Tag: rotational motion of a rigid body and moment of inertia

Questions Related to rotational motion of a rigid body and moment of inertia

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.

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

  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.

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Correct Option: A,C,D
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.

A body is moving on a rough horizontal plate in a circular path being tide to a nail (at the centre) by a string, while the body is in motion the friction force of the body

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

  1. changes direction

  2. changes magnitude

  3. changes both magnitude and direction

  4. none of he above

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Correct Option: A
Explanation:

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

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.

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

  1. $\dfrac {3L}{4}$

  2. $\dfrac {L}{4}$

  3. $\dfrac {L}{2}$

  4. $L$

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Correct Option: D
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 .

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

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

  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.

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Correct Option: B
Explanation:

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

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 

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

  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. $\dfrac{1}{2} \dfrac{I _bI _t}{(I _t+I _b)}\omega^2 _1$

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Correct Option: D
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)  } $

 

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:

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

  1. Is zero

  2. Goes on increasing

  3. Goes on decreasing

  4. None of these

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Correct Option: A
Explanation:

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

The shape of the orbit a planet depends on:

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

  1. angular momentum

  2. total energy

  3. both angular momentum and total

  4. None of the above

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Correct Option: A
Explanation:

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

A man standing on a platform holds weight in his outstreached arms. The system rotates freely about a central vertical axis. If he now draws the weights inward close to his body 

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

  1. The angular velocity of the system will increase

  2. The angular momentum of the system will increase

  3. The kinetic energy of the system will increase

  4. He will have to expend some energy to draw the weights in

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Correct Option: D
Explanation:
$\tau=0=\cfrac{dl}{dt}$
$\Rightarrow L=constant$, Angular momentum remains constant.
By drawing arms in, $I$ is decreased.
$L=I\omega$
For $L$ to remain constant, $\omega$ increases.
$KE=\cfrac{L^{2}}{2I}$, $I$ decreases.
$\Rightarrow KE \,\,increases$
The energy spent in drawing weight is converted into KE of rotation

A force $\vec { F } =\alpha \hat { i } +3\hat { j } +6\hat { k }$ is acting at a point $\vec { r } =2\hat { i } -6\hat { j } -12\hat { k }$. The value of $\alpha$ for which angular momentum about origine is conserved is

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

  1. $1$

  2. $-1$

  3. $2$

  4. Zero

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Correct Option: B
Explanation:

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

$\therefore$ Torque $=r \times F$
$\Rightarrow (2\hat i-6\hat j- 12\hat k ) \times (\alpha \hat i + 3 \hat j + 6 \hat k)= (-36 +36) \hat i + (-12 \alpha -12 ) \hat j + ( 6+6\alpha ) \hat k=0$   $\Rightarrow \alpha =-1$

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:

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

  1. $ ml\sqrt { gl } $

  2. $ ml\sqrt { 2gl } $

  3. $ml\sqrt { \dfrac { gl }{ 2 } } $

  4. $ ml\sqrt { 3gl } $

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Correct Option: B
Explanation:

By conservation of energy

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