A uniform cylindrical turntable of radius 2.00 m and mass 27.8 kg rotates counte

Question

A uniform cylindrical turntable of radius 2.00 m and mass 27.8 kg rotates counterclockwise in a horizontal plane with an initial angular speed of 4 pi rad/s. The fixed turntable bearing is frictionless. A lump of clay of mass 2.44 kg and negligible size is dropped onto the turntable from a small distance above it and immediately sticks to the turntable at a point 1.90 m to the east of the axis. (a) Find the final angular speed of the day and turntable. magnitude 10.84 rad/s direction (b) Is mechanical energy of the turntable-clay system constant in this process? Yes No What, if any, is the change in internal energy? Your response differs from the correct answer by more than 100% (c) Is momentum of the system constant in this process? Yes No What, if any, is the amount of impulse imparted by the bearing? magnitude kg middot m/s direction

Explanation / Answer

given

R = 2 m
M = 27.8 kg

wi = 4*pi rad/s

m = 2.44 kg

r = 1.9 m

a) Apply conservation of angular momentum

Lf = Li

If*wf = Ii*wi

(0.5*M*R^2 + m*r^2)*wf = (0.5*M*R^2)*wi

==> wf = (0.5*M*R^2)*wi/(0.5*M*R^2 + m*r^2)*wf

= (0.5*27.8*2^2)*4*pi/(0.5*27.8*2^2 + 2.44*1.9^2)

= 10.85 rad/s

direction : counterclockwise

b) No

change in intenral energy = (1/2)*Ii*wi^2 + (1/2)*If*wf^2

= 0.5*(0.5*27.8*2^2)*(4*pi)^2 - (1/2)*((0.5*27.8*2^2 + 2.44*1.9^2)*(10.85)^2

= 599 J

c) No

Impulse imparted by the bearing = change in momentum of the clay

= m*(vf - vi)

= m*r*wf - 0

= 2.44*1.9*10.85

= 50.3 kg.m/s

direction : North

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