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A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is slowly heated until it reaches its melting temperature, at which point it flattens into a uniform disk of thickness D 2 . By what factor is the angular velocity changed? (Give your answer as a factor of ωi.)

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ajeigbeibraheem

Answer:

0.6

Explanation:

The volume of a sphere = [tex]\frac{4}{3} \pi (\frac{D}{2})^3[/tex]

Therefore [tex]\pi * r^2 * (\frac{D}{2} ) = \frac{4}{3} \pi (\frac{D}{2})^3[/tex]

r of the disc = [tex]1.15(\frac{ D}{2} )[/tex]

Using conservation of angular momentum;

The [tex]M_i[/tex] of the sphere = [tex]\frac{2}{5} m \frac{D}{2}^2[/tex]

[tex]M_i[/tex] of the disc = [tex]m*\frac{ \frac{1.15*D}{2}^2 }{2}[/tex]

[tex]\frac{wd}{ws} = \frac{\frac{2}{5}m * \frac{D}{2}^2}{ m * \frac{(\frac{`.`5*D}{2})^2 }{2} }[/tex]

= 0.6

The factor where the angular velocity changed should be considered as the 0.6.

Calculation of the change in the angular velocity:

Since

volume of sphere =  4/3 pi*(D/2)^3

Now

pi*r^2*(D/2) =4/3 pi*(D/2)^3

Now

r of disc =1.15(D/2)

So,

Mi of sphere =2/5 m(D/2)^2

Now

mi of disc =m*(1.15D/2)^2 /2

Idwd = Isws

So,

= wd/ws =Is/Id

=[2/5 m(D/2)^2]/[m(1.15D/2)^2 /2]

=0.6

Hence, The factor where the angular velocity changed should be considered as the 0.6.

Learn more about velocity here: https://brainly.com/question/1303810

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