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