posted by HALP on .
A thin uniform stick having a mass of 0.680 kg and a length of 1.50 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 12.7 N towards the East.
(a) Find the magnitude of the resulting acceleration of the center of mass of the stick.
(b) While the 12.7-N force is being applied, find the horizontal force on the stick by the hinge.
(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.
M=Torque about hinge = 12.7 * .75 = 9.525
I about hinge = (1/3) m l^2 = (.68/3)2.25
angular acceleration alpha = M/I = 9.525/.51 = 18.7 radians/s^2
resulting linear acceleration of center East = .75 alpha = 14 m/s^2 (part a)
we better get the same acceleration East from the sum of forces on the stick
12.7east + force from hinge F = .68 (14)
12.7 + F = 9.52
F = -3.18 or 3.18 N west
Now the sweet point:
M = 12.7 x
alpha = 12.7 x/.51 = 24.9 x
a of center = .75 alpha = 18.7 x meters/s^2
F = m a
18.7 x = F/m but F is just 12.7 because the hinge force is zero so
18.7 x = 12.7/.68
x = 1 meter, the end of the stick
Try this with a mass distribution like a baseball bat or tennis racquet and the result will make more sense :)
Oh, one meter is not the end of the stick. It is 2/3 of the way out.
Which is about how far out on a bat you want to hit the ball without stinging your hand.