Respuesta :
Answer:
[tex]\frac{\omega_{SS}}{\omega_{S}} \approx 1.698[/tex]
Explanation:
Let assume that both the spherical shell and the sphere are rigid bodies. The kinetic energy due to rolling is:
[tex]K = \frac{1}{2}\cdot I_{g}\cdot \omega^{2}[/tex]
The moments of inertial for the spherical shell and the sphere are, respectively:
Spherical Shell
[tex]I_{g, SS} = \frac{2}{3}\cdot m \cdot r_{SS}^{2}[/tex]
[tex]I_{g,SS} = 9.830\cdot m \times 10^{-4}[/tex]
Sphere
[tex]I_{g,S} = \frac{2}{5}\cdot m \cdot r_{S}^{2}[/tex]
[tex]I_{g,S} = 2.384\cdot m \times 10^{-3}[/tex]
Given that both rigid bodies have the same kinetic energy and the same mass, then:
[tex]\frac{1}{2}\cdot (9.830\cdot m\times 10^{-4})\cdot \omega_{SS}^{2} = \frac{1}{2}\cdot (2.834\cdot m\times 10^{-3})\cdot \omega_{S}^{2}[/tex]
[tex](9.830\cdot m\times 10^{-4})\cdot \omega_{SS}^{2} = (2.834\cdot m\times 10^{-3})\cdot \omega_{S}^{2}[/tex]
[tex]\frac{\omega_{SS}^{2}}{\omega_{S}^{2}} = 2.883[/tex]
The ratio of the spherical shell's angular speed to the sphere's angular spped is:
[tex]\frac{\omega_{SS}}{\omega_{S}} \approx 1.698[/tex]
The ratio of the spherical shell's angular speed to the sphere's angular speed be;
(ω_ss)/(ω_s) = 1.557
We are given;
Radius of spherical shell; r = 3.84 cm = 0.0384 m
Radius of sphere; r = 7.72 cm = 0.0772 m
Formula for rotational kinetic energy is;
KE = ½Iω²
Now, moment of inertia for spherical shell is;
I_ss = ⅔mr²
While moment of inertia for solid sphere is;
I_s = (2/5)mr²
Thus;
I_ss = ⅔m(0.0384²)
I_s = (2/5)m(0.0772²)
We are told that their kinetic energy is equal. Thus;
½(⅔m(0.0384²))(ω_ss)²= ½((2/5)m(0.0772²))(ω_s)²
m will cancel out to give;
⅓(0.0384²)(ω_ss)² = (1/5)(0.0772²)(ω_s)²
0.00049152(ω_ss)² = 0.001191968(ω_s)²
Thus;
(ω_ss)²/(ω_s)² = 0.001191968/0.00049152
(ω_ss)²/(ω_s)² = 2.425065
(ω_ss)/(ω_s) = √2.425065
(ω_ss)/(ω_s) = 1.557
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