3. A Frisbee is thrown into the air with an angular velocity that is not exactly

Question

3. A Frisbee is thrown into the air with an angular velocity that is not exactly along the axis of symmetry, so that it has a definite wobble. Air resistance exerts a fric- tional torque of -c on the rotation of the Frisbee. Let's define the body coordinates (x1,r2,3) to align with the principal axes of the frisbee. The axis of symmetry that is perpendicular to the plane of the frisbee is x3, and has principal moment I3. The r and a2 axes are in the plane of the frisbee, and due to symmetry the principal moments 112112. (a) (1 pt) Use the Euler equations to show that ws, the component of , along the direction of symmetry axis, decreases exponentially with time. (b) (2 pts) Use the Euler equations to show that a, the angle between the r3 axis and ö decreases with time if I3 is larger than /12, which is the case for a flat object like a frisbee. In this way, the air resistance causes the wobble of the frisbee to gradually diminish. (Hint: Express tan a in terms of the components of .)

Explanation / Answer

given

rotational torque onthe frisbee due to wind resistance = T

T = -cw' ( where w' is a vector, not aligned to x1, x2, x3 unit vectors)

now,

let w' = w1*x1 + w2*x2 + w3*x3

where x1, x2 and x3 are unit vectors

then

T = -cw1*x1 - cw2*x2 - cw3*x3

but from newtons second law

T = I1*alpha1*x1 + I2*alpha2*x2 + I3*alpha3*x3

hence

comparing we get

I3*alpha3 = -c*w3

alpha3 = d(w3)/dt

hence

I3*dw3/dt = -cw3

I3*dw3/w3 = -c*dt

integrating from wo to w, t = 0 to t

then

I3*ln(w/wo) = -c*t

hence

w3 = w3o*e(-c*t/I3)

hence we can see the angular velocity in x3 direction, w3 decreases exponentially form the initial value of w3o

b. now,

w' . x3 = |w'|*cos(alpha)

cos(alpha) = w3/sqrt(w1^2 + w2^2 + w3^2)

now, d(cos(alpha))/dt = -sin(alpha)*d(alpha)/dt

now, if d(alpha)/dt < 0

then

-sin(alpha)*d(alpha)/dt > 0

now,

w3 = w3o*e^(-ct/I3)

similiarly

w2 = w2o*e^(-ct/I2)

w1 = w1o*e^(-ct/I1)

hence

for d(cos(alpha))/dt < 0

d(w3/sqrt(w1^2 + w2^2 + w3^2))/dt > 0

w3/sqrt(w1^2 + w2^2 + w3^2) = w3o*e^(-ct/I3)/sqrt(w3o^2*e^(-2ct/I3) + w2o^2*e^(-2ct/I2) + w1o^2*e^(-2ct/I1))

d(w3/sqrt(w1^2 + w2^2 + w3^2))/dt > 0

sqrt(w1^2 + w2^2 + w3^2)d(w3/dt) - w3(w1*d(w1/dt) + w2*d(w2/dt) + w3*d(w3/dt))/sqrt(w1^2 + w2^2 + w3^2) > 0

sqrt(w1^2 + w2^2 + w3^2)d(w3/dt) > w3(w1*d(w1/dt) + w2*d(w2/dt) + w3*d(w3/dt))/sqrt(w1^2 + w2^2 + w3^2)

(w1^2 + w2^2 + w3^2)dw3/dt > w3(w1*dw1/dt + w2*dw2/dt + w3*dw3/dt)

(w1o^2e^(-2ct/I1) + w2o^2*e^(-2ct/I2) + w3o*e^(-2ct/I3))w3o(-2c/I3)*e^(-2ct/I3) > w3o*e^(-2ct/I3)(w1o^2*e^(-2ct/I1)/I1 + w2o^2*e^(-2ct/I2)/I2 + w3o^2*e^(-2ct/I3)/I3)c

(w1o^2*e^(-2ct/I1)[1/I1 - 2/I3] + w2o^2*e^(-2ct/I2)[1/I2 - 2/I3] + w3o^2*e^(-2ct/I3)[1/I3 - 2/I3] > 0

hence for I3 > I1, I3 > I2 we have d(alpha)/dt < 0

hence alpha decreases with time

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