JEE Previous Year Questions: Rotational Motion - JEE MCQ

The mass B is moving in a circular path centred at A. The centripetal force (mlω²) F'. Therefore a force F' acts on a (the hinge) which is equal to mlω2 . The same is the case for mass C. Therefore the net force in the hinge is

(b) The force F acting on B will provide a torque to the system. This torque is

The total force acting on the system along xdirection is F + (F_net)_x This force is responsible for giving an acceleration ax to the system. 

Therefore 

Fnet)y remains the same as before 

The net force acting on a particle undergoing uniform circular motion is centripetal force which always passes through the centre of the circle. The torque due to this force about the centre is zero, therefore, is conserved about O.

A massless rod will not apply a force perpendicular to its length.
∴ Velocity of the particle immediately after impact = v. And velocity of the other particle immediately after impact = 0. Angular velocity of moving end with the stationary end is ω = v/l.

In case of pure rolling bottom most point is the instantaneous centre of zero velocity.
Velocity of any point on the disc, v = rω, where r is the distance of point from O
r_Q ​> r_C​ > r_P​
∴ v_Q​ > v_C​ > v_P​

According to the principle of conservation of angular momentum,
Iω = constant
As I is doubled, w becomes half.
Now K.E. of rotation, 
Since I is doubled and ω is halved, therefore
K.E. will become half i.e. k/2.

For horizontal and vertical equilibrium F = N and f = mg. For rotational equilibrium about the centre C, line of normal force (N) must pass below C. This will ensure that torque due to f gets balanced by torque due to N.

The moment of inertia of complete disc about a perpendicular axis passing through center O is,

The mass of cut out disc if radius R/3 is,

Now, using the theorem of parallel axis, the moment of inertia of cut out disc about the perpendicular axis passing through center O is,

The moment of inertia of residue disc is, 

Angular momentum is an axial vector, so its direction is perpendicular to plane of motion which is not going to change because of change in speed. So only its direction remains same but its magnitude will vary.

Moment of inertia of solid sphere of mass M and radius R about an axis passing through the centre of mass is 
Let the radius of the disc is r.
Moment of inertia of circular disc of radius r and mass M about an axis passing through the centre of mass and perpendicular to its plane = 
Using theorem of parallel axes, moment of inertia of disc about its edge is:

Here, mg sin θ drags the cylinder downwards. Thus, the friction acts upwards and hence opposes translation. Since it is the friction alone which creates a torque about CM (centre of mass) for rotation (anticlockwise), it supports rotation.


On decreasing the value of θ, the frictional force f also decreases.

Using conservation of energy, we can write the following:
E_A ​= mgh_A​ + K​_A(1)
E_B​ = K_B​(2)
E_C ​= mgh_C ​+ K_C​(3)

Since E_A ​= E_B ​= E_C​, we have the following:
K_B​ > K_A​
K_B ​> K_C​
From Eqs. (1) and (3), we get
E_A ​− E_C​ = 0
That is,
mg (h_A​ − h_C​) + [K_A​ − K_C​] = 0

From kinetic energy K_B​, only translational part gets converted into potential energy at point C, we get h_C ​< h_A​. Therefore, K_C ​> K_A​

The potential energy stored in the spring gets converted into the rotational kinetic energy of the discs.
For Disc A:

Simplifying:

For Disc B:

Simplifying:

Dividing Equation (1) by Equation (2):

This simplifies to:

Taking the square root on both sides:

The ratio 

Therefore, the correct answer is:

Option C: √2.

The effect moment of inertia is 31.
Then from the conservation of angular momentum, we have

and the torque on the first disc is

Substituting for from first equation, we get

We need to calculate the loss of kinetic energy when two discs with moments of inertia III and 2I2I2I rotate together on a common axis after being imparted with different angular velocities.

Step 1: Initial Kinetic Energy of Each Disc
Disc A:

• Moment of inertia I
• Angular velocity 2ω
• Rotational kinetic energy:
  

Disc B:

• Moment of inertia 2I
• Angular velocity ω
• Rotational kinetic energy:
  

Total Initial Kinetic Energy:
 
Step 2: Final Kinetic Energy After Interaction

When the discs come together and rotate with a common angular velocity, the angular momentum is conserved.

• Initial Angular Momentum:
  

• Final Angular Momentum (since the discs rotate together):
  

• Using the conservation of angular momentum:
  

• Final Kinetic Energy:
  

Step 3: Loss of Kinetic Energy

The loss in kinetic energy is:

Final Answer:

The loss of kinetic energy is which corresponds to Option B:

Concept:

• Law of conservation of mechanical energy: It states that the total mechanical energy in a system (i.e., the sum of the potential plus kinetic energies) remains constant as long as the only forces acting are conservative forces.

Formula:

K.E_i + P.Ei = K.E_f  + P.E_f

where, K.E_i = initial kinetic energy, P.E_i = initial potential energy, K.E_f = final kinetic energy, P.E_f = final potentila energy.

Explanation:

The kinetic energy of the rolling object is converted into potential energy at the height

h=  3v²/4g

So by the law of conservation of mechanical energy, we have

Hence the object is Disc.

The velocity of centre of mass remains constant if no external force acts on the body. 

External torque (τ = r × F) can be zero if F is finite but vectors r × F are parallel to each other. Hence statement 1 is false. Also Linear momentum is always conserved.

acceleration = a ∙ [(gsinθ) / {1 + (I_CM / mR²)}]

For hollow cylinder I_CM = mR² = Ih

Solid cylinder I_CM = [(mR²) / ∝] = I_s 

i.e I_h > I_s 

hence a_h < a_s 

hence solid cylinder will reach bottom first.

Also

by
principle of conservation of energy.

KE at bottom = PE = mgh

hence KE is same for each cylinder at bottom (m, h same)

Since it is a inertial frame the Newton's laws of motion are applicable.

So when the net force is zero, the change in linear momentum must be zero.

but we cannot conclude anything about rotational motion of the body.

 is the velocity of centre of the sphere, then

(b) is the correct opton.

(c ) is the correct option.

Under the influence of the force of stick (2N), the point of contact O of the ring with ground tends to slide. But the frictional force f₂ does not allow this and cretes a torque which starts rolling the ring. A frictjion forve f₁ also acts between the ring & the stick. Applying F ≠ t = m a in the horizontal direction. We get

At time t,
x = vt

Moment of inertia of the system about rotation axis (through O) is I = I₀​ + m(vt)²
Where I₀ ​= MI of the rod.
∴ Angular momentum at time t is
L = Iω = [I₀ ​+ mv²t²]ω
L is changing as I is increasing. An external torque is needed to change L. [If there is no torque, the rod will slow down as the MI increases.]

∴ Torque increases linearly with time. For t > T, no torque is needed.