System Of Particles Rotational Motion

As no external force is acting on the system, the centre of mass must be at rest i.e

Apply conservation of angular momentum.

Here,

or
....(i)
Here,
Substituting in E (i), we get

Further,
Correct options are (a), (c) and (d).

Since, it is a square lamina

and (by symmetry)
From perpendicular axes theorem.
Moment of inertia about an axis perpendicular to square plate and passing from is

or
Hence,
Rather we can say
Therefore, can be obtained by adding any two i.e.

After collision between C and A, C stops while A moves with speed of C i.e. v [in head on elastic collision, two equal masses exchange their velocities]. At maximum compression, A and B will move with same speed v/2 (From conservation of linear momentum).

Let x be the maximum compression in this position.

KE of A-B system at maximum compression



or

From conservation of mechanical energy in two positions shown in above figure

or

Work done
If x is the distance of mass 0.3 kg from the centre of mass, we will have,

For work to be minimum, the moment of inertia (/) should be minimum, or
or
or

Let v' be the velocity of second fragment. From conservation
of linear momentum,
2 m (v cos
v ' = 3 cos

|v| = v = constant and |r | = r (say)
Angular momentum o f the particle about origin 0 w ill be given by
Therefore, angular momentum o f particle about origin will remain constant. The direction o f r x v also remains the same (negative z).
NOTE: Angular momentum of a particle moving with constant velocity about any point is always constant e.g.
Angular momentum of the particle shown in figure about origin 0 will be
L = m vh = constant

i.e
This relation implies that is perpendicular to both A and L Therefore, option (a) is correct.
(c) Here,
Differentiating w.r.t. time, we get


But since,

Therefore, from E (i) or magnitude of L i.e. L does not change with time,
(b) So far we are confirm about two points
(1) and
(2) | L | = L is not changing with time, therefore, it is a case
when direction of L is changing but its magnitude is constant and x is perpendicular to L at all points.
This can be written as
If
Here, a = positive constant
Then,
So, that
Now, A is constant vector and it is always perpendicular to x.
Thus, A can be written as A = A
we can see that L A = 0 i.e. L A also.
Thus, we can say that component of L along A is zero or component of L along A is always constant.
Finally we conclude that , A and L are always mutually perpendicular.

Since, it is head on elastic collision between two identical spheres, they will exchange their linear velocities i.e., A comes to rest and B starts moving with linear velocity v. As there is no friction anywhere, torque on both the spheres about their centre of mass is zero and their angular velocities remain unchanged. Therefore,
and

..........(i)
We may write
Angular momentum about O = Angular momentum about
CM + Angular momentum o f CM about origin

'

NOTE that in this case [ Figure (a) ] both the terms in E (i)
i.e have the same direction A. That is why we haveused . We will use =/o)~MRv if they are in opposite direction as shown in figure (b).

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

At the critical condition, normal reaction N will pass through point P. In this condition (about P) the block will topple when or

Therefore, the minimum force required to topple the block is

Mass of the whole disc = 4M

Moment of inertia of the disc about the given axis


Moment of inertia of quarter section of the disc

Mass of the ring M = L. Let R be the radius of the ring, then
L = 2n R or R =
Moment of inertia about an axis passing through O and parallel to XX 'will be

Therefore, moment of inertia about XX' (from parallel axis theorem) will be given by

Substituting values of M and R

Velocity of point O is is in the direction shown in figure. In vector form But

Since, there is no external torque, angular momentum will remain conserved. The moment o f inertia will first decrease till the tortoise moves from A to C and then increase as it moves from C and D. Therefore, (0 w ill initially increase and then decrease.
Let R be the radius o f platform, m the mass o f disc and M is the mass o f platform.
Moment o f inertia when the tortoise is at A

and moment o f inertia when the tortoise is at B

Here ,
From conservation o f angular momentum

Substituting the values we can see that variation o f is non-linear.

In uniform circular motion the only force acting on the particle is centripetal ( towards centre). Torque o f this force about the centre is zero. Hence, angular momentum about centre remain conserved.

Let be the angular velocity of the rod. Applying, angular impulse = change in angular momentum about centre of mass of the system

Direction of hence the direction of angular momentum remains the same.
Correct answer is (d).

or

In case of pure rolling,
(upwards)

Therefore, as 0 decreases force of friction will also decrease.

Since,
Angle between F and p should be

Initial momentum of the system
Final momentum should also be zero
Option (b) is allowed because if we put
W'H t>e zero. Similary, we can check other options.
Correct options are (a) and (d).

Condition of sliding is

or
or ...(i)
Condition of toppling is
Torque of

or .....(ii)
With increase in value of , condition of sliding is satisfied first.

and

- 2 =


= 0 +
= (1) (3) = 3 kg -m/s
( 5 ) (1) = 5 kg - m / s
J
J
Correct options are (a) and (c)

The data is incomplete. Let us assume that friction from ground on ring is not impulsive during impact.
From linear momentum conservation in horizontal direction,
we have


Here, v is the velocity of CM of ring after impact. Solving the above equation, we have v = 0
Thus CM becomes stationary.
Correct option is (a).
Linear impulse during impact
(i) In horizontal direction

(ii) In vertical direction
Writing the equation (about CM)
Angular impulse = Change in angular momentum
We have,

Solving this equation co comes out to be positive or anti-clockwise. So just after collision rightwards slipping is taking place.
Hence, friction is leftwards.
Therefore, option (c) is also correct.

R =

Applying momentum conservation just before and just after the collision
(0.01) (v) = (0.2) (20) + (0.01) (100)
v = 500 m/s

or (about axis of rod)

Here m = mass of insect

Now or
i.e. the graph is straight line passing through origin.
After time T,L = constant
or

Angular momentum of a particle about a point is given by
For
= constant
Direction of is always upwards. Therefore, complete is constant, both in magnitude as well as direction.
For

Magnitude o f will remain constant but direction o f keeps on changing.

In case of pure rolling,
as its moment of inertia is less. Therefore, Q reaches first with more linear speed and more translational kinetic energy.
Further, or
as

Condition of translational equilibrium


Solving

Applying torque equation about corner (left) point on the floor

Solving

edge length : (x)


Now,
mass of cube :

Given:
- Mass of bob, m = 0.1 kg
- Length of string, L = 10 m
- Height of suspension point from floor, H = 9 m
- Impulse imparted, P = 0.2 kg·m/s

Step 1: Velocity of the bob immediately after impulse
Using the impulse-momentum theorem:
P = m × v ⟹ v = P / m = 0.2 / 0.1 = 2 m/s

Step 2: Geometry at the moment of lift-off
The bob lifts off when the string becomes taut. At that moment:
- The string is stretched to full length L = 10 m
- The suspension point is 9 m above the floor
So, the bob must be horizontally separated by:
x = √(L² − H²) = √(100 − 81) = √19 ≈ 4.36 m

Step 3: Angular momentum just before lift-off
At the moment of lift-off, the velocity of the bob is horizontal, and the position vector from the point of suspension to the bob makes an angle θ with the vertical.
We use the formula:
J = m × v × L × sin(θ)
Here, sin(θ) = horizontal distance / string length = √19 / 10
So,
J = 0.1 × 2 × 10 × (√19 / 10) = 0.1 × 2 × √19 ≈ 0.1 × 2 × 4.36 = 0.872 kg·m²/s

the angular momentum about the point of suspension using the perpendicular distance to the direction of velocity (which is vertical height H = 9 m), we use:
J = m × v × H = 0.1 × 2 × 0.9 = 0.18 kg·m²/s