Tag: physics

Questions Related to physics

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

Show answer
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

Show answer
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

Show answer
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

Show answer
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

Show answer
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 } $

Show answer
Correct Option: B
Explanation:

By conservation of energy

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

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

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. $14.29\ rad\ s _1$

  2. $17.3\ rad\ s\ _1$

  3. $12.4\ rad\ s\ _1$

  4. $9.82\ rad\ s _1$

Show answer
Correct Option: A
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 $

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

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. $\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) } $

Show answer
Correct Option: A
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
$\begin{array}{l} { P _{ 2 } }=-P \ Energy={ E _{ 1 } }\Rightarrow \dfrac { { { P^{ 2 } } } }{ { 2{ m _{ 1 } } } } d{ E _{ 2 } }=\dfrac { { { { \left( { -P } \right)  }^{ 2 } } } }{ { 2{ m _{ 2 } } } }  \ net\, \, energy\Rightarrow { E _{ 1 } }+{ E _{ 2 } }=\dfrac { { { P^{ 2 } } } }{ { 2{ m _{ 2 } } } } +\dfrac { { { P^{ 2 } } } }{ { 2{ m _{ 2 } } } } \Rightarrow \dfrac { { { P^{ 2 } } } }{ { 2\left( { { m _{ 1 } }+{ m _{ 2 } } } \right)  } }  \end{array}$

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:

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. $40$units

  2. $60$units

  3. $0$

  4. $40\sqrt{2}$ units

Show answer
Correct Option: B

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

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. $1.2rad$ ${s}^{-1}$

  2. $0.4rad$ ${s}^{-1}$

  3. $2.0rad$ ${s}^{-1}$

  4. $7.0rad$ ${s}^{-1}$

Show answer
Correct Option: D
Explanation:

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

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