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 40 Kg mass hanging at the end of a rope of length L oscillates in a vertical plane with an angular amplitude $ \theta _o $. If the breaking strength of the rope is 80 Kg. Wt What is the maximum angular amplitude so that the mass can oscillates without the rope breaking ?

  1. $ 90^o $

  2. $ 60^o $

  3. $ 30^o $

  4. $ 45^o $


Correct Option: A

Assertion (A): Even though a planet revolve around the sun in an elliptical orbit, the angular momentum of planet is constant.
Reason (R) : Any force other than mutual gravitational force is absent between the planet and the sun.

  1. Both A and R are true and R is correct explanation of A

  2. Both A and R are true and R is not correct explanation of A

  3. A is true and R is false

  4. A is false and R is true


Correct Option: A
Explanation:

Angular momentum of planet revolving around the sun is constant because the line of force of the gravitational force passes through the  sun . Therefore torque about the sun is zero.
$\therefore$  Angular momentum will remain constant.  And there is no other force acting on the planet, which results into zero tangential force.

The law of conservation of angular momentum is obtained from Newton's II law in rotational motion when:

  1. External torque is maximum

  2. External torque is minimum

  3. External torque is zero

  4. External torque is constant


Correct Option: C
Explanation:

Newton's 2nd law says that External Force is directly proportional to the rate of change of momentum. Similarly, external torque is directly proportional to the rate of change of angular momentum. So if angular momentum is conserved i.e., the rate of change of angular momentum is zero then external torque is zero.

The energy of electron in an excited hydrogen atom is -3.4 eV. Its angular momentum according to Bohr's theory will be:

  1. $\cfrac{h}{\pi }$

  2. $\cfrac{h}{2\pi }$

  3. $\cfrac{3h}{2\pi }$

  4. $\cfrac{2h}{\pi }$


Correct Option: A
Explanation:

$E=\dfrac{-13.6}{n^{2}}eV$

$\dfrac{-13.6}{n^{2}}=\ 3.4 $
$\Rightarrow    n = 2$
So, its angular momentum $=\dfrac{nh}{2\pi }$ $=\dfrac{2h}{2\pi }$ $=\dfrac{h}{\pi }$

Under the action of a central force, there is a conservation of-

  1. Angular momentum only

  2. Mechanical energy only

  3. Angular momentum and mechanical energy

  4. Neither angular momentum nor mechanical energy


Correct Option: C
Explanation:

Under the action of a central force, there is a conservation ofAngular momentum and mechanical energy.

Four particles of masses in the ratio 1:2:3:4 are rotating in concentric circles of radii proportional to their masses with the same angular velocity. What is the angular momentum of the system

  1. $4mw^2r$

  2. $10mw^2r$

  3. $30mw^2r$

  4. $20mw^2r$


Correct Option: C
Explanation:

The angular momentum of a single particle is $mw^2r$. If there are 4 particles, we can write,
$L _{tot}=mw^2r+2mw^2(2r)+3mw^2(3r)+4mw^2(4r)=30mw^2r$

The correct option is (c)

Two identical particles of mass m move in a circle of radius r , 180 degrees out of phase at an angular speed  $\omega$ about the z-axis in a plane parallel to but a distance h above the x-y plane (Figure 19.14). Find the magnitude and the direction of the angular momentum L relative to the origin.

  1. $2mr^2w$

  2. $mr^2w$

  3. zero

  4. $mr^2w/2$


Correct Option: A

Two identical particles of mass m are separated by a distance of 1 metre each and are rotating with a constant speed of 5 m/s . What will be their angular momentum about an axis distant 0.8 m from the first particle on the line joining them

  1. 5m

  2. 3m

  3. 2m

  4. m


Correct Option: B

A particle is in uniform circular motion in a   horizontal plane. Its angular momentum is constant when the origin is taken at

  1. centre of the circle

  2. any point on the circumference of the circle.

  3. any point inside the circle.

  4. any point outside the circle.


Correct Option: A