Askiitians Tutor Team
Last Activity: 5 Months ago
To solve the problem of a point mass colliding with a disc and determining the angular velocity of the system after pure rolling begins, we need to consider the principles of conservation of momentum and the conditions for rolling motion. Let's break this down step by step.
Understanding the Collision
In this scenario, we have a point mass \( m \) moving with an initial velocity \( u \) towards a stationary disc of mass \( m \) and radius \( r \). The collision is perfectly inelastic, meaning the point mass will stick to the disc upon impact. This type of collision conserves momentum but not kinetic energy.
Applying Conservation of Linear Momentum
Before the collision, the total linear momentum of the system is solely due to the point mass:
- Initial momentum of the point mass = \( mu \)
- Initial momentum of the disc = \( 0 \) (since it is at rest)
After the collision, the combined mass (point mass + disc) moves with a common velocity \( V \). The total mass after the collision is \( 2m \). Therefore, we can write the conservation of momentum as:
Initial momentum = Final momentum
Thus, we have:
\( mu = (2m)V \)
From this, we can solve for \( V \):
\( V = \frac{u}{2} \)
Finding the Angular Velocity
Next, we need to relate the linear velocity \( V \) to the angular velocity \( \omega \) of the disc. The point where the disc contacts the ground must have zero velocity for pure rolling to occur. The relationship between linear velocity and angular velocity is given by:
\( V = r\omega \)
Substituting \( V \) into this equation gives:
\( \frac{u}{2} = r\omega \)
From this, we can solve for \( \omega \):
\( \omega = \frac{u}{2r} \)
Considering the Moment of Inertia
However, we also need to account for the moment of inertia of the system. The moment of inertia \( I \) of the disc about its center is:
\( I_{disc} = \frac{1}{2}mr^2 \)
When the point mass sticks to the edge of the disc, we need to add its contribution to the moment of inertia. The point mass contributes:
\( I_{point} = mr^2 \)
Thus, the total moment of inertia \( I_{total} \) after the collision is:
\( I_{total} = I_{disc} + I_{point} = \frac{1}{2}mr^2 + mr^2 = \frac{3}{2}mr^2 \)
Final Angular Velocity Calculation
Now, we can use the conservation of angular momentum about the center of the disc. The initial angular momentum \( L_{initial} \) of the point mass is:
\( L_{initial} = m \cdot u \cdot r \)
After the collision, the angular momentum \( L_{final} \) is given by:
\( L_{final} = I_{total} \cdot \omega \)
Setting these equal gives:
\( mu \cdot r = \frac{3}{2}mr^2 \cdot \omega \)
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\( u \cdot r = \frac{3}{2}r^2 \cdot \omega \)
Dividing both sides by \( r \) (assuming \( r \neq 0 \)) leads to:
\( u = \frac{3}{2}r\omega \)
Solving for \( \omega \) gives:
\( \omega = \frac{2u}{3r} \)
Final Result
After evaluating the options provided, we find that the angular velocity of the system after pure rolling starts is:
None of the options directly match \( \frac{2u}{3r} \), so it seems there may be a discrepancy in the provided choices.
In summary, the key steps involved conservation of momentum, understanding the moment of inertia, and applying the relationship between linear and angular velocities. If you have any further questions or need clarification on any part of this process, feel free to ask!
