Use conservation of energy. When the ball hits the ground its kinetic energy will equal its starting potential energy:
thus
On rebound its speed will be v1 = 0.8v0. It will rise to a height h1 where its kinetic energy is converted to potential energy:
thus
(10 points) If the motion of a one
dimensional oscillator is not sinusoidal, is the
oscillator harmonic (i.e. is the motion governed
by an equation of the form mx[ddot] + kx = 0)?
Absolutely not. The general
solution for any harmonic oscillator is a sinusoidal
function. The only way to get non-sinusoidal motion is to
not have an harmonic oscillator.
(a) (5 points) If a
simple harmonic oscillator has energy E and amplitude A,
what's the value of the spring constant k?
Energy is the sum of kinetic
plus potential energies for a simple harmonic oscillator.
At a maximum of its displacement, the kinetic energy of
the oscillator is zero, so all of its energy is potential
energy: E = (1/2)kA2, from which we find k =
2E/A2.
(b) (5 points) If this oscillator has a damped frequency ωd, and it takes n
complete oscillations for the amplitude to decrease to
1/e of its initial value, what is the mass of the
oscillator.
e-nβTd = e-1 therefore,
β = 1/nTd = ωd/2nπ. But ω0²
= ωd² + β² = ωd²(1 + 1/4n²π²) = k/m. Solving for m yields:
m = 2E(Aωd)-2(1 + 1/4n²π²)-1.
This result is expressed in terms of known quantities.
(10 points) An electron moves in a region
of space with a constant magnetic field, B
= Bk, and a constant electric field, E
= Ej. Write down the x, y, and z
equations of motion for the electron.
For a charge q moving in electric
and magnetic fields, the Lorentz force is given by F
= q(E + v×B).
For the given fields this becomes:
F = q(Ej + y[dot]Bi
- x[dot]Bj) = mr[ddot] so
the components are
mx[ddot] = y[dot]B
my[ddot] = qE - x[dot]B
mz[ddot] = 0
(10 points) A particle with mass m moves in one dimension subject to a force F = -kx - 2(mk)1/2x[dot].
Set up the equation of motion and
show that
x = Ate-βt
+ Be-βt
is a solution, where A and B are constants of
integration and β²
= k/m.
The equation of motion can
be written:
mx[ddot] + 2(mk)½x[dot] + kx = 0.
The derivatives of the proposed solution are:
x[dot] = -βAte-βt + (A - βB)e-βt and x[ddot]
= β²Ate-βt - β(2A - βB)e-βt.
Upon inserting these expressions into the
equation of motion, and canceling out the common
factor of e-βt, we have:
At{mβ² -2β(mk)½ + k} + mβ(βB -
2A) + 2(mk)½(A - βB) + kB = 0
The expression in curly braces, multiplied by t
must be zero independent of the remaining part.
That expression gives us β² = k/m. Then the
remaining expression becomes:
kB -2kB + kB -2(mk)½A +2(mk)1/2A
= 0.
This is obviously satisfied for any values of A
and B, so we have shown that the proposed
solution works. Additionally, since this solution
contains two independent constants of integration
for a second order differential equation, it must
be a complete solution.
Given that β =
1sec-1 and x[dot](t=1sec) = -0.736m/s and
x(t=1sec) = 0.736m, determine the values of the
constants of integration A and B.
x(t=1sec) = (A+B)e-1 = 0.736m and
x[dot](t=1sec) = -Be-1 = -0.736m. Therefore, B = 2.00m
and A = 0m/s.
(10 points each) A truck at rest has its rear door fully open. The truck accelerates forward at constant rate A, and the door begins to swing shut. The door is uniform, with mass M, height h, and width w, and negligible thickness. Neglect air resistance.
Where is the center of mass of
the door, and what is its moment of intertia
about the hinge (the hinge is along one edge of
the door)?
The center of mass of the
door is at its geometric center, h/2 from the top
or bottom edge, and w/2 from the left or right
edge. The moment of inertia is, using the
parallel axis theorem, (1/12)Mw² +
M(w/2)² = (1/3)Mw².
Find the instantaneous angular
velocity of the door about its hinge when it has
swung through 90°.
The truck is uniformly
accelerating horizontally, so in a frame moving
with the truck, it appears as though there is a
horizontal component to gravity with a value of
-A. Conservation of energy then gives: MAw/2 =
1/2 Iω² or ω = (3A/w)½.
(10 points) A small block of mass m sits
on a disk rotating with angular velocity ω. The coefficient of friction
between the block and the disk is μ.
At an instant of time the block is located a distance x0
from the center of rotation, with an arbitrary velocity.
Write down all the apparent forces (including fictitous
forces) acting on the block as seen from a reference
frame rotating with the disk.
Place the origin or coordinates at
the center of rotation, and fix the x axis to point to
the block. Then the coordinate system rotates with the
angular velocity of the disk, ω = &omega k.
The block is located at r = x0i,
and let's assume it has velocity v = vxi
+ vyj. The forces acting on
the block are:
Centrifugal force: Fcentf = mω²x0i
Coriolis force: Fcor = -2mω×v = 2mω(vyi - vxj)
Friction: Ff = -μmgv/|v|.
(10 points) A bead slides along a circular loop of wire without friction. The loop sits vertically and rotates about an exis passing through its center, with constant angular velocity ω. The loop has radius a, and gravity points downward. The angle from the axis of rotation to the bead is . There are 3 angles for which the bead is in equilibrium, that is, F = 0.
Set up the equations of motion in
an appropriate coordinate system, and find the 3
equilibrium angles.
Let's choose a coordinate
system fixed to the bead. This will then look
similar to the Earth based problems we worked. In
this system, ω = ωsinλ j + ωcosλ k.
(The difference from earlier problems is due to
the definition of λ.) A0
points towards the axis of rotation and is given
by A0 = ωa²sinλ(cosλ j
- sin&lambda k).
For the bead, r = 0, and v
= vj. So, the
transverse and centrifugal forces are zero, and
the coriolis force is Fc
= -2mω×v = -2mvωa²sin²&lambda i. The remaining
forces are the force of gravity, and the force of
the wire on the bead. Fg
= -mg(sinλ j
+ cosλ k). The x and z
components of the forces must be canceled by the
force the wire exerts on the bead. The resulting
force is:
Fy = -mωa²sinλ cos&lambda -
mgsinλ = -msinλ(ωa²cosλ + g).
This force is zero when sinλ=0 at λ=0° or 180°, or when λ = cos-1(-g/ωa²).
If the loop rotates too slowly, one of the
equilibrium angles disappears (two of the angles
merge into one). What is the rotation rate where
this occurs?
If the loop rotates too
slowly such that g/ωa²>1,
then the third angle listed above merges with the
180° solution. This occurs for ω<g/a².