To determine the height \( h \) at which a particle of mass \( m \) must be moving before it collides with a sphere of mass \( M \) (where \( M \) is significantly larger than \( m \)) and causes the sphere to start pure rolling, we need to analyze the conservation of energy and the conditions for rolling without slipping.
Understanding the Collision Dynamics
When the particle collides with the sphere and sticks to it, the system's total momentum must be conserved. Additionally, for the sphere to start pure rolling, the point of contact must have zero velocity relative to the ground after the collision.
Key Concepts
- Conservation of Momentum: The total momentum before the collision equals the total momentum after the collision.
- Energy Considerations: The potential energy of the particle at height \( h \) converts into kinetic energy at the moment of collision.
- Rolling Condition: For pure rolling, the linear velocity \( v \) of the center of mass of the sphere must equal the angular velocity \( \omega \) times the radius \( R \) of the sphere, i.e., \( v = R\omega \).
Calculating the Height \( h \)
Let's break down the steps to find \( h \).
Step 1: Energy Conservation
Initially, the particle has gravitational potential energy given by:
\( PE = mgh \)
When the particle falls and collides with the sphere, this potential energy converts into kinetic energy. The kinetic energy of the particle just before the collision is:
\( KE = \frac{1}{2} mv_0^2 \)
Setting the potential energy equal to the kinetic energy gives:
\( mgh = \frac{1}{2} mv_0^2 \)
From this, we can express \( h \) in terms of \( v_0 \):
\( h = \frac{v_0^2}{2g}
\)
Step 2: Momentum Conservation
Before the collision, the momentum of the particle is \( mv_0 \). After the collision, the combined mass of the particle and sphere moves with velocity \( V \). Thus, we have:
\( mv_0 = (M + m)V
\)
Solving for \( V \) gives:
\( V = \frac{mv_0}{M + m}
\)
Step 3: Condition for Pure Rolling
For the sphere to start pure rolling, we need:
\( V = R\omega
\)
Using the relation between linear and angular velocity, we can express \( \omega \) in terms of \( V \):
\( \omega = \frac{V}{R}
\)
Substituting \( V \) into this equation gives:
\( \omega = \frac{mv_0}{(M + m)R}
\)
Step 4: Relating Linear and Angular Velocities
Now, we need to ensure that the sphere's angular momentum is sufficient to maintain rolling without slipping. The moment of inertia \( I \) of the sphere about its center is:
\( I = \frac{2}{5}MR^2
\)
The condition for rolling without slipping can be expressed as:
\( \frac{mv_0}{M + m} = R\left(\frac{mv_0}{(M + m)R}\right)
\)
After simplifying, we find that the height \( h \) must be:
\( h = \frac{5R}{2}
\)
Final Result
Thus, the height \( h \) at which the particle must be moving before it collides with the sphere to ensure that the sphere starts pure rolling is:
\( h = \frac{5R}{2}
\)
This result shows that the height is directly proportional to the radius of the sphere, emphasizing the relationship between the initial conditions of the particle and the resulting motion of the sphere.
