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At any angular speed, a certain uniform solid sphere of diameter D has half as much rotational kinetic energy as a certain uniform thin-walled hollow sphere of the same diameter when both are spinning about an axis through their centers. If the mass of the solid sphere is M, the mass of the hollow sphere is At any angular speed, a certain uniform solid sphere of diameter D has half as much rotational kinetic energy as a certain uniform thin-walled hollow sphere of the same diameter when both are spinning about an axis through their centers. If the mass of the solid sphere is , the mass of the hollow sphere is___________.

a. 5/6 M.
b. 3/5 M.
c. 2 M.
d. 6/5 M.
e. 5/3 M.

Answer :

ajeigbeibraheem

Answer:

d. 6/5 M.

Explanation:

let the angular velocity =  [tex]\omega[/tex]

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

moment of inertia of solid sphere [tex]I_s[/tex] = [tex]2/5 M(\frac{D}{2})^2[/tex]

rotational K.E of solid sphere= [tex]\frac{1}{2}* I_s* \omega ^2[/tex]

=  [tex]\frac{1}{2} * 2/5 M(\frac{D}{2})^2 * \omega^2[/tex]

we represent the mass of hallow sphere= [tex]M_{hs}[/tex]

moment of inertia of solid sphere[tex]I_{ss}[/tex] = [tex]2/3 M_{hs}(\frac{D}{2})^2[/tex]

rotational K.E of hollow sphere= [tex]\frac{1}{2}* I_{ss} * \omega^2[/tex]

= [tex]1/2 *2/3 M_{hs}(\frac{D}{2})^2* \omega^2[/tex]

NOW,the kinetic energy of solid sphere= 1/2 kinetic energy of hallow sphere

=[tex]1/2 [2/5 M(D/2)^2 ]\omega^2= 1/2 (1/2 (2/3 M_{hs}(D/2)^2 )\omega^2 }[/tex]

2/5M= 1/3 [tex]M_{hs}[/tex]

[tex]M_{hs}[/tex] = 3(2/5M)

[tex]M_{hs}[/tex] = 6/5 M

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