Rotating Nucleus Physics Problem According to a crude model, the ^9Be nucleus is

Question

Rotating Nucleus Physics Problem

According to a crude model, the ^9Be nucleus is a uniform solid sphere of mass 1.49 Times 10^-26 kg and radius 2.50 Times 10^-15 m which spins about its diameter. The angular momentum of this spinning berylium nucleus is 1.58 Times 10^-34 J s. According to this model, what is the angular velocity of the berylium nucleus? What is the linear velocity of a point on its equator? (Aside: Compare your result to the speed of light, c = 3.00 Times 10^8 m/s) What is the rotational kinetic energy? What is the ratio of the rotational kinetic energy to the relativistic rest-mass energy?

Explanation / Answer

Given that L = 1.58*10^-34 J-s

m = 1.49*10^-26 kg

and r = 2.5*10^-15 m

we know that L = I*w

I is the moment of inertia = 0.5*m*r^2 = 0.5*1.49*10^-26*(2.5*2.5)*10^-30 = 4.65625*10^-56 kg*m^2

angular velocity is w = L/I = (1.58*10^-34)/(4.65625*10^-56) = 3.39*10^21 rad/s

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linear velocity v = r*w = 2.5*10^-15*3.39*10^21 = 8.475*10^6 m/s

comparing with the speed of light ,then v = 0.02825*c

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Krot = 0.5*I*w^2 = 0.5*4.65625*10^-56*(3.39*3.39)*10^42 = 26.755*10^-14 J

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Krot/E = (26.75*10^-14)/(1.49*10^-26*2.9979*2.9979*10^16) = 2*10^-4

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