The angular velocity in rpm of the bowling ball is equal to 71.32 rpm.
Given the following data:
- Diameter = 22 cm to m = 0.22 m
- Angular momentum = 0.21 [tex]kgm^2/s[/tex]
Radius = [tex]\frac{Diameter}{2} = \frac{0.22}{2} =0.11\;m[/tex]
To determine the angular velocity in rpm of the bowling ball:
First of all, we would calculate the moment of inertia of the bowling ball by using the formula:
[tex]I = \frac{2}{5} mr^2[/tex]
Where:
- I is the moment of inertia
- m is the mass of an object.
Substituting the given parameters into the formula, we have;
[tex]I = \frac{2}{5} \times 5.8 \times 0.11^2\\\\I=2.32 \times 0.0121\\\\I=0.0281\;Kgm^2[/tex]
To find the angular velocity in rpm:
Mathematically, angular momentum is given by the formula:
[tex]L=I\omega\\\\\omega = \frac{L}{I} \\\\\omega = \frac{0.21}{0.0281}\\\\\omega = 7.47 \;rad/s[/tex]
Converting rad/s to rpm:
[tex]\omega = 7.47 \times \frac{60}{2\pi} \\\\\omega = \frac{448.2}{6.284}\\\\\omega = 71.32 \;rpm[/tex]
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