Physics CHALLENGE Question?

Question

A small black rubber ball (radius 0.1 m) is placed at the top of a large white hemispherical dome (radius 10 m). The ball is slightly displaced and allowed to fall/roll/slide down the side of the dome. Calculate the arc length of the skid marks left on the dome’s surface.

Coefficient of friction (static & kinetic), μ = 1
Mass moment of inertia of a sphere, I = 2/5 * m*r^2

Answer 1

http://i15.tinypic.com/61myqa1.jpg

Before the ball begins sliding, we have constraint
ω = v/r . . . . . (1)

The ball will start sliding, when friction becomes insufficient to maintain this constraint.
F = μN = N

****** Force of friction
Lets first find what force of friction is needed.
In the (non-inertial, but non-roatating) frame of reference of center of the ball angular momentum is
M = Iω = mv/r
torque = Fr = dM/dt = I dω/dt = I/r dv/dt
F = 2/5m dv/dt
dv/dt = 5/2 F/m


Tangential acceleration dv/dt comes from the second law of newton (from now on in motionless frame of reference):
m dv/dt = mg sin(α) - F
m 5/2 F/m = mg sin(α) - F
F = 2/7 mg sin(α)


****** Normal reaction
Kinetic energy of the ball at all times:
K = 1/2mv² + 1/2Iω²
which becomes before sliding due to (1)
K = 7/10 mv²

Potential energy of the ball:
P(α) = mg(R+r)(cos(α)-1)

Conservation of energy:
K = - P(α)
7/10 mv² = mg(R+r)(1-cos(α))
v² = 10/7 g(R+r)(1-cos(α))

Centripetal acceleration:
ma = mg cos(α) - N
N = m[g cos(α) - v²/(R+r)]
N = mg [17/7 cos(α) - 10/7]


****** Beginning of skidding
F = μN = N
2/7 mg sin(α) = mg [17/7 cos(α) - 10/7]
17 cos(α) - 2 sin(α) = 10
cos(α + arccos(17/√293)) = 10/√293

α0 = arccos(10/√293) - arccos(17/√293) = 0.829786 rad.
cos(α0) = (170 + 2√193)/293
sin(α0) = (17√193 - 20)/293
At this time
v(α0)² = 10/7 g(R+r)(1-cos(α)) =
= 10/7 g(R+r) (123 - 2√193)/293 =
0.4642375 g(R+r)

******* Dynamic equation of skidding ball:
Now F = N at all times.

m dv/dt = mg sin(α) - F = mg sin(α) - N
dv/dt = g[sin(α) - cos(α)] + v²/(R+r)


This ODE is to be intgrated like conservation of energy.
Multiply both sides by v, now instead of forces we have powers:
vdv/dt = v g[sin(α) - cos(α)] + v v²/(R+r)
1/2 d[v²] = (R+r) dα g[sin(α) - cos(α)] + dα v²
-----------> substitution u = v²
du/dα = 2(R+r)g [sin(α) - cos(α)] + 2u


******* Explicit expression for u:
Genaral solution of this linear equation is
u(α) = Uo exp(2α) + √(8/5)(R+r)g cos(α + arccos(1/√10))

Initial condition when the ball begins skidding is
u(α0) =
0.4642375 g(R+r) =
Uo exp(2α0) - 0.61533(R+r)g

Uo = 0.205356 g(R+r)

********* Ball separates from the spere
when N = 0.
N = m[g cos(α) - v²/(R+r)] = 0
(R+r)g cos(α) = u

(R+r)g cos(α) = u(α) =
= Uo exp(2α) + √(8/5)(R+r)g cos(α + arccos(1/√10))

And finally we have equation to solve:
3/√5 cos(α - arctan(2)) = 0.205356 exp(2α)
which has the solution
α1 = 0.93063

Between angles α0 = 0.829786 and α1 = 0.93063 the ball is skidding.



Answer:
L = R(α1 - α0) = 1.01 meter

Interestingly, the answer does not depend on the size of small ball r, even when it's not small.

Answer 2

Uh, don't you think it's time to put down the pen and paper and run a test?