Two 2.00 kg balls are attached to the ends of a thin rod of length 50.0 cm and n

Question

Two 2.00 kg balls are attached to the ends of a thin rod of length 50.0 cm and negligible mass the rod is free to rotate in a vertical plane without friction about a horizontal axis through is center. With the rod initially horizontal (Fig. 11-52), a 50.0 g wad of wet putty drops onto one of the balls, biting it with a speed of 3.00 ms and then sticking to in. (a) What is the angular speed of the system just after the putty wad tits? b) What is the ratio of the kinetic energy of the system after the collision to that of the putty wad just before? (c) Through what angle will the system rotate be force

Explanation / Answer

a)

m = mass of each ball = 2 kg

r = distance of each ball from the center = 50/2 = 25 cm = 0.25 m

I = moment of inertia of the balls about the rotation axis = 2m r2 = 2 (2) (0.25)2 = 0.25 kgm2

Wi = initial angular velocity of two balls before collision = 0 rad/s

M = mass of mud putty = 0.050 kg

v = speed of putty = 3 m/s

Wf = final angular velocity of the combination = ?

using conservation of angular momentum

I Wi + M v r = (I + Mr2) Wf

(0.25) (0) + (0.05) (3) (0.25) = (0.25 + (0.05) (0.25)2) Wf

Wf = 0.14 rad/s

b)

KEf = final KE after collision = (0.5) (I + Mr2) W2f = (0.5) (0.25 + (0.05) (0.25)2) (0.14)2 = 0.0025 J

KEi = initial KE before collision = (0.5) IWi2 + (0.5) Mv2 = (0.5) (0.25) (0)i2 + (0.5) (0.05) (3)2 = 0.225 J

Ratio = 0.0025/0.225 = 0.011

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