. A spool of radius R is free to rotate about a fixed point P. A massless rod of

Question

. A spool of radius R is free to rotate about a fixed point P. A massless rod of length L is rigidly attached to the spool. A weight of mass mp is attached to the end of the rod. A string is wrapped many times around the spool. The end of the string is attached to block of mass mb. Gravity acts downward. Let Ip denote the moment of inertia of the spool about point P.

One may derive a nonlinear differential equation for in the form: (Ip + mp*L2 + mb*R2)¨ = R*mb*g mp*g Lsin() c*

Suppose that R = 0.3m, L = 1.0m, mb = mp = 3.0kg, Ip = 2.0kg m2, g = 9.81m/s2 .

(a) Find all equilibria in the range < .

(b) For c = 0, linearize the differential equation at each of the equilibria

Explanation / Answer

Given equation

(Ip + mp*L^2 + mb*R^2)*(theta)" = R*mb*g - mp*g*Lsin(theta) - c*(theta)'

Given

R = 0.3 m

L = 1 m

mb = 3 kg = mp

Ip = 2 kg m^2

g = 9.81 m/s^2

so the equation becomes

(2 + 3*1^2 + 3*0.3^2)*theta" = 0.3*3*9.81 - 3*9.81*1*sin(theta) - c*theta'

5.27*theta" = 8.829 - 29.43sin(theta) - c*theta'

a) for equilibrium theta" = 0

theta' = 0

and, hence

8.829 = 24.43sin(theta)

theta = 21.18 deg 158.58 deg

b) for c = 0

5.27*theta" = 8.829 - 29.43sin(theta)

let theta = phio + phi ( where phio is equilibrium angle)

then

5.27*phi" = 8.829 - 29.43sin(phio + phi)

sin(a + b) = sin(a)cos(b) + sin(b)cos(a)

5.27phi" = 8.829 - 29.43(sin(phio)cos(phi) + sin(phi)cos(phio))

for small phi, cos(phi) = 0 and sin(phi) = phi

5.27phi" = 8.829 - 29.43(phi*cos(phio)) = 8.829 - 29.43cos(phio)*phi

here phio = 21.18 deg or 158.8 dfeg

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