Tag: physics

Questions Related to physics

When tall buildings are constructed on earth, the duration of day night 

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. slightly increases

  2. slightly decreases

  3. has no change

  4. none of these

Show answer
Correct Option: A
Explanation:

Moment of inertia $I$ inerease. Therefore, $\omega $ decreases (as $I \omega = $ constant). Hence time period increases
$\displaystyle \left( as: T=\frac{2 \pi}{\omega} \right) $

If the ice at the poles melts and flows towards the equator, how will it affect the duration of day-night?

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. Increase

  2. Decrease

  3. Remain same

  4. none of these

Show answer
Correct Option: A
Explanation:

 Mass will move towards equator (away from axis) So, I will increase. Therefore $\omega$ will decrease
(as $I \omega =$ constant) Hence, time period will increase
$\displaystyle \left( as T=\frac{2 \pi}{\omega} \right) $
Duration of day-night increase

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?

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. True

  2. False

Show answer
Correct Option: B
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.\ $

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 :

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. $\displaystyle \frac{\omega\, M}{(M\, +\, m)}$

  2. $\displaystyle \frac{\omega M}{(M\, +\, 2m)}$

  3. $\displaystyle \frac{\omega M}{M\, -\, 2m}$

  4. $\displaystyle \frac{\omega(M\, +\, 3m)}{M}$

Show answer
Correct Option: B
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 ^ {'} $
$\Rightarrow MR^2 \times \omega= \dfrac { MR^2 \times \omega ^{'}}{ M + 2m}$,
$\Rightarrow \omega ^{'} = \dfrac {\omega M}{M+2m}$

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?

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. 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

Show answer
Correct Option: A
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

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, ${\omega}^{\prime}$?

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. ${\omega}^{\prime}=\dfrac{\omega}{3}$

  2. ${\omega}^{\prime}=\dfrac{3\omega}{4}$

  3. ${\omega}^{\prime}=\omega$

  4. ${\omega}^{\prime}=3\omega$

  5. ${\omega}^{\prime}=\dfrac{2\omega}{3}$

Show answer
Correct Option: A
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 :      $L _i = L _f$
$\therefore$   $\dfrac{1}{2}mr^2 w = \dfrac{3}{2}mr^2 w'$                     $\implies w ' =\dfrac{w}{3}$

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.

Show answer
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

Show answer
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. $L$

Show answer
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.

Show answer
Correct Option: B
Explanation:

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