A uniform rod of mass 2.65×102 kg and length 0.450 m rotates in a horizontal pla

Question

A uniform rod of mass 2.65×102 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.250 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 4.60×102 m on each side from the center of the rod, and the system is rotating at an angular velocity 35.0 rev/min . Without otherwise changing the system, the catches are released, and the rings slide outward along the rod and fly off at the ends.

Explanation / Answer

here

given mass = m = 2.65 x 10-2 = 0.0265 kg

length = 0.45 m;

M1 = M2= M = 0.25 kg

r = 4.6x 10-2 m = 0.046 m

W1 = 35 rev/min = = 35*2pi/60 = 3.66 rad/sec

Let us consider the rings at the end of the rods as point masses,

so the moment of inertia of the system will be:

MOI Inet = 1 = m L^2/12 + 2 M r^2

I1 = 1/12 * 0.0265 * 0.45 * 0.45 + 2 * 0.25 * 0.046 * 0.046

I1 = = 0.0015 kg m^2

Now

Second MOI I2 = 1/12 m L^2 + 2 M (L/2)^2

I2 = 1/12 *0.0265 * 0.45* 0.45 + 2 * 0.25 * 0.225 * 0.225

I2 = 0.0257 kg-m2

From conservation of angular momentum :

L(i) = L(f)

I1 W1 = I2 W2

W2 = I1 W1 /I2

W2 = 0.0015 * 3.66 / 0.0257

W2 = 0.213 rad/s

hence, W2 = 0.213*60/2pi = 2.03 Rpm

for b0th parts the answer is same as above

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