Question

Two equal masses of 3.0 kg are at the ends of a 1.00 kg mass and 120.0 cm-long rod. The rod spins at 30.0 rpm about an axis through its midpoint. Suddenly, a compressed gas expands the rod out to a length of 180 cm. What is the rpm of the system after the expansion? Find speed of masses before and after expansion. Please help with formulas.

Answer 1

Using angular momentum conservation

L1 = L2

I1w1 = I2w2

here, I1 = Initial Moment of inertia of system = I_rod1 + 2*m*r1^2

I2 = final Moment of inertia of system = I_rod2 + 2*m*r2^2

I_rod1 = Initial moment of inertia of rod about midpoint = M*L^2/12

I_rod2 = final moment of inertia of rod about midpoint = M*L'^2/12

L = initial length of rod = 120.0 cm = 1.20 m

L' = final length of rod = 180.0 cm = 1.80 m

M = mass of rod = 1.00 kg

m = masses = 3 kg

r1 = initial distance from midpoint = L/2 = 60.0 cm = 0.60 m

r2 = final distance from midpoint = L'/2 = 90.0 cm = 0.90 m

I2 = 139 + 27*1.9^2 = 236.47

w1 = Initial Angular Speed = 30 rpm

w2 = final angular speed = ?

So,

w2 = w1*(I1/I2)

Using above values:

w2 = 30*(1*1.2^2/12 + 2*3*0.60^2)/(1*1.8^2/12 + 2*3.0*0.90^2)

w2 = 13.33 rpm = rpm of the system after the expansion

Now, speed of masses before expansion will be,

v1 = w1*r1 = (30 rpm)*0.60 = (30*2*pi/60 rad/s)*0.60

v1 = 1.88 m/s

Similarly, speed of masses after expansion will be,

v2 = w2*r2 = (13.33 rpm)*0.90 = (13.33*2*pi/60 rad/s)*0.90

v2 = 1.26 m/s

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