JEE Advanced (Single Correct MCQs): Rotational Motion - JEE MCQ

Since the objects are placed gently, therefore no external torque is acting on the system. Hence angular momentum is constant.

i.e., I₁w₁ = I₂ w₂

The moment of inertia of the system about axis of rotation O is

I = I₁ + I₂ = 0.3x² + 0.7 (1.4 – x)²         = 0.3x² + 0.7 (1.96 + x² – 2.8x)

= x² + 1.372 – 1.96x

The work done in rotating the rod is converted into its rotational kinetic energy.

For work done to be minimum

As the spheres are smooth there will be no friction (no torque) and therefore there will be no transfer of angular momentum. Thus A, after collision will remain with its initial angular momentum. i.e., ω _A =ω

KEYCONCEPT. The disc has two types of motion namely translational and rotational. Therefore there are two types of angular momentum and the total angular momentum is the vector sum of these two.
In this case both the angular momentum have the same direction (perpendicular to the plane of paper and away from the reader).

LT = angular momentum due to translational motion.

L_R = angular momentum due to rotational motion about C.M.

L = MV × R + I_cmω

I_cm = M.I. about centre of mass C.

(v = Rω in case of rolling motion and surface at rest)

Net torque about O is zero.
Therefore, angular momentum (L) about O will be conserved, or L_i = L_f

 Note :  When we are giving an angular acceleration to the rod, the bead is also having an instantaneous acceleration a = Lα. This will happen when a force is exerted on the bead by the rod. The bead has a tendency to move away from the centre. But due to the friction between the bead and the rod, this does not happen to the extent to which frictional force is capable of holding the bead.
The frictional force here provides the necessary centripetal force. If instantaneous angular velocity is ω then

By applying

We get ω =αt

The applied force shifts the normal reaction to one corner as shown. At this situation, the cubical block starts topping about O. Taking torque about O

Moment of inertia about the diameter of the circular loop (ring) = 1/2 MR²

Using parallel axis theorem

The moment of inertia of the loop about XX' axis is

Where M = mass of the loop and R = radius of the loop Here M =Lρ and 

The M.I. about the axis of rotation is not constant as the perpendicular distance of the bead with the axis of rotation increases.
Also since no external torque is acting.

Since, I increases, ω decreases.

The mass distribution of this sector is same about the axis of rotation as that of the complete disc about the axis. Therefore the formula remains the same as that of the disc.

Imagine the cylinder to be moving on a frictionless surface. In both the cases the acceleration of the centre of mass of the cylinder is g sin q. This is also the acceleration of the point of contact of the cylinder with the inclined surface. Also no torque (about the centre of cylinder) is acting on the cylinder since we assumed the surface to be frictionless and the forces acting on the cylinder is mg and N which pass through the centre of cylinder. Therefore the net movement of the point of contact in both the cases is in the downward direction as shown. Therefore the frictional force will act in the upward direction in both the cases.

Note :  In general we find the acceleration of the point of contact due to translational and rotational motion and then find the net acceleration of the point of contact.
The frictional force acts in the opposite direction to that of net acceleration of point of contact.

Since there is no external torque, angular momentum remains conserved. As moment of inertia initially decreases and then increases, so ω will increase initially and then decreases.

Note : The M.I. of the system decreases when the tortoise move from A to B and then increases from B to A.
So the variation of w is nonlinear.

Change in angular momentum of the system = angular impulse given to the system about the centre of mass (Angular momentum)_f – (Angular momentum)_i

Let the system starts rotating with the angular velocity ω.
Moment of Inertia of the system about its axis of rotation [centre of mass of the system]

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, L  is conserved
about O.

KEYCONCPET :  (K.E.)_rotation =L²/2I

Here , L = constant

∴ (K.E.)_rotational × I = constant.
When I is doubled, K.E._rotational becomes half.

Note : In pure rolling, th e poin t of conta ct is t he instantaneous centre of rotation of all the particles of the disc. On applying v = rω

We find ω is same for all the particles then v α r.
Farther the particles from O, higher is its velocity.

The cubical block is in equilibrium.
For translational equilibrium (a) ∑F_x = 0 ⇒ F = N

(b) ∑F_y = 0 ⇒ f = mg

For Rotational Equilibrium

Where τ_c = torque about c.m.
Torque created by frictional force (f) about C = f^× a in clockwise direction.
There should be another torque which should counter this torque. The normal reaction N on the block acts as shown. This will create a torque N × b in the anticlockwise direction.
Such that f × a = N × b Note :  The normal force does not act through the centre of the body always. The point of application of normal force depends on all the forces acting on the body.

Let σ be the mass per unit area

  The total mass of the disc

=σ×πR² = 9M 

The mass of the circular disc cut

Let us consider the above system as a complete disc of mass 9M and a negative mass M super imposed on it.
Moment of inertia (I1) of the complete disc = 1/2 9MR² about a n a xis pa ssi n g th r ough O and perpendicular to the plane of the disc.
M.I. of the cut out portion about an axis passing through O' and perpendicular to the plane of disc

∴ M.I. (I₂) of the cut out portion about an axis passing through O and perpendicular to the plane of disc

[Using perpendicular axis theorem] ∴ The total M.I. of the system about an axis passing through O and perpendicular to the plane of the disc is I = I₁ + I₂

Since v is changing (decreasing), L is not conserved in magnitude. Since it is given that a particle is confined to rotate in a circular path, it cannot have spiral path.
Since the particle has two accelerations a_c and a_t therefore the net acceleration is not towards the centre.

The direction of L remains same even when the speed decreases.

For solid sphere

For solid disc

.

From (i) and (ii),

By the concept of energy conservation

This is the formula of the moment of inertia of the disc.

This is the case of vertical motion when the body just completes the circle. Here v  = √5gL              ....(i)

Applying energy conservation, Total energy at A = Total energy at P

  Therefore, the value of q lies in the range

The system is made up of five bodies (three circles and two straight lines) of uniform mass distribution. Therefore we assume the system to be made up of five point masses where the mass of each body is considered  at its geometrical centre.

The y-coordinate of the centre of mass is

The angular momentum of the mass m about O is mr²ω and is directed toward +z direction for all locations of m.
The angular momentum of mass m about P is mvl and is directed for the given location of m as shown in the figure.
The direction of remains changing for different locations of m.

We know that 

From the situation it is clear that the moment of inertia for (rod + insect) system is increasing.

Let at any instant of time 't', the insect is at a distance x from O. At this instant, the moment of inertia of the system is

From (i) & (ii)

When the insect stops moving, ur does not change and therefore τ becomes constant.

Considering the normal reaction of the floor and wall to be N and with reference to the figure.

By vertical equilibrium.

By horizontal equilibrium

Taking torque about A we get

1.6g × AB = N × x