Consider a cylindrical turntable whose mass is M andradius is R , turning with a

Question

Consider a cylindrical turntable whose mass is M andradius is R, turning with an initial angular speed1. (Use the following variables asnecessary: M, m, v, R, and omega_1 for1.) a) A parakeet of mass m, afterhovering in flight above the outer edge of the turntable, gentlylands on it and stays in one place on it, as shown below. What isthe angular speed of the turntable after the parakeet lands?

b) Becoming dizzy, the parakeet jumps off (not flies off) with avelocity v relative to the turntable. The direction of v is tangentto the edge of the turntable and in the direction of its rotation.What will be the angular speed of the turntable afterwards? Expressyour answer in terms of the two masses m and M,the radius R, the parakeet speed and the initial angularspeed 1.
a) A parakeet of mass m, afterhovering in flight above the outer edge of the turntable, gentlylands on it and stays in one place on it, as shown below. What isthe angular speed of the turntable after the parakeet lands?

b) Becoming dizzy, the parakeet jumps off (not flies off) with avelocity v relative to the turntable. The direction of v is tangentto the edge of the turntable and in the direction of its rotation.What will be the angular speed of the turntable afterwards? Expressyour answer in terms of the two masses m and M,the radius R, the parakeet speed and the initial angularspeed 1.

Explanation / Answer

Moment of inertia of the turntable(Ii)=1/2MR2 When the parakeet lands at the edge, it adds its MOI to thesystem's. If=1/2MR2+mR2 Initial angular momentum of the system=Ii1=1/2MR21 Final angularmomentum=Iff=R2(0.5M+m)f By conservation of angular momentum, these two values areequal. 1/2MR21=R2(0.5M+m)f f=0.5M1/(0.5M+m)      OR      =M1/(M+2m) The bird jumps off, imparting angular momentum to the system as itflies away. This ang momentum has the value +mvR Again.. Initial angular momentum =Iff=R2(0.5M+m)f                Final angular momentum=1/2MR23 +mvR Equate them and calculate for 3: R2(0.5M+m)f=1/2MR23+mvR R(M+2m)(M1)/2(M+2m)-mv=1/2MR3 (MR1-2mv)/2=1/2MR33=1-(2mv/MR)

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