003 10.0 points A uniform 140 g rod with length 54 cm rotates in a horizontal pl

Question

003 10.0 points A uniform 140 g rod with length 54 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 40 g beads are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches at positions 12 cm on each sides of the center, at which time the system rotates at an angular speed of 23 rad/s. Suddenly, the catches are released and the small beads slide outward along the rod. Find the angular speed of the system at the instant the beads reach the ends of the rod. Answer in units of rad/s.

Explanation / Answer

initial moment of inertia I1 = (1/12)*Mrod*L^2 + 2*Mbead*r1^2

I1 = (1/12)*0.14*0.54^2 + 2*0.04*0.12^2

I1 = 0.004554 kg m^2

initial angular speed w1 = 23 rad/s

final moment of inertia I2 = (1/12)*Mrod*L^2 + 2*Mbead*r2^2

I2 = (1/12)*0.14*0.54^2 + 2*0.04*0.27^2

I2 = 0.009234 kg m^2

final angular speed w2 = ?

from angular momentum intial angular momentum = final angular momentum

Li = Lf

I1*w1 = I2*w2

0.004554*23 = 0.009234*w2

w2 = 11.34 rad/s

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016)

initial energy Ei = m*g*h

final energy at the bottom of incline Ef = (1/2)*I*w^2 + (1/2)*m*v^2

I = (1/2)*m*r^2

w = v/r

Ef = (1/2)*(1/2)*m*v^2 + (1/2)*m*v^2 = (3/4)*m*v^2

from energy conservation

Ef = Ei

(3/4)*m*v^2 = m*g*h

v = sqrt(4gh/3)

v = sqrt(4*9.81*4.6/3)

v = 7.75 m/s

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