A thin uniform stick having a mass of 0.610 kg and a length of 1.36 m is at rest

Question

A thin uniform stick having a mass of 0.610 kg and a length of 1.36 m is at rest, hanging vertically from a strong, fixed hinge at the uppermost end of the stick. The stick is free to swing back and forth in an Up-Down-East-West plane. Suddenly, the midpoint of the stick is struck by a hammer; the force on the stick by the hammer is 15.2 N towards the East. (a) Find the magnitude of the resulting acceleration of the center of mass of the stick. _________ m/s2 (b) While the 15.2-N force is being applied, find the horizontal force on the stick by the hinge. _________ N (c) There is some point at which the hammer can strike the stick such that the horizontal force on the stick by the hinge will be zero, even though the stick is being struck by the hammer at that instant. Calculate the distance from the hinge to the location of this special point. At _________ m below the hinge.

Explanation / Answer

M=Torque about hinge = 15.2 * 0.68 = 10.37
I about hinge = (1/3) m l^2 = (0.61/3)*1.85 = 0.38

angular acceleration alpha = M/I = 10.37/0.38 = 27.6 radians/s^2

resulting linear acceleration of center East = 0.68 alpha = 18.75 m/s^2 (--> Answer part a)

we better get the same acceleration East from the sum of forces on the stick
15.2east + force from hinge F = 0.61 (18.75)

15.2 + F = 11.44
F = -3.76 or 3.76 N west (--> Answer part B)

Torque
M = 15.2 x
alpha = 15.2 x/0.38 = 40 x
a of center = 0.68 alpha = 27.2 x meters/s^2

F = m a
27.2 x = F/m but F is just 15.2 because the hinge force is zero so
27.2 x = 15.2/0.61
x = 0.92 meter, the end of the stick (--> Answer part C)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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