }^{2}}\text{  } \\ \left( \because \omega \text{=}\dfrac{v}{r}\text{ and }I=\dfrac{1}{2}m{{r}^{2}} \right) \\ $  Conservation of energy, gives: $ \\ \Rightarrow 0=\dfrac{7}{4}m{{v}^{2}}-mgd \\ \Rightarrow v=\sqrt{\dfrac{4}{7}gd} \\ $ The usual kinematic result, $v=\sqrt{2ad}$, holds here so, we obtain, $a=\dfrac{2g}{7}$

 A weight hangs from a string having no mass and swings around in a horizontal circle, as shown in figure. The length of the string is then very slowly increased (or decreased). Let $\theta $, $l$ , $r$ and $h$ be defined as shown in the figure. Assuming that $\theta $ is very close to $\dfrac{\pi }{2}$, how does $h$ depend on $l$? $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/6504e07e-e56e-4d4a-8366-92d34b71b1ff594616852023763211.png"/>

 For determining the frequency of the circular motion, use the relation of force $F=ma$. Also, find the relation between tension in the string and horizontal radial force to obtain a constant relation $\dfrac{{{l}^{2}}}{\sqrt{h}}$. 

 Consider the angular momentum relative to the support point $P$. The motion is circular as the length of the string changes very slowly. Now, let ${{\omega }_{l}}$ be the frequency of the circular motion when the string has length ${{L}_{Z}}$. $ \\ $ Thus, it can be said that, $ \\ $ ${{L}_{Z}}=m{{r}^{2}}{{\omega }_{l}}$, is constant. $ \\ $ The frequency ${{\omega }_{l}}$ can be obtained by using $F=ma$, for the circular motion. $ \\ $ The tension in the string is $ \\ \dfrac{mg}{\cos \theta } \\ $ So, the horizontal radial force is $ \\ mg\tan \theta  \\ \therefore mg\tan \theta =mr{{\omega }_{l}}^{2}=m\left( l\sin \theta  \right){{\omega }_{l}}^{2} \\  \omega l=\sqrt{\dfrac{g}{l\cos \theta }}=\sqrt{\dfrac{g}{h}} \\ $ Thus, the constant value of ${{L}_{z}}$ is given by: $ \\ {{L}_{z}}=m{{r}^{2}}\sqrt{\dfrac{g}{h}} \\ $ For $\theta =\dfrac{\pi }{2}$, we have, $r=l$ $ \\ $   This holds at all times, so the quantity $\dfrac{{{r}^{2}}}{\sqrt{h}}$ is constant $ \\ $ $\dfrac{{{l}^{2}}}{\sqrt{h}}$ is constant. $ \\ $ $\therefore h\propto {{l}^{4}}$, which means that $h$ grows very quickly as you let the string out when $\theta =\dfrac{\pi }{2}$