MP HW14 solution (due Apr 18st) PHY211 spring 2014
Prepared by Ganapathi Kamath gkamathh@syr.edu
Problem: Hoop on a Ramp A circular hoop of mass m, radius r,
and infinitesimal thickness rolls without slipping down a ramp
inclined at an angle θ with the horizontal
a) acceleration a
Ihoop = m r2
Moment of inertia of a Hoop
ma = Σ F = mg sin(θ) - Fr
Concept: Since there is no slip, the frictional force
Fr = μmg cos(θ)
a =αr
ma = mg sin(θ) - μmg cos(θ)
τ = I α = Fr r
m r2 α = μmg cos(θ) r = Fr r
∴ ma = m r α = μmg cos(θ) = Fr
… goes into causing rotational motion.
This angular acceleration eq holds under no slip.
Newton’s II Law gives acceleration of the hoop.
Torque equations.
… equating and carefully canceling only an r
… we get relation between a and Fr .
ma = mg sin(θ) - ma
a = ½ g sin(θ)
b) minimum coefficient of (static) friction μmin
If slip exists, it means the frictional force is insufficient to generate the torque necessary to cause
angular acceleration to keep up with the linear acceleration.
r α = μg cos(θ)
a=r α = μming cos(θ)
Min friction that causes sufficient α (ang. accel.)
μmin = a / g cos(θ)
= ½ tan(θ)
c) maximum angle θmax, for μ ≥ 0.9
θmax = tan-1(2μ)
= 60.945°
nb. For icy roads slipping happens at smaller θ
Problem: 12.34) Solid Sphere : d=0.076m
a) angular velocity ω=?
I = 2/5 m r2
m=0.4Kg
ma = mg sin(θ) – Fr
= mg sin(θ) – μmg cos(θ)
τ=I α = Fr r
2/5 m r2 (a/r) = μmg cos(θ) r
θ=18°
s=2m
Moment of inertia of a Sphere
Newton’s II Law gives acceleration of the Sphere
Torque equations
Eliminate μ, Solve for a
2/5 m a = μmg cos(θ)
ma = mg sin(θ) – 2/5 ma
a = 5/7 g sin(θ)
b) K=KR+ KL
ωf = vf / r = (√(2as)) / r
vf2= v02 + 2as
= 2 (√(2as)) / d
v= ωr
= 77.408 rad/s
KR/K = ? What fraction of the total kinetic energy is rotational
KR/K = (½ Iω2) / (½ Iω2 + ½ mv2)
= ( (2/5) r2 ω2) / ( (2/5) r2 ω2 + v2)
v=ωr
= ( (2/5)) / ( (2/5) + 1)
= 2/5 / 7/5 = 2/7 = 0.286
No units, this is a ratio
Problem: Energy of a Spacecraft
Mass and Radius : Me, Re
R=∞
speed on crash = ?
T=Ui+Vi =Uf+Vf
0+0 =-GMem/ Re + ½ mse2
R= α
se = √ (2GMe /Re)
speed on at αRe= ?
T=Ui+Vi =Uf+Vf
0+0 =-GMem / αRe + ½ msα2
sα = √ (2GMe / αRe )
sα = se /√α
To check , confer with intuition
when α=∞ then sα=0
when α=1 then sα=se
Problem 13.13
What is the escape speed from Uranus?
G = 6.67384 * 10-11 m3kg-1s-2
MU=86.81* 1024 kg RU=25.559*106 m
vr=√ ( 2 G MU / RU )
<< req. Expression for Escape Speed
= 21292 m/s
Ans in MP=22.3km/s (different constants ?)
Problem Orbiting Satellite
Find the kinetic energy of this satellite.
R1=Radius of the planet
R2=Radius of orbit of the satellite around the planet
map= FG = G Mplanet msatellite / R22 Newtons II Law, FG causing centripetal acceleration
ap = vt2/R2
mvt2/R2= G Mplanet msatellite / R22
K=½ mplanet vr2= ½ G Mplanet msatellite / R2
= G (4/3πR13ρ) m / R2
= 2πR13ρG msatellite / 3R2
Find U, the gravitational potential energy of the satellite
U= - G Mplanet msatellite / R2
Gravitational Potential Energy
3
U= - G (4/3πR1 ρ) msatellite / R2
What is the ratio of the kinetic energy of this satellite to its potential energy?
K/U= -½
Explanation1: The satellite and planet form a binary system. In general, for a many body system, the
potential energy released in descent from ∞ is shared between the kinetic energies of multiple bodies
interacting through force laws.
FYI/For the curious: Explanation2: This is a special case of a general and powerful theorem of
advanced classical mechanics: The Virial Theorem http://en.wikipedia.org/wiki/Virial_theorem. The
theorem applies to the time average of the kinetic and potential energies of any one or multiple objects
moving over any closed (or almost closed) path that returns very close to itself provided that all objects
interact via potentials with the same power law dependence on their separation. The Virial theorem is
used
to
detect
the
presence
of
unseen
matter
such
as
Dark
Matter.
http://en.wikipedia.org/wiki/Dark_matter
Let nF be the order of the distance of separation in the power law equations of forces: F α rnF
Ex1. Gravitational Force is an inverse squared law (nF = -2) FG = G Mplanet msatellite / R22
Ex2. Spring Force is linear in distance (nF = 1) F = -kx1
A notion related to the forces in the system Gvirial is introduced
The Virial theorem says for such power law systems: dGvirial/dt = 2K – (nF+1)U
Angular brackets <z> means time average: <z>= ∫z dt / T. <dGvirial/dt>= 2<K> – (nF+1) <U>
Under Equilibrium, one sets <dGvirial/dt> = 0 and we get 0 = 2<K> – (nF+1) <U>
So, one deduction from the Virial theorem is that <K>/<U> = (nF+1)/2
So, in case of gravitational/electrostatic forces, nF = -2: <K>= -1/2 <U>
<K>/<U> = -1/2
Similarly, in case of springs, nF = 1: <K>/<U> = 1
nb. in the Wikipedia article(Apr 28th 2014), K=T, U=VTot and n is order of the distance of separation in
the power law equations of potentials : n=nF+1 V α rn (Equations for potential are higher by 1
order than the eqs for force because they are integrals of the respective force over distance)
The Virial theorem and the Law of conservation of energy are two different things.
Virial theorem is a statement about the time integrals of Kinetic and Potential energy under
equilibrium.
Law of conservation of Energy is a statement about the indestructibility of the sum of Kinetic and
Potential and other forms of energies at any point in time.
The virial theorem can be applied here despite being a statement on time integrals because a satellite in
equilibrium, in a constant height circular orbit has constant potential and kinetic energies. And Hence,
time integrals (the time averages) become the same as the constant instantaneous values.
<K>= ∫K dt / T =K
<U>= ∫U dt / T = U
∴ K/U = -1/2
The <K>/<U> = -1/2
remains the same even when the satellite is in a stable equilibrium orbit with
perturbations, though K and U are themselves no longer constant and K/U ≠ -1/2 .
This virial theorem argument can’t be used to show something similar for the spring because, the
Kinetic and Potential Energies of the spring change with time. One can show that for the spring
K= ½ mv2 = - ½ m A2 ω2 sin2(ωt) <K>= ∫K dt / T = ¼ k A2 = ½ ETot
U= ½ kx2 = ½ m A2 cos2(ωt) <U>= ∫U dt / T = ¼ k A2 = ½ ETot
So in case of springs, nF = 1: <K>/<U> = 1
but K/U=ω2 tan2(ωt)
0 ≤ K/U < ∞
Problem 13.16
Orbital velocity on a different planet, whose mass is three times larger, and gp is one fourth
G = 6.67384 * 10-11 m3kg-1s-2
Mp = 3 Me = 3 * 5.972 *1024 Kg
gp = 0.25 g
Re = 6.378 *106 m
mg= FG = G M m / R2
g= G Me / Re2
gp= G Mp / Rp2
gp/g= ( Mp/Me )( Re2 / Rp2 )
0.25= 3 * (Re / Rp)2
Rp= Re √ ( 3 / 0.25 )
= Re 2√ 3
= 22.094 * 106 m
Orbital velocity
vt=√ ( 2 G Mp / Rp )
=10.4 km/s
Newtons II Law, FG causing centripetal acceleration
Acceleration due to gravity
Divide one equation by the other
<< req. Expression for Orbital speed