A uniform rod of length 60.0 cm and mass m1 = 1.50 kg is pivoted at a horizontal

Question

A uniform rod of length 60.0 cm and mass m1 = 1.50 kg is pivoted at a horizontal, frictionless axle at P, which is 15.0 cm from the lower end of the rod, as shown, and the rod is rotating in a vertical plane; at the instant the rod is vertical, it has an angular velocity of 3.20 rad/s counterclockwise, and exactly at this same instant, two small point masses which were travelling to the right strike the rod with speeds each of 4.00 m/s and stick to the rod. Mass m2 = 2.00 kg sticks at 15.0 cm from the upper end of the rod, and mass m3 = 3.00 kg sticks to the lower end of the rod, as shown.

Explanation / Answer

(a) The moment of inertia of the rod about point P before collision which is given as :

using an equation, Ip = m (d/2)2 + Icm

Ip = m (d/2)2 + (1/12) m d2

Ip = (1/3) m d2                                                                           { eq.1 }

where, m1 = mass of the rod = 1.5 kg

d = length of rod = 60 cm = 0.6 m

inserting the values in eq.1,

Ip = (1/3) (1.5 kg) (0.6 m)2

Ip = 0.18 kg.m2

(b) Magnitude of the rod's angular velocity after the collision which is given as :

using conservation of energy,

(1/2) m v02 = (1/2) m vf2 + (1/2) Ip wf2

m v02 - m vf2 = Ip wf2

m (v02 - vf2) = [(1/3) m d2] wf2                                                                (from eq.1)

(v0 - vf) (v0 + vf) = [(1/3) d2] wf2    

wf2 = 3 [(v0 - vf) (v0 + vf)] / d2 { eq.2 }

where, v0 = initial speed of rod = 4 m/s

vf = final speed of rod = 0 m/s

inserting the values in eq.2,

wf2 = 3 (4 m/s)2 / (0.6 m)2

wf2 = (48 m2/s2) / (0.36 m2)

wf2 = 133.3 rad2/s2

wf = 11.5 rad/s

And after the collision, the rod rotates clockwise about its pivot point P with angular speed, wf.

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