System of Particles and Rotational Motion - Online Test

Q1. A solid sphere of mass M and radius R spins about an axis passing through its centre making 600 rpm. Its kinetic energy of rotation is
Answer : Option B
Explaination / Solution:

Kinetic  energy of rotation 

K =

 revolution per second

angular velocity  rad/sec

moment of inertia about an axis passing through centre of solid sphere


Kinetic  energy of rotation 


Q2. A child swinging on a swing in a sitting position stands up. Then, the time period of the swing will
Answer : Option C
Explaination / Solution:

As the child stand up then Centre of mass is shifted in upwards direction from the sitting position of child, thus decreasing the effective length of  the swing, so as the time period is proportion to square root of length, thus time period decreases.

time period 


Time period of swing decreases as effective length decreases.


Q3. Angular momentum is
Answer : Option D
Explaination / Solution:

Axial vector represent rotational effect and are always along the axis of rotation. Direction of angular momentum always along the axis of rotation in accordance with right hand screw rule. Hence angular momentum is an axial vector.


Q4. Angular momentum of a body is defined as the product of
Answer : Option A
Explaination / Solution:


if r is perpendicular to p (in case of circular motion)

L = mvr


Q5. The radius of gyration of a circular loop of radius r and mass m rotating about its diameter as axis is
Answer : Option C
Explaination / Solution:

Moment of inertia of ring or circular loop about axis passing through centre of mass and perpendicular to plane is MR2 ,

Applying perpendicular axis theorm 


Here I = Moment of inertia about axis passing through diameter

As  moment of inertia in terms of radius of gyration (k)  is Mk2

I = Mk2


Q6. In an orbital motion, the angular momentum vector is
Answer : Option D
Explaination / Solution:


Hence direction of angular momentum  is perpendicular to both  and  that is perpendicular to the orbital plane.


Q7. The moment of inertia of a circular disc of mass M and radius R rotating about an axis passing through its edge and perpendicular to the plane of the disc is
Answer : Option C
Explaination / Solution:

Moment of inertia of a disc about axis passing through Centre of mass and perpendicular to plane


if I = Moment of inertia of a disc about axis passing through its edge and perpendicular to plane

applying parellal axis theorm


 


Q8. The moment of inertia of a circular ring about an axis passing through its centre and normal to its plane is 200 gm . Then its moment of inertia about a diameter is
Answer : Option D
Explaination / Solution:

applying perpendicular axis theorm


I = moment of inertia about a axis passing through diameter

I = 100 gm cm2


Q9. A dancer on ice spins faster when she folds her hands. This is due to
Answer : Option A
Explaination / Solution:

if L = const.

then 

When a dancer is spinning, she has a certain angular momentum. When the dancer folds her arms, the distance of all the points of her body decreases with respect to axis of rotation so that her moment of inertia decreases. Thus, in order to conserve angular momentum, the speed of rotation has to increase and hence the dancer spins faster.


Hence if moment of inertia decrease then kinetic energy will increase.


Q10. A uniform rod of mass 2 kg is 1 meter long. Its radius of gyration about an axis through its one end and perpendicular to its length in SI units is
Answer : Option C
Explaination / Solution:

Moment of inertia of rod about an axis through its one end and perpendicular to its length


M = 2 Kg

L = 1 m

Moment of inertia of rod  in terms of radius of gyration

I = Mk2