In a proposed rocket design, the nose cone is a parabolic solid, with a circular

Question

In a proposed rocket design, the nose cone is a parabolic solid, with a circular base. 6 m in diameter, and with an axis of rotational symmetry running through the center of the circle and the vertex of the parabola. (The vertical cross section of the "cone" is a parabola.) The height of the solid from base to vertex is 12 m and it has uniform density of 2700 kg m^-3. (a) Where is the center of mass of the nose cone? (Define axes and give cinfinityrdinates.) (b) What is the moment of inertia of the nose cone about its axis of rotational symmetry?

Explanation / Answer

For convenience lets take upward base-parabola

y =Ax2, putting x=3,y=12, A becomea 9/12 = 0.75

y=0.75x2

x2 = 1.33y

Lets take a disc of thickness dy at y

volume = 3.14x2 dy =  3.14*1.333ydy =4.1867 y dy

mass = d*4.1867 y dy where d is density

ycm of the element = y

COM = integral of [d*4.1867 y *ydy] / integral of [d*4.1867 y dy]

=  integral of [y *ydy] from 0 to 12 / integral of [y dy] from 0 to 12

= y3/3 / y2/2 = 2/3y =12*2/3 = 8

Now take the figure given in question, i.e inverted parabola,

so center of mass = 12-8 = 4m

Hence COM = (0, 4m, 0) where nose is in +y direction and base is in x-z plane

b)moment of inertia of a disc = 0.5 MR2

since moment of inertia of inverted parabola will be same, so

moment of inertia of a disc element = 0.5 *d*4.1867 y dy*x2

   = 0.5 *2700*4.1867 y2 dy

=5652 y2 dy

moment of inertia of the solid = integral from 0 to 12m of [5652 y2 dy]

=5652 y3 /3 = 1884*12*12*12 = 3255552 kgm2

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.