A thin rod of mass 0.620 kg and length 1.22 m is at rest, hanging vertically fro

Question

A thin rod of mass 0.620 kg and length 1.22 m is at rest, hanging vertically from a strong, fixed hinge at its top end. Suddenly, a horizontal impulsive force (13.2î) N is applied to it. Find everything in vector/ i +j form)

a) Suppose the force acts at the bottom end of the rod. Find the acceleration of its center of mass.

b) Find the force the hinge exerts.

(e) Where can the impulse be applied so that the hinge will exert no horizontal force? This point is called the center of percussion.
(down from the top)

Explanation / Answer

About its end, the moment of inertia
I = mL²/3 = 0.620kg * (1.22m)² = 0.922 kg·m²

(a) torque = I*
but also
= F*L = 13.2N * 1.22m = 16.10 N·m
so
0.922kg·m² * = 16.10 N·m, and
= 17.46 rad/s²

At the CM:
acceleration a = *L/2 = 17.46rad/s² * 1.22m/s² / 2 = 10.65 m/s²

(b) net force Fnet = m*a = 0.620kg * 10.65m/s² = 6.60 N
so the force exerted by the hinge is
Fh = F - Fnet = (13.2 - 6.60)N = 6.6 N

(c) Now = F * L/2 = 8.05N·m
and = 8.05N·m / 0.92kg·m² = 8.75 rad/s²
So a = 8.75m/s² * 1.56m / 2 = 5.33 m/s²

(d) Fh = 13.2N - 0.620*5.33m/s² = 9.89 N

(e) We need the point where
a = F / m = 13.2N / 0.620kg = 21.29 m/s²
= a / L/2 = 21.29m/s² / (1.22m / 2) = 34.9 rad/s²
0.92kg·m² * 34.9rad/s² = 13.2 N * d
d = 2.43m

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