Step 1: Understanding the Concept:
When an impulsive force is applied to an object, it produces both a linear change in momentum and an angular change in momentum.
For a sphere to "roll without slipping" immediately, the linear velocity of the center of mass (\(v\)) and the angular velocity (\(\omega\)) must satisfy the kinematic constraint \(v = R\omega\).
Since the surface is smooth, friction does not act during the impact to provide any additional torque or force.
Therefore, the relationship between translation and rotation is purely determined by where the impulse is applied relative to the center.
Key Formula or Approach:
1. Linear Impulse: \(P = \int F dt = Mv\)
2. Angular Impulse: \(\tau_{imp} = P \times h = I\omega\)
3. Moment of Inertia for a solid sphere: \(I = \frac{2}{5}MR^2\)
4. Rolling Condition: \(v = R\omega\)
Step 2: Detailed Explanation:
Let a horizontal impulse \(P\) be applied at a height \(h\) above the center of mass.
1. Linear Motion:
The impulse causes the center of mass to gain velocity \(v\).
\[ P = Mv \quad \text{--- (Equation 1)} \]
2. Rotational Motion:
The impulse \(P\) acting at a distance \(h\) from the center creates an angular impulse.
\[ P \cdot h = I\omega \]
Substitute the moment of inertia for a uniform solid sphere:
\[ P \cdot h = \left( \frac{2}{5}MR^2 \right) \omega \quad \text{--- (Equation 2)} \]
3. Combining Equations for Pure Rolling:
For rolling without slipping immediately, we substitute \(\omega = \frac{v}{R}\) into Equation 2:
\[ P \cdot h = \frac{2}{5}MR^2 \left( \frac{v}{R} \right) \]
\[ P \cdot h = \frac{2}{5}MRv \]
From Equation 1, we know that \(Mv = P\). Substituting this back:
\[ P \cdot h = \frac{2}{5}R \cdot P \]
By canceling \(P\) from both sides (as \(P \neq 0\)):
\[ h = \frac{2}{5}R \]
This specific height is the "sweet spot" where the imparted rotation perfectly matches the imparted translation for immediate rolling.
Step 3: Final Answer:
The height \(h\) must be \(\frac{2R}{5}\).