Physics

Rotational and Circular Motion

235 Questions

Rotational and circular motion examines the dynamics of objects moving in circular paths or rotating around an axis. Key concepts include angular momentum, torque, moment of inertia, and centripetal force. This is a highly scoring topic in the physics section of competitive exams.

Angular momentumCentripetal forceMoment of inertiaRolling objectsGyroscopic effect

Rotational and Circular Motion Questions

Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

A particle moves on a circular path with decreasing speed. Choose the correct statement.

  1. Angular momentum remains constant.
  2. Acceleration $\vec{a}$ is towards the centre.

  3. Particle moves in a spiral path with decreasing radius.

  4. The direction of angular momentum remains constant.
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

A particle of mass 2 kg located at the position $ \left( \hat { i } +\hat { j }  \right) m $ has a velocity $ 2\left( \hat { i } -\hat { j } +\hat{ k } \right) ms^{-1} $ . Its angular momentum along z-axis in $ kgm^2 s^{-1}  $ is

  1. $-8$
  2. $+8$
  3. $-4$

  4. zero
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

A cylinder is rolling down a rough inclined plane. Its angular momentum about the point of contact remains constant. Is this statement true or false?

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$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.\ $

Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

A uniform thin circular ring of mass 'M' and radius 'R' is rotating about its fixed axis, passing through its centre and perpendicular to its plane of rotation, with a constant angular velocity $\omega$. Two objects each of mass m, are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity of :

  1. $\displaystyle \frac{\omega\, M}{(M\, +\, m)}$
  2. $\displaystyle \frac{\omega M}{(M\, +\, 2m)}$
  3. $\displaystyle \frac{\omega M}{M\, -\, 2m}$
  4. $\displaystyle \frac{\omega(M\, +\, 3m)}{M}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

MOI of the ring after attaching the masses $I^{'}= (M+2m) R^2$
Let angular momentum after the attaching the masses $\omega ^{'}$
Since there is no external torque,  so we use conservation of angular momentum.
$I \times \omega= I ^ {'} \times \omega ^ {'} $
$\Rightarrow MR^2 \times \omega= \dfrac { MR^2 \times \omega ^{'}}{ M + 2m}$,
$\Rightarrow \omega ^{'} = \dfrac {\omega M}{M+2m}$

Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

A student is rotating on a stool at an angular velocity $'\omega'$ with their arms outstretched while holding a pair of masses. The frictional effects of the stool are negligible.
Which of the following actions would result in a change in angular momentum for the student?

  1. A clockwise torque of $50 Nm$ and a counterclockwise torque of $25 Nm$ are both applied to the students arms by fellow students

  2. The student brings the masses closer to their body

  3. The student stretches the masses further away from their body

  4. A second student steps onto the stool with the first student
  5. A clockwise torque of $50 Nm$ and a counterclockwise torque of $50 Nm$ are both applied to the students arms by fellow students
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
When no external force is acting, the total angular momentum of the system remains constant. So in this case, angular momentum will change only is an external force is applied on the system thus option (A) is correct
Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

A solid cylinder of mass, $m$, and radius, $r$, is rotating at an angular velocity, $\omega$ when a non-rotating hoop of equal mass and radius drops onto the cylinder.
In terms of its initial angular velocity, $\omega$, what is its new angular velocity, ${\omega}^{\prime}$?

  1. ${\omega}^{\prime}=\dfrac{\omega}{3}$
  2. ${\omega}^{\prime}=\dfrac{3\omega}{4}$
  3. ${\omega}^{\prime}=\omega$
  4. ${\omega}^{\prime}=3\omega$
  5. ${\omega}^{\prime}=\dfrac{2\omega}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Thus initial angular momentum     $L _i = \dfrac{1}{2}mr^2 w$
Final moment of inertia of the system      $I' = I _{cylinder}+ I _{hoop} = \dfrac{1}{2}mr^2+ mr^2   = \dfrac{3}{2}mr^2$ 
Final angular momentum     $L _f =I'w' = \dfrac{3}{2}mr^2 w'$

Using conservation of angular momentum :      $L _i = L _f$
$\therefore$   $\dfrac{1}{2}mr^2 w = \dfrac{3}{2}mr^2 w'$                     $\implies w ' =\dfrac{w}{3}$

Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

A bar of length l carrying a small mass m at one of its ends rotates with a uniform angular speed $\omega$ in a vertical plane about the mid-point of the bar. During the rotation, at some instant of time when the bar is horizontal, the mass is detached from the bar but the bar continues to rotate with same $\omega$. The mass moves vertically up, comes back and reches the bar at the same point. At that place, the acceleration due to gravity is g.

  1. This is possible if the quantity $\dfrac{{\omega}^2 \ell}{2\pi g}$ is an integer
  2. The total time of flight of the mass is proportional to ${\omega}^2$
  3. The total distance travelled by the mass n air is proportional to ${\omega}^2$

  4. The total distance travelled by the mass in air and its total time of flight are both independent on its mass.
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation
The whole system is possible only if $\dfrac{\omega^2L}{2mg}$ is an integer, where the total distance travelled by the mass is proportional to $\omega^2$ and the distance travelled by the mass in air and its total time of flight are both independent on its mass.
The law of conservation of angular momentum states that '' When the net external torque acting on a system about a given axis is zero, the total angular momentum of the system about the axis remains constant''.
Hence, this question has multiple solutions.
Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

A dancer is rotating on smooth horizontal floor with an angular momentum $L$. The dancer folds her hands so that her momentof inertia decreases by $25$%. The new angular momentum is.

  1. $\dfrac {3L}{4}$
  2. $\dfrac {L}{4}$
  3. $\dfrac {L}{2}$
  4. $L$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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.

                $L=I\omega$
when $I$ decreases  , $\omega$ increases .
Hence , total angular momentum would be L .

Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

Two men of equal masses stand at opposite ends of the diameter of a turntable disc of a certain mass, moving with constant angular velocity. The two men make their way to the middle of the turntable at equal rates. In doing so will

  1. kinetic energy of rotation has increased while angular momentum remains same.

  2. kinetic energy of rotation has decreased while angular momentum remains same.

  3. kinetic energy of rotation has decreased but angular momentum has increased.

  4. both, kinetic energy of rotation and angular momentum have decreased.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

A circular disk of moment of inertia $I _t$ is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed $\omega _i$. Another disk of moment of inertia $I _b$ is dropped co-axially onto the rotating disk.Initially the second disk has zero angular speed.Eventually both the disks rotate with a constant angular speed $\omega _p$ .The energy lost by the initially rotating disc due to friction is 

  1. $\dfrac{1}{2} \dfrac{I^2 _b}{(I _t+I _b)}\omega^2 _1$
  2. $\dfrac{1}{2} \dfrac{I^2 _t}{(I _t+I _b)}\omega^2 _1$
  3. $\dfrac{I _b-I _t}{(I _t+I _b)}\omega^2 _1$
  4. $\dfrac{1}{2} \dfrac{I _bI _t}{(I _t+I _b)}\omega^2 _1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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)  } $

 

Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

Two particles each of mass m move in opposite direction along Y-axis. One particle moves in positive direction with velocity v while the other particle moves in negative direction with speed 2v. The total angular momentum of the system with respect to origin is:

  1. Is zero

  2. Goes on increasing

  3. Goes on decreasing

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

A point sized sphere of mass $'m'$ is suspended from a point using a string of length $'l'$. It is pulled to a side till the string is horizontal and released. As the mass passes through the portion where the string is vertical, magnitude of its angular momentum is:

  1. $ ml\sqrt { gl } $
  2. $ ml\sqrt { 2gl } $
  3. $ml\sqrt { \dfrac { gl }{ 2 } } $
  4. $ ml\sqrt { 3gl } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By conservation of energy

$\dfrac{1}{2}mv^2=mgl$
$v=\sqrt{2gl}$
Angular momentum$=mvr=$$ml\sqrt{2gl}$

Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

A disc of mass $100\ g$ and radius $10\ cm$ has a projection on its circumference. The mass of projection is negligible. A $20\ g$ bit of putty moving tangential to the disc with a velocity of $5\ m\ s^{-1}$ strikes the projection and sticks to it. The angular velocity of disc is

  1. $14.29\ rad\ s _1$
  2. $17.3\ rad\ s\ _1$
  3. $12.4\ rad\ s\ _1$
  4. $9.82\ rad\ s _1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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 $

Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

A particle of mass $5kg$ is moving with a uniform speed $3\sqrt{2}$ in $XOY$ plane along the line $Y=X+4$. The magnitude of its angular momentum about the origin is:

  1. $40$units
  2. $60$units
  3. $0$
  4. $40\sqrt{2}$ units
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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.

Multiple choice physics rotational motion of a rigid body and moment of inertia angular momentum (l) and conservation of angular momentum angular momentum in case of rotation about a fixed axis law of conservation of angular momentum

A circular platform is mounted on a vertical frictionless axle. Its radius is $r=2m$ and its moment inertia is $I=200kg$ ${m}^{2}$. It is initially at rest. A $70kg$ man stands on the edge of the platform and begins to walk along the edge at speed ${v} _{0}=10{ms}^{-1}$ relative to the ground. The angular velocity of the platform is

  1. $1.2rad$ ${s}^{-1}$
  2. $0.4rad$ ${s}^{-1}$
  3. $2.0rad$ ${s}^{-1}$
  4. $7.0rad$ ${s}^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

Let angular velocity of the platform be $w.$
$L _i= L _f$
$0 = Iw - mv _o r$
$0 = 200 (w) - 70 (10 ) (2)$
$\implies w= 7    rad/s$