A block of mass m = 3 kg is attached to a spring (k = 35 N/m) by a rope that hangs over a pulley of mass M = 6 kg and radius R = 6 cm, as shown in the figure. Treating the pulley as a solid homogeneous disk, neglecting friction at the axle of the pulley, and assuming the system starts from rest with the spring at its natural length, answer the following.
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Block
T is tension between block and pulley)
F = m a
m g - T = m a
T = m(g-a)
T - spring force = Torque/R
T - k x = Torque/R
Torque = I alpha (I is in moment of inertia table for M and R)
T -k x = I alpha/R
a = alpha R
T - k x = I a/R^2
m(g-a) - kx = (I/R^2)a
m g = k x + ma +(I^2/R) a
m g = k x +(m+I^2/R^2)d^2x/dt^2
(This is a spring mass system with equilibrium at x = mg/k and 2 pi f=sqrt(k/(m+I^2/R^2)
but onward
split x into Xo + A sinwt
then mg = kXo for steady result
then
x(t) time function = A sin wt
then
a = d^2x/dt^2 = -Aw^2 sin wt = -w^2 x
0 = k x(t) - (m+I^2/R^2)w^2 x(t)
w^2 = k/(m+I^2/R^2)
like we could already see
so we have a vibration at frequency w/2pi about the point mg/k
