A uniform rod of length l is pivoted at point A. It is struck by a horizontal force which delivers an impulse J at a distance x from point A as shown in figure, impulse delivered by pivot is zero, if x is equal to :

Idea (Center of Percussion)

If the blow is delivered at a special point (the center of percussion), the required impulse at the pivot to keep it instantaneously at rest is zero. We will compute that point using impulse–momentum relations and the kinematic constraint that the pivot point has zero velocity.

Notation

• Mass of rod: m
• Length of rod: l
• Distance of center of mass from pivot: r_c = l/2 (uniform rod)
• Moment of inertia about pivot A: I_A = (1/3) m l² (uniform rod about one end)
• Angular velocity just after the impulse: ω

Step 1: Linear impulse–momentum

Let the impulsive reaction at the pivot be R. Along the direction of the applied impulse:

J + R = m v_C,

where v_C is the velocity of the center of mass immediately after the strike. We want the pivot impulse to be zero, i.e., R = 0, hence

v_C = J / m.

Step 2: Angular impulse–momentum about the pivot

The angular impulse about A due to J is J × x. Therefore,

I_A ω = J x  →  ω = (J x) / I_A.

Step 3: Kinematic constraint (pivot is instantaneously at rest)

The velocity of the center of mass due to rotation about A is v_C = ω r_c. Since the pivot is fixed in space (its instantaneous velocity is zero), this relation must hold. Combining with Step 1:

J / m = ω r_c = (J x / I_A) r_c.

Cancel J (nonzero impulse) and solve for x:

1 / m = (x r_c) / I_A  →  x = I_A / (m r_c).

Step 4: Substitute for a uniform rod

I_A = (1/3) m l²,   r_c = l/2.

Hence
x = ((1/3) m l²) / (m · (l/2)) = (1/3) · (l² / (l/2)) = (1/3) · 2l = (2/3) l.

Answer

x = (2/3)   l.

The point at distance (2/3)l from the pivot is the rod’s center of percussion. A blow delivered there produces no impulsive reaction at the pivot.