Acircular bird feeder 19.0 cm in radius has rotational inertia 0.130 kg·m2. it's suspended by a thin wire and is spinning slowly at 24.2 rpm. a 173–g bird lands on the feeder's rim, coming in tangent to the rim at 1.20 m/s in a direction opposite the feeder's rotation. what's the rotation rate (in rpm) after the bird lands? (you do not need to enter any units.)

N₂=20.05 rpm

Explanation:

Given that

R= 19 cm

I=0.13 kg.m²

N₁ = 24.2 rpm

ω₁= 2.5 rad/s

m= 173 g = 0.173 kg

v=1.2 m

Initial angular momentum L₁

L₁ =  Iω₁  - m v r       ( negative sign because bird coming opposite to motion of the wire motion)

Final linear momentum L₂

L₂=  I₂ ω₂

 I₂ = I + m r²

The is no any external torque that is why angular momentum will be conserve

L₁ = L₂

Iω₁  - m v r =  I₂ ω₂

Iω₁  - m v r =  ( I + m r²) ω₂

Now by putting the all values

Iω₁  - m v r =  ( I + m r²) ω₂

0.13 x 2.5 - 0.173 x 1.2 x 0.19 =  ( 0.13 + 0.173 x  0.19²) ω₂

0.325  - 0.0394 = 0.136 ω₂

ω₂ = 2.1 rad/s

N₂=20.05 rpm

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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