A thin rod of mass 0.570 kg and length 1.62 m is at rest, hanging vertically fro

Question

A thin rod of mass 0.570 kg and length 1.62 m is at rest, hanging vertically from a strong, fixed hinge at its top end. Suddenly, a horizontal impulsive force (14.2i cap) N is applied to it. Suppose the force acts at the bottom end of the rod. Find the acceleration of its center of mass, a vector= Find the horizontal force the hinge exerts. F vector = Suppose the force acts at the midpoint of the rod. Find the acceleration of this point. a vector = Find the horizontal hinge reaction force. F vector = Where can the impulse be applied so that the hinge will exert no horizontal force? This point is called the center of percussion. m (from the top)

Explanation / Answer

here,

m = 0.57 kg

L = 1.62 m

About its end, the moment of inertia

I = mL^2/3 = 0.570kg * (1.62m)^2 = 1.496 kg·m^2

(a)

torque t = I*alpha

but also

t = F*L = 14.2N * 1.62m = 23 N·m

so

1.496 kg·m^2 * alpha = 23 N·m,

and

alpha = 15.38 rad/s^2

At the CM:

acceleration a = alpha*L/2

a = 13.8rad/s^2 * 1.62 m/s^2 / 2 = 12.46 m/s^2

(b)

net force Fnet = m*a = 0.570kg * 12.46 m/s^2 = 7.1 N

so the force exerted by the hinge is

Fh = F - Fnet = (14.2 - 7.1)N = 7.1 N

(c)

Now t = F * L/2 = 11.5 N·m

and alpha = 15.38 N·m /
1.496 kg·m^2 = 10.28 rad/s^2

So a = 10.28 m/s^2 * 1.62m / 2 = 8.33 m/s^2

(d)

Fh = 14.2 N - 0.810 * 10.28 m/s^2 = 5.87 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.