Answer:
2.969 rad/s
Explanation:
Let's R be the radius of the spinning disk. By law of conservation of angular momentum we have before and after removing the bowl from disk
[tex]A_1 = A_2[/tex]
[tex]I_1\omega_1 = I_2\omega_2[/tex]
Where the mass moment of inertia for disk is [tex]MR^2/2[/tex] and for the bowl of fruit at edge of the disk is [tex]mR^2[/tex]. Assume the bowl is a point mass spinning at radius R
[tex]I_1 = \frac{MR^2}{2} + mR^2 = 2.65R^2 + 4R^2 = 6.65R^2[/tex]
After removing the bowl, the moment of inertia is only the disk
[tex]I_2 = \frac{MR^2}{2} = 2.65R^2[/tex]
Therefore we can substitute in for the equation of angular momentum conservation:
[tex]6.65R^2\omega_1 = 2.65R^2*7.45[/tex]
Divide both sides by [tex]R^2[/tex] and we get
[tex]6.65\omega_1 = 19.742[/tex]
[tex]\omega_1 \approx 2.969 rad/s[/tex]
