A metal rod of length L = 6.1 m is free to rotate about a frictionless hinge at
ID: 1400514 • Letter: A
Question
A metal rod of length
L = 6.1 m
is free to rotate about a frictionless hinge at one end (see figure below). The rod is initially at rest and oriented horizontally and then released so that it swings to the bottom.
(a) What are all the forces acting on the rod? (Select all that apply.)
-Friction
-Frictionforce from the hingegravity
-Force
Which force or forces do work on the rod as it swings? (Select all that apply.)
-Friction
-Frictionforce from the hingegravity
-Force
(b) Is the mechanical energy of the rod conserved?
-Yes
-No
(c) Make a sketch showing the rod in its initial orientation and when it is at the bottom of its swing.
(e) If the speed of the end of the rod when it is at its lowest point is vf,
what is the final mechanical energy of the rod? Express your answer in terms of vf.
(Use the following as necessary: L, I for the moment of inertia of the rod, and vf. Do not substitute numerical values; use variables only.)
Ef = ?
(f) Find vf.
vf = ? m/s
Explanation / Answer
a) friction from hinge
Gravity
b) Gravity Force
c) Yes
d) for potential energy, First we need its centre of mass.
its centre of mass is at middle of rod.
it goes to the lowet point when rod is vertical downwards.
that point PE = 0
from that, its height is L/2.
so initial PE = mgL/2
initially it is at rest so initial K.E. = 0
inital M.E. = 0 + mgL/2 = mgL / 2
e) final PE = 0 (we have taken in part d)
K.E. = Iw^2 / 2
w = v / r = vf / L
I = mL^2 / 3
K.E. = m L^2 vf^2 / 6L^2 = m vf^2 / 6
final ME = m vf^2 / 6
f) from energy conserrvation,
inital ME = final ME
mgL / 2 = m vf^2 / 6
vf = sqrt ( 3gL)