A uniform rod of mass 3.00x 10 2 kg and length 0.390 m rotates in a horizontal p

Question

A uniform rod of mass 3.00x 10 2 kg and length 0.390 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.210 kg, are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.00x10-2 m on each side from the center of the rod, and the system is rotating at an angular velocity 27.0 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. Part A What is the angular speed of the system at the instant when the rings reach the ends of the rod? ANSWER Wsystem -2.36 rev/min Correct Part B What is the angular speed of the rod after the rings leave it ANSWER wrod 101.6 rev/min Incorrect; Try Again; 19 attempts remaining

Explanation / Answer

I_rod = I1 = m L^2 / 12 = 3.8025 x 10^-4 kg m^2

For ring:

initially , I2 = 2 (0.210)(5 x 10^-2)^2 = 1.05 x 10^-3

w1 = 27 rev/min

finally, I3 = 2(0.210)(0.390/2)^2 = 0.0160 kg m^2

Applying momentum conservation,

(I1 + I2) w1 = (I1 + I3) w

w = 2.36 rev/min

(B) ring leaves rod with same speed.

so w_rod = w_system = 2.36 rev/min

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along the incline,

F_net = m a

m g si27 - f = m a

and torque = I alpha

r f = (n m r^2)(a/r)

f = n m a

putting in previous equation,

m g sin27 = m a (1 + n)

a = g sin27 / (1 + n )

f = (n / (n + 1)) m g sin27

and N = m g cos27

u = f/N = (n / (n + 1)) tan27

solid cylinder, n = 0.5

u = 0.17

hollow cylinder, n = 1

u = 0.25

solid sphere, n = 0.4

u = 0.15  

hollow sphere, n =2/3

u = 0.204

Ans: u = 0.25 ....Ans

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