Solution:

If We take Ring and 4 small blocks as one system net Torque will be zero, Using Principal of conservation of angular momentum.

\[ I_{1}\omega= \left( I_{1}+ I_{2} \right)\omega_{1} \]

\[ MR^{2}\omega= \left( MR^{2}+ 4mR^{2} \right)\omega_{1} \]

\[ \omega_{1}= \left( \frac{M}{M+4m} \right)\omega \]

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Solution:

Angular Momentum = momentum × ( Perpendicular distance of momentum from axis of rotation )

Angular Momentum = mv cos (45º) × h = mvh/√2

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Solution:

\[ L = I \omega \]

and

\[ K = \frac{1}{2} I \omega^{2} \]

Squaring L and Dividing it with K we get,

\[ I=  \frac{L^{2}}{2K} \]

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Solution:

Distance of line 

\[ ax+by+c=0 \]  from point (x1,y1) is given by 

\[ d = \left( \frac{ax_{1}+ by_{1}+c}{\sqrt{a^{2}+b^{2}}} \right) \]

So, distance of direction of velocity from origin is d= 2√2 

Angular momentum = Perpendicular distance of momentum × momentum = 2√2 × 5 ×3√2= 60 Unit

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Solution:

As object is in circular motion angular momentum

\[ \vec L=I\vec\omega \]

Direction of omega is along the axis so, L will have direction along axis OC. So both magnitude

and direction of angular momentum L is constant.

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Solution:

Taking rectangle and the object as one system angular momentum is conserved.

so,  mva= ( m(√5a/2)² + m((2a)²+ a²)/12) ω

\[ \omega = \frac{3u}{5a} \]

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Solution:

For Rolling L = MvR + Iω = M(ωR)R + (MR²/2)ω = (3/2)MR²ω

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Solution:

We can solve this problem using the principle of conservation of angular momentum since no external torque acts on the system.

Step 1: Initial Angular Momentum

The moment of inertia of a thin circular ring about its axis is:

$$I_{\text{initial}} = M R^2$$

The initial angular momentum is given by:

$$L_{\text{initial}} = I_{\text{initial}} \cdot \omega = (M R^2) \cdot \omega$$

Step 2: Final Moment of Inertia

When two objects of mass m are attached to the ring, assuming they are symmetrically placed on the ring, their contribution to the moment of inertia is:

$$I_{\text{added}} = 2m R^2$$

Thus, the new total moment of inertia becomes:

$$I_{\text{final}} = M R^2 + 2m R^2 = (M + 2m) R^2$$

Step 3: Applying Conservation of Angular Momentum

Since no external torque acts on the system:

$$L_{\text{initial}} = L_{\text{final}}$$

$$(M R^2) \cdot \omega = (M + 2m) R^2 \cdot \omega'$$

Canceling

$R^2$

from both sides:

$$M \omega = (M + 2m) \omega'$$

Solving for the new angular velocity

$\omega'$

:

$$\omega' = \frac{M \omega}{M + 2m}$$

Final Answer:

$$\omega' = \frac{M \omega}{M + 2m}$$

This shows that the angular velocity decreases after attaching the masses, as expected due to an increase in the moment of inertia.

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Solution:

The angular momentum $L$ of a particle about the origin is given by:

$$L = m v r \sin\theta$$

where:

• $m = 5$ g (mass of the particle),
• $v = 3\sqrt{2}$ cm/s (speed of the particle),
• $r = 2\sqrt{5}$ cm (perpendicular distance from the origin),
• $\theta = 90^\circ$ (since the velocity is along a straight line parallel to the x-axis, the perpendicular distance is directly used).

Since $\sin 90^\circ = 1$, the equation simplifies to:

$$L = m v r$$

Substituting the given values:

$$L = (5) \times (3\sqrt{2}) \times (2\sqrt{5})$$ $$L = 5 \times 3 \times 2 \times \sqrt{10}$$ $$L = 30 \sqrt{10} \text{ g-cm²/s}$$

Thus, the magnitude of the angular momentum is:

$$\mathbf{30\sqrt{10} \text{ g-cm²/s}}$$

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Solution:

The angular momentum of a projectile about the point of projection when it reaches its maximum height is given by:

$$L = m v_x r$$

where:

• 
  $m$ 

  = mass of the projectile

• 
  $v_x$ 

  = horizontal component of velocity

• 
  $r$ 

  = perpendicular distance from the point of projection (which is the maximum height $h$ 

  )

Step 1: Horizontal Component of Velocity

The initial velocity components are:

$$v_x = v \cos 45^\circ = \frac{v}{\sqrt{2}}$$

$$v_y = v \sin 45^\circ = \frac{v}{\sqrt{2}}$$

Since there is no acceleration in the horizontal direction (ignoring air resistance),

$v_x$

remains constant throughout the motion.

Step 2: Maximum Height

Using the kinematic equation:

$$v_y^2 = u_y^2 - 2 g h$$

At maximum height, the vertical velocity becomes zero, so:

$$0 = \left(\frac{v}{\sqrt{2}}\right)^2 - 2 g h$$

Solving for

$h$

:

$$h = \frac{v^2}{2 g} \cdot \frac{1}{2} = \frac{v^2}{4g}$$

Step 3: Angular Momentum Calculation

The angular momentum at maximum height is:

$$L = m v_x h$$

Substituting values:

$$L = m \left(\frac{v}{\sqrt{2}}\right) \left(\frac{v^2}{4g}\right)$$

$$L = m \cdot \frac{v}{\sqrt{2}} \cdot \frac{v^2}{4g}$$

$$L = \frac{m v^3}{4g \sqrt{2}}$$

Thus, the magnitude of the angular momentum about the point of projection at maximum height is:

$$\frac{m v^3}{4g \sqrt{2}}$$

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