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A Texas cockroach of mass 0.17 kg runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has radius 15 cm, rotational inertia 5.0 $ 10%3 kg"m2 , and frictionless bearings. The cockroach’s speed (relative to the ground) is 2.0 m/s, and the lazy Susan turns clockwise with angular speed v0 ! 2.8 rad/s. The cockroach finds a bread crumb on the rim and, of course, stops.
(a) What is the angular speed of the lazy Susan after the cockroach stops?
(b) Is mechanical energy conserved as it stops?

Respuesta :

Answer:

a) [tex]w_f[/tex] = 4.192 rad/s

b) The mechanical energy for the cockroach is not conserved as it stops

Explanation:

1) Data given and notation

m = 0.17 kg (mass of the Texas cockroach)

I = [tex]5x10^{-3}\frac{kg m^2}{s}[/tex] (rotational inertia for the Lazy Susan)

v = 2m/s (cockroachs speed)

r = 15 cm = 0.15 m (radius for the Lazy Susan)

wi = 2.8 rad/s (initial angular speed for the Lazy Susan)

wf = ? rad/s (final angular speed for the Lazy Susan)

vf =0 m/s (Final speed of the cockroach)

Lis= Initial angular moment of the lazy susan

Lic= Initial angular moment of the cockroach

Li= Initital angular moment for the system Lazy Susan-Cockroach

If= final inertia for the disk

Lfs= final angular momentum for the disk (lazy Susan)

Lfc= final angular momentum for the cockroach =0

Part a

We can begin calculating the initial angular momentum's for lazy Susan and the cockroach

[tex]L_{is}=I w_i=5x10^{-3}kgm^2 x2.8rad/s= 0.014\frac{kgm^2}{s}[/tex]

Since the initial angular momentum of the cockroach is on the axle of the disk, the formula for this case is:

[tex]L_{ic}=-mvr=-0.17kgx(2m/s)x0.15m=-0.051\frac{kgm^2}{s}[/tex]

After this since we have a system composed by the lazy Susan and the cockroach, the total initial angular momentum would be the sum of the inidivual angular moments

[tex]L_i=L_{is} +L_{ic}=0.014-0.051=-0.037\frac{kgm^2}{s}[/tex]

When the cockroach stops, we can use this formula to calculate the final inertia for the disk

[tex]I_f=I+mr^2=5x10^{-3}\frac{kgm^2}{s}+0.17Kgx(0.15m)^2=0.008825\frac{kgm^2}{s}[/tex]

With this info we can calculate the final angular momentum for the disk

[tex] L_{fs}=I_f w_f=0.008825\frac{kgm^2}{s} w_f[/tex]   (1)

We can have another equation using conservation of angular momentum for the system:

[tex]L_i=L_{fs}+L_{fs}[/tex]   (2)

And since Lfc =0 kgm^2/s, we can repplace equation (1) into equation (2)

[tex]-0.037\frac{kgm^2}{s}=0.008825w_f+0[/tex]  

Solving for wf we got

[tex]w_f=\frac{-0.037\frac{kgm^2}{s}}{0.008825}=-4.192\frac{rad}{s}[/tex]

Based on this the angular speed of lazy Susan would be -4.192rad/s when the cockroach stops, and the minus sign is because is on the opposite direction of the initial angular speed of lazy Susan.

Part b

The mechanical energy for the cockroach is not conserved as it stops.

The reason is because part of the energy os the system is converted into internal energy, and the balance is not satisfied.

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