Respuesta :
Answer:
Angular momentum, [tex]L=2.57\ kg-m^2/s[/tex]
Explanation:
It is given that,
Diameter of a basketball, d = 31.7 cm
Radius of the basketball, r = 15.85 cm
Angular speed of the basketball, [tex]\omega=220\ rad/s[/tex]
Mass of the basketball, m = 700 g = 0.7 kg
The moment of inertia of the spherical shell,
[tex]I=\dfrac{2}{3}mr^2[/tex]
[tex]I=\dfrac{2}{3}\times 0.7\ kg\times (15.85\times 10^{-2}\ m)^2[/tex]
[tex]I=0.0117\ kg-m^2[/tex]
We know that the angular momentum of the ball is given by the product of moment of inertia and the angular velocity. It is given by :
[tex]L=I\times \omega[/tex]
[tex]L=0.0117\ kg-m^2 \times 220\ rad/s[/tex]
[tex]L=2.57\ kg-m^2/s[/tex]
So, the magnitude of the angular momentum of the ball is [tex]2.57\ kg-m^2/s[/tex]. Hence, this is the required solution.
The magnitude of the angular momentum of the spherical shell basketball is 2.58kg.m²/s.
Given the data in the question;
- Mass of basket ball; [tex]m = 700g = 0.7kg[/tex]
- Diameter; [tex]D = 31.7cm = 0.317m[/tex]
- Radius; [tex]R = \frac{Diameter}{2} = \frac{0.317m}{2} = 0.1585m[/tex]
- Angular velocity; [tex]w = 220rad/s[/tex]
Magnitude of the angular momentum; [tex]L = \ ?[/tex]
Angular momentum
Angular momentum is simply the property of the rotational inertia of an object about an axis.
It is expressed as;
[tex]L = Iw[/tex]
Where;
- w is angular velocity
- [tex]I[/tex] is moment of inertia of a spherical shell ( [tex]I = \frac{2}{3}mR^2[/tex] )
Hence
[tex]L = I w\\\\L = (\frac{2}{3}mR^2)w[/tex]
To determine the magnitude of the angular momentum of the ball, we substitute our given values into the expression above.
[tex]L = (\frac{2}{3}mR^2)w\\ \\L = (\frac{2}{3} * 0.7kg * (0.1585m)^2)220rad/s\\\\L = 0.0117237167kgm^2 * 220rad/s\\\\L = 2.58kgm^2/s[/tex]
Therefore, the magnitude of the angular momentum of the spherical shell basketball is 2.58kg.m²/s.
Learn more about moment of inertia and angular velocity: https://brainly.com/question/13031713
