Answer:
25.8 rad/s
Explanation:
C = circumference of the hollow sphere = 0.749 m
r = radius of the sphere
Circumference of the hollow sphere is given as
C = 2π r
0.749 = 2 (3.14) r
r = 0.12 m
m = mass of the basketball = 0.624 kg
Moment of inertia of basketball is given as
[tex]I = \left ( \frac{2}{3} \right )mr^{2}[/tex]
[tex]I = \left ( \frac{2}{3} \right )(0.624)(0.12)^{2}[/tex]
I = 5.99 x 10⁻³ kg-m²
w = angular speed
KE = Kinetic energy of the ball = 1.99 J
Kinetic energy of the ball is given as
KE = (0.5) I w²
1.99 = (0.5) (5.99 x 10⁻³) w²
w = 25.8 rad/s
The angular speed at which the ball will rotate [tex]w=28.8\frac{rad}{sec}[/tex]
It is given that
C = circumference of the hollow sphere = 0.749 m
r = radius of the sphere
Circumference of the hollow sphere is given as
[tex]C=2\pi r[/tex]
[tex]0.749=2(3.14)r[/tex]
[tex]r=0.12m[/tex]r
m = mass of the basketball = 0.624 kg
The moment of inertia of basketball is given as
[tex]I=\dfrac{2}{3} mr^2[/tex]
[tex]I=\dfrac{2}{3} (0.624)(0.12)^2[/tex]
[tex]I=5.99\times10^{-3}\ kg-m^2[/tex]
w = angular speed
KE = Kinetic energy of the ball = 1.99 J
The kinetic energy of the ball is given as
[tex]KE=0.5(Iw^2)[/tex]
[tex]1.99= (0.5)(5.99\times10^{-3})w^2[/tex]
[tex]w=28.8\frac{rad}{sec}[/tex]
Thus the angular speed at which the ball will rotate [tex]w=28.8\frac{rad}{sec}[/tex]
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