Answer:
0.19 rev
Explanation:
We can solve the problem by using the law of conservation of angular momentum. In fact, if we assume there are no external torques acting on the diver, the angular momentum must be conserved:
[tex]L=I\omega[/tex]
where
L is the angular momentum
I is the moment of inertia
[tex]\omega[/tex] is the angular velocity
- When the diver is tucked,
[tex]I = 3.6 kg m^2[/tex] is her moment of inertia
She makes 2 revolutions (so, [tex]4 \pi rad[/tex]) in t = 1.0 s, so her angular velocity is
[tex]\omega=\frac{4\pi}{1.0 s}=4.0 rad/s[/tex]
So her angular momentum is
[tex]L=(3.6 kg m^2)(4.0 rad/s)=14.4 kg m^2 /s[/tex]
- When the diver is not tucked,
The angular momentum is conserved, so [tex]L=14.4 kg m^2 /s[/tex]
the moment of inertia is [tex]I=18 kg m^2[/tex]
So the angular velocity is
[tex]\omega=\frac{L}{I}=\frac{14.4 kg m^2/s}{18 kg m^2}=0.8 rad/s[/tex]
So in a time of t = 1.5 s, the angular displacement is
[tex]\theta=\omega t=(0.8 rad/s)(1.5 s)=1.2 rad[/tex]
Converting into revolutions,
[tex]2 \pi rad : 1 rev = 1.2 rad : x[/tex]
[tex]x=\frac{(1 rev)(1.2 rad)}{2\pi rad}=0.19 rev[/tex]
If she hadn’t tucked at all, the number of revolutions she would have made is 0.6 rev.
The initial angular speed of the diver will be determined by applying the principle of conservation of angular momentum as shown below;
[tex]I_i\omega_i = I_f \omega _f\\\\\omega_i = \frac{I_f \omega _f}{I_i}[/tex]
where;
ωf = 2 rev/1.0 s = 2 rev/s = 2 x 2π rad = 12.57 rad/s
[tex]\omega _i = \frac{3.6 \times 12.57}{18} \\\\\omega _i = 2.514 \ rad/s[/tex]
θ = ωt
θ = 2.514 x 1.5
θ = 3.771 rad
θ = 3.771 / 2π = 0.6 rev
Thus, If she hadn’t tucked at all, the number of revolutions she would have made is 0.6 rev.
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