Answer :
Answer:
Explanation:
ω = angular velocity = 2π n = 2π x 72 / 60
= 7.536 rad /s
mass of head = 90 x .07 = 6.3 kg
moment of inertia of head = 2 /5 m R²
= .4 x 6.3 x .08²
= .016128 kg m²
moment of inertia of trunk + legs
= 1/2 m R²
= .5 x .8 x 90 x .12²
= .5184 kg m²
moment of inertia of arms
= 1/3 m L²
= 1 / 3 x .13 x 90 x .60²
= 1.404 kg m²
Total moment of inertia
I = 1.938 kg m²
kinetic energy = 1/ 2 I ω ²
where I is moment of inertia and ω is angular velocity
= .5 x 1.9338 x 7.536²
=55 J approx .
The definition of kinetic energy and moment of inertia allows finding the result for the kinetic energy of the dancer with arms extended is:
- The kinetic energy is: KE = 56.3J
Given parameters
- Frequency is: f = 72 rad / min (1 min / 60s) = 1.2 rad / s
- The mass is m = 90 k
- Head radius is r₁ = 8.0 cm = 0.08 m
- The leg and trunk are cylinders of radius r₂ = 12.0 cm = 0.12 m
- The length of the arms, approximated as rods L = 60.0 cm = 0.600 m
- The dancer spins with outstretched arms
- The percentage of the mass is:
* Head 7%
* Arms 13%
* Trunk and legs 80%
To find.
- Kinetic energy
The kinetic energy is the energy due to the movement, in this case the movement is rotational, therefore the expression is:
KE = ½ I w²
The angular velocity is related to the frequency.
w = 2π f
w = 2π 1.2
w = 7.540 s
The moment of inertia is a scalar, therefore it is a quantity that can be added, the total moment of inertia of the dancer is the sum of the moments of inertia of each part with respect to the axis of rotation of the person.
[tex]I_{toal}= I_{head}+ I_{trunk}+I_{arms}[/tex]
The moments of inertia with respect to the centers of mass are tabulated.
Sphere I = 2/5 mr²
Cylinder I = ½ m r²
Rod I = ⅓ m r²
The axis of rotation of the head and the trunk are in the axis of rotation of the person, therefore their moment of inertia is those corresponding to the center of mass.
At the end of the arms it is at a distance of D = [tex]\frac{r_2}{2}[/tex] from the axis of rotation of the dancer, therefore to find the moment of inertia we must use the theorem of parallel axes, see attached.
[tex]I_{arms} = I_{cm} + M D^2[/tex]
[tex]I_{arms}[/tex] = ⅓ m L² + m D²
[tex]I_{arms} = m_{arms} ( \frac{L^2}{3} + D^2)[/tex]
The masses of each part of the body are:
[tex]m_{head}[/tex] = m 0.07
[tex]m_{trunk}[/tex] = m 0.80
[tex]m_{arms}[/tex] = m 0.13
Let's find the total moment of inertia.
[tex]I_{total} = \frac{2}{5} \ 0.07m \ r_1^2 + 0.13m\ ( \frac{L^2}{3} + D^2) + \frac{1}{2} \ 0.80m \ r_2^2[/tex]
[tex]I_{total} = m ( 0.028 \ r_1^2 + 0.13 (\frac{L^2}{3} + D^2) + 0.40 \ r_2^2}[/tex]
Let's calculate.
[tex]I_{total} = 90\ ( 0.028 \ 0.08^2 + 0.13\ ( \frac{0.6^2}{3} + 0.06^2) + 0.40 \ 0.12^2 )[/tex]
[tex]I_{total}= 90 \ 0.0220 \\ \\I_{total} = 1.9806 \ Kg m^2[/tex]
we substitute in the kinetic energy
KE = ½ 1.9806 7,540²
KE = 56.3 J
In conclusion using the definition of kinetic energy and moment of inertia we can find the result for the kinetic energy of the dancer with the extended arms is:
- The kinetic energy is: KE = 56.3 J
Learn more here: brainly.com/question/15246709
