# Find the increase of moment of inertia of a steel sphere of diameter 10 cm about it's diameter when it's temperature is raised from 15C to 65C.Density of steel at 15C = 7,7 gm.c.c. The coefficient...

Find the increase of moment of inertia of a steel sphere of diameter 10 cm about it's diameter when it's temperature is raised from 15C to 65C.

Density of steel at 15C = 7,7 gm.c.c.

The coefficient of linear expansion of steel 0.000012/C

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### 2 Answers

The moment of inertia of a solid sphere is 2mr^2/5, where m is the mass of the sphere and r is the radius.

The mass of sphere = volume at 15 deg * density at 15deg C

= (4/3)*10^3 * 7.7= 32253.68458gram

When heated radius increases.

The coefficient of linear expansion of steel sphere is a -0.000012 /C.

So the new radius r2= 10+10*0.000012)(65-15) = 10.006

Therefore the new moment of inertia = 2mr2^2/5

So the increase in MI = m(r2^2-r1)^2 /5 = 2m(10.006^2-10^2)/5 = 0.0480144m= 0.0480144*32253.68458gram =1548.641313kgm^2

We'll write the momentof inertia as a function of temperature.

The radius of sphere is r0, at 0 degrees Celsius. The mass of sphere is m.

The moment of inertia of the sphere about the diameter is:

(2/5)*m*rt^2

When the temperature rises from 0C to tC, the nwe radius is:

rt = r0(1+a*t)

The moment of inertia is:

It = (2/5)*m*rt^2

It = (2/5)*m*[r0(1+a*t)]^2

We'll expand the square:

It = (2/5)*m*r0^2*(1 + 2a*t + a^2*t^2)

We'll neglect the quantity a^2*t^2. We'll remove the brackets:

It = (2/5)*m*r0^2 + (4/5)*m*r0^2*a*t

It = I0 + I0*2*a*t

We'll calculate the moment of inertia at t = 65 C

I65 = I0(1 + 2*a*65)

We'll calculate the moment of inertia at t = 15 C

I15 = I0(1 + 2*a*15)

I65/I15 = 1 + 100*a

I65 = I15(1 + 100*a)

I65 - I15 = 100*a*I15

I15 = 2*4*5^5*7.7*pi/5*3

I15 = 8*7.7*625pi/3

I65 - I15 = 8*7.7*625pi*0.0012/3

**I65 - I15 = 48.40 gm cm^2**