Mechanics Problem 23

Position and velocity of a point on the disk:

\[ x = r \cos \theta \\ y = r \sin \theta \\ \dot{x} = -r \sin \theta \, \dot{\theta} \\ \dot{y} = r \cos \theta \, \dot{\theta}\]

Since the friction force opposes the velocity, define a unit vector opposite the velocity.

v=x˙2+y˙2ux=x˙/vuy=y˙/v v = \sqrt{\dot{x}^2 + \dot{y}^2} \\ u_x = - \dot{x} / v \\ u_y = - \dot{y} / v

Let the mass of the disk be M M and the radius of the disk be R R . Derive the mass and normal force for a patch of the disk as follows:

ρ=MπR2dA=rdrdθdm=ρdAdN=dmg \rho = \frac{M}{\pi R^2} \\ dA = r \, dr \, d \theta \\ dm = \rho \, dA \\ dN = dm \, g

If x>0 x > 0 , the infinitesimal vector force on a patch of disk is:

dFx=μ1dNuxdFy=μ1dNuydF_x = \mu_1 \, dN \, u_x \\ dF_y = \mu_1 \, dN \, u_y

If x0 x \leq 0 , the infinitesimal vector force on a patch of disk is:

dFx=μ2dNuxdFy=μ2dNuydF_x = \mu_2 \, dN \, u_x \\ dF_y = \mu_2 \, dN \, u_y

The total force is then:

Fx=02π0RdFxFy=02π0RdFy F_x = \int_0^{2 \pi} \int_0^R dF_x \\ F_y = \int_0^{2 \pi} \int_0^R dF_y

This could probably be done analytically, but I did it numerically instead. The answer is independent of the particular values of MM , RR , and θ˙\dot{\theta} . The initial acceleration of the center of mass is approximately 0.35 0.35 .

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import math
import random

Num = 5000

u1 = 0.78
u2 = 0.67
pi = 22.0/7.0
g = 10.0

m = 1.0 + 2.0*random.random()
R = 1.0 + 2.0*random.random()
A = pi*(R**2.0)
rho = m/A

thetad = 1.0 + 5.0*random.random()

dtheta = 2.0*math.pi/Num
dr = R/Num

###################################

Fx = 0.0
Fy = 0.0

r = dr

while r <= R:

    theta = 0.0

    while theta <= 2.0*math.pi:

        x = r*math.cos(theta)
        y = r*math.sin(theta)

        xd = -r*math.sin(theta)*thetad
        yd = r*math.cos(theta)*thetad

        v = math.hypot(xd,yd)

        ux = -xd/v
        uy = -yd/v

        dA = r*dr*dtheta
        dm = rho*dA
        dN = dm*g

        if x > 0.0:
            dF = u1*dN
        else:
            dF = u2*dN

        dFx = dF*ux
        dFy = dF*uy

        Fx = Fx + dFx
        Fy = Fy + dFy

        theta = theta + dtheta

    r = r + dr

###################################

xdd = Fx/m
ydd = Fy/m

a = math.hypot(xdd,ydd)

print Num
print ""
print xdd
print ydd
print ""
print a

###################################

#>>> 
#5000

#3.93753857887e-13
#-0.349929953951

#0.349929953951
#>>> ================================ RESTART ================================
#>>> 
#5000

#3.93609652909e-13
#-0.349929953951

#0.349929953951
#>>> ================================ RESTART ================================
#>>> 
#10000

#5.7533637902e-13
#-0.349964988488

#0.349964988488
#>>> ================================ RESTART ================================
#>>> 
#10000

#5.75344675709e-13
#-0.349964988488

#0.349964988488
#>>> 

Note by Steven Chase
2 weeks, 3 days ago

No vote yet
1 vote

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@Talulah Riley Here is my solution for Problem 23

Steven Chase - 2 weeks, 3 days ago

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@Steven Chase Thank you so much for the solution . By the way check your last 8 hour notifications.

Talulah Riley - 2 weeks, 3 days ago

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@Talulah Riley Hey; try my latest problem.

Krishna Karthik - 2 weeks, 3 days ago

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Tricky problem. Nice solution; I can't think of doing it by hand, however.

Krishna Karthik - 2 weeks, 3 days ago

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Your line number 18 is wrong. The question says that we should use the approximation π=227\pi = \frac{22}7, But you took π\pi as 3.1415926533.141592653\ldots.

The same issue arise in line number 31.

Pi Han Goh - 2 weeks, 3 days ago

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Ah yes, old habits die hard

Steven Chase - 2 weeks, 3 days ago

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