A simple pendulum is shown a mass of 0.4 kg suspended from the ceiling by means
ID: 1511742 • Letter: A
Question
A simple pendulum is shown a mass of 0.4 kg suspended from the ceiling by means of a string of length 3.3 m. Assume that the acceleration due to gravity is g=10 m/s^2, and that there is no friction or air resistance.
a) Suppose you were to release the pendulum from rest, starting from an angle 36.2 degrees with respect to the vertical, as shown. What will be the speed of the pendulum at the instant it swings through its lowest point (that is, when its momentarily hanging vertically)? (Answer in m/s)
b)Assuming the same situation as in part (a) what will be the speed of the pendulum at the instant it reaches an angle of 30 degrees above the vertical? (Answer in m/s)
c) Assume you start with the pendulum again at 36.2 degrees with respect to the vertical, but rather than releasing it from rest, you give it a push downward. It swings to the otherside, and reaches a maximum angle of 52.1 degrees with respect to the vertical. What must have been the initial speed of the pendulum just after you pushed it? (Answer in m/s)
d) Suppose you start with the pendulum hanging vertically, at rest. You then give it a push so that it starts swinging with a speed of 4.3 m/s. What maximum angle (in degrees) will it reach, with respect to the vertical, before falling back down? (answer in degrees)
e) Same situation as in part (d), but now, due to friction at the pivot point, 1 joule of the pendulum's initial kinetic energy is lost during the upward swing. Then again, what maximum angle (in degrees) will it reach, with respect to the vertical, before falling back down? (Answer in degrees)
Explanation / Answer
a)
initial potential energy Ui = m*g*L*(1-costheta)
initial kinetic energy ki = 0
total initial energy Ei = ui + ki
final potential energy Uf = 0
final kinetic energy kf = 0.5*m*v^2
total final energy Ef = uf + kf
from energy conservation
Ei = Ef
m*g*l*(1-costheta) = 0.5*m*v^2
0.4*9.8*3.3*(1-cos36.2) = 0.5*0.4*v^2
v = 3.53 m/s
++++++++++++++++
(b)
initial potential energy Ui = m*g*L*(1-cos36.2)
initial kinetic energy ki = 0
total initial energy Ei = ui + ki
final potential energy Uf = m*g*l*(1-cos30)
final kinetic energy kf = 0.5*m*v^2
total final energy Ef = uf + kf
from energy conservation
Ei = Ef
m*g*l*(1-cos36.2) = m*g*l*(1-cos30) + 0.5*m*v^2
0.4*9.8*3.3*(1-cos36.2) = (0.4*9.8*3.3*(1-cos30)) + (0.5*0.4*v^2)
v = 1.95 m/s
++++++++
(c)
initial potential energy Ui = m*g*L*(1-cos36.2)
initial kinetic energy ki = 0.5*m*u^2
total initial energy Ei = ui + ki
final potential energy Uf = m*g*l*(1-cos30)
final kinetic energy kf = 0
total final energy Ef = uf + kf
from energy conservation
Ei = Ef
m*g*l*(1-cos36.2) + (0.5*m*u^2) = m*g*l*(1-cos52.1)
0.4*9.8*3.3*(1-cos36.2) + (0.5*0.4*u^2) = (0.4*9.8*3.3*(1-cos52.1))
u = 3.53 m/s
____________________
(d)
initial potential energy Ui = m*g*L*(1-cos36.2)
initial kinetic energy ki = 0.5*m*u^2
total initial energy Ei = ui + ki
final potential energy Uf = m*g*l*(1-cos30)
final kinetic energy kf = 0
total final energy Ef = uf + kf
from energy conservation
Ei = Ef
m*g*l*(1-cos36.2) + (0.5*m*u^2) = m*g*l*(1-costheta)
0.4*9.8*3.3*(1-cos36.2) + (0.5*0.4*4.3^2) = (0.4*9.8*3.3*(1-costheta))
theta = 58.6 degrees
+++++++++++++++++++
(e)
initial potential energy Ui = m*g*L*(1-cos36.2)
initial kinetic energy ki = 0.5*m*u^2
total initial energy Ei = ui + ki
final potential energy Uf = m*g*l*(1-cos30)
final kinetic energy kf = 0
total final energy Ef = uf + kf
from energy conservation
Ei - W = Ef
m*g*l*(1-cos36.2) + (0.5*m*u^2) - w = m*g*l*(1-costheta)
0.4*9.8*3.3*(1-cos36.2) + (0.5*0.4*4.3^2) - 1 = (0.4*9.8*3.3*(1-costheta))
theta = 53.2 degrees
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.