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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 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 

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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

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 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

  1. $1$

  2. $-1$

  3. $2$

  4. Zero
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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 stationary body explodes into two fragments of masses ${m} _{1}$ and ${m} _{2}$. If momentum of one fragment is $p$, the minimum energy of explosion is

  1. $\cfrac { { p }^{ 2 } }{ 2\left( { m } _{ 1 }+{ m } _{ 2 } \right) } $

  2. $\cfrac { { p }^{ 2 } }{ 2\left( \sqrt { { m } _{ 1 }{ m } _{ 2 } } \right) } $

  3. $\cfrac { { p }^{ 2 }\left( { m } _{ 1 }+{ m } _{ 2 } \right) }{ 2{ m } _{ 1 }{ m } _{ 2 } } $

  4. $\cfrac { { p }^{ 2 } }{ 2\left( { m } _{ 1 }-{ m } _{ 2 } \right) } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Initially body is stationary (zero velocity)

using conservation of momentum
$\left( A \right) O=P+{ P _{ 2 } },{ P _{ 2 } }$ is momentum of mass ${m _2}$ after explosion
$\begin{array}{l} { P _{ 2 } }=-P \ Energy={ E _{ 1 } }\Rightarrow \dfrac { { { P^{ 2 } } } }{ { 2{ m _{ 1 } } } } d{ E _{ 2 } }=\dfrac { { { { \left( { -P } \right)  }^{ 2 } } } }{ { 2{ m _{ 2 } } } }  \ net\, \, energy\Rightarrow { E _{ 1 } }+{ E _{ 2 } }=\dfrac { { { P^{ 2 } } } }{ { 2{ m _{ 2 } } } } +\dfrac { { { P^{ 2 } } } }{ { 2{ m _{ 2 } } } } \Rightarrow \dfrac { { { P^{ 2 } } } }{ { 2\left( { { m _{ 1 } }+{ m _{ 2 } } } \right)  } }  \end{array}$

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 $w.$
$L _i= L _f$
$0 = Iw - mv _o r$
$0 = 200 (w) - 70 (10 ) (2)$
$\implies w= 7    rad/s$