Chapter 11: Q62P (page 325)

During a jump to his partner, an aerialist is to make a quadruple somersault lasting a time $t = 1 .87 s .$ For the first and last quarter-revolution, he is in the extended orientation shown in Figure, with rotational inertia $I_{1} = 19.9 kg . m^{2} .$ around his center of mass (the dot). During the rest of the flight he is in a tight tuck, with rotational inertia $I_{2} = 3 .93 k g . m^{2}$ .What must be his angular speedv₂around his center of mass during the tuck?

Short Answer

Expert verified

Angular speed around the center of mass is $ω = 3.23 rev/s.$

Step by step solution

Step 1: Given

$\begin{array}{l} I_{1} = 19.9 kg . m^{2} \\ I_{2} = 3.93 kg . m^{2} \\ t = 1.87 s \end{array}$

Determining the concept

Find the final angular speed using law of conservation of angular momentum.According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formula is as follow:

$Initial angular momentum = Final angular momentum$

Determining the angular speed around the center of mass

According to law of conservation of angular momentum:

$Initialangularmomentum = Finalangularmomentum$

$\begin{matrix} L_{1} = L_{2} \\ I_{1} ω_{1} = I_{2} ω_{2} \\ ω_{1} = I_{2} \frac{ω_{2}}{I_{1}} \end{matrix}$

$Totaltime (t) = t_{1} + t_{2}$

$\begin{matrix} t = \frac{θ_{1}}{ω_{1}} + \frac{θ_{2}}{ω_{2}} \\ t = \frac{θ_{1}}{I_{2} \frac{ω_{2}}{I_{1}}} + \frac{θ_{2}}{ω_{2}} \\ 1.87 = \frac{0.5}{3.93 \times \frac{ω_{2}}{19.9}} + \frac{3.5}{ω_{2}} \end{matrix}$

By solving above equation for $ω_{2}$ ,

$ω_{2} = 3.23 rev / s$

Hence,angular speed around the center of mass is $ω = 3.23 rev/s.$

Therefore, by using the total time and conservation of momentum principle, the final angular speed can be found.

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Short Answer

Step by step solution

Step 1: Given

Determining the concept

Determining the angular speed around the center of mass

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