Chapter 11, Problem 054 The figure shows an overhead view of a ring that can rot

Question

Chapter 11, Problem 054 The figure shows an overhead view of a ring that can rotate about its center like a merry-go-round. Its outer radius R2 is 1.0 m, its inner radius R1 is R2/2, its mass M is 8.1 kg, and the mass of the crossbars at its center is negligible. It initially rotates at an angular speed of 7.4 rad/s with a cat of mass m = M/4 on its outer edge, at radius R2. By how much does the cat increase the kinetic energy of the cat-ring system if the cat crawls to the inner edge, at radius R1? Number Units

Explanation / Answer

Consider the following -

M.I (Moment of Inertia) of the ring I1
M.I of the cat initially I2
MI of the cat finally I3

Initial angular momentum (I1+ I2) i
Final angular momentum (I1+ I3) f
(I1+ I3) f = (I1+ I2) i

Initial k.e = 0.5 (I1+ I2) i²
Final k.e = 0.5 (I1+ I3) f²
And increase in energy = 0.5 (I1+ I3) f² - 0.5 (I1+ I2) i²

Now put the values -  

I1 = (1/2) M (R1² + R2²) = (5/8) M R2²
I2 = mR2² = (M/4) R2²
I1 + I2 = (5/8) M R2² + (M/4) R2 ² = (7/8) M R2² = (7/8)* 8.1*1.0² = 7.09 kg.m²
I1 + I3 = (5/8) M R2² + (M/4) (R2/2) ² = (11/16) M R2² = (11/16) 8.1*1.0² = 5.57 kg.m²
so -
f = (I1+ I2) i/ (I1+ I3) = (7.09 x 7.4) / 5.57 = 9.42 rad/s.

Increase in energy = 0.5 {(I1+ I3) f² - (I1+ I2) i²}
Increase in energy = 0.5 {5.57*9.42² -7.09*7.4²} = 0.5*{494.26 - 388.25} = 53.00 J

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