Compound pendulum — distance to the centre of percussion The centre of percussion lies below the centre of gravity (C.G.) of a body. If h is the distance from the pivot to the C.G. and k_G is the radius of gyration about the C.G., then the distance from the C.G. to the centre of percussion equals:

Introduction / Context:
The centre of percussion (C.P.) is the point where an impulsive force produces pure rotation about the pivot with zero initial reaction at the pivot. This idea is central to analyzing impacts on compound pendulums and tools like bats, hammers, and machine links.

Given Data / Assumptions:

• Rigid body pivoted at a point at distance h from its centre of gravity (C.G.).
• Radius of gyration about the C.G. is k_G.
• Impulsive force acts at the centre of percussion.

Concept / Approach:
For a compound pendulum, the distance from the pivot to the centre of percussion is L_cp = I_p / (m * h), where I_p is the mass moment of inertia about the pivot. Using the parallel-axis theorem, I_p = I_G + m h^2 = m k_G^2 + m h^2. The distance from the C.G. to the C.P. is then L_cp − h = (m(k_G^2 + h^2) / (m h)) − h = k_G^2 / h.

Step-by-Step Solution:

Verification / Alternative check:
Special cases: as h increases, the C.P. moves farther from the C.G.; if k_G → 0 (point mass), the C.P. coincides with the C.G., giving zero lever arm—consistent with intuition.

Why Other Options Are Wrong:

• Expressions like h/k_G, h^2/k_G, or h*k_G have wrong dimensions.
• k_G / h^2 also has incorrect dimensions; the distance must have units of length.

Common Pitfalls:
Mixing up the distance from the pivot with the distance from the C.G.; forgetting to use the parallel-axis theorem.

Final Answer:
k_G^2 / h