Problem 3 A spherical globe (a model of the Earth) is made from a very light wei

Question

Problem 3 A spherical globe (a model of the Earth) is made from a very light weight material so its mass distribution can be ignored. The radius of the globe is R. Four point masses, all of mass m, are glued to the surface of the globe. The locations are: North Pole, South Pole, and two other points whose colatitude and longitude (spherical and coordinates on the globe) are (Bo, /2) and (-6,-7/2), where ° is an angle between 0 and . Thus, the center of mass of the resulting mass distribution is at the center of the globe. The center of the globe is at the origin of a Cartesian coordinate system At time t 0 the North Pole is on the positive z axis, and the Prime Meridian intersects the equator on the positive r axis (a) Compute the moment of inertia tensor at time t = 0 (b) If at t = 0 the vector points in the 2 direction, what is the angular momentum? If we have steady rotation with this w (that is, if w is constant), then at time t = what is the inertia tensor? What is the angular momentum vector at that time? (c) Repeat part (b) with the w vector pointing in the r direction (d) If w wn, where n = (0, sin cosy), calculate the inertia tensor as a function of time. Show that the angular momentum is a constant parallel to w. For this part of the problem, you can use a different Cartesian coordinate system. The calculations are simpler if you define a new coordinate system with the same r-axis and its z-axis parallel to n

Explanation / Answer

3. given radius of model = R

location of 4masses = North pole, South Pole, and at (thetao, pi/2) and (pi - thetao, -pi/2)

so considering z axis going from south to north pole, in cartesian coordiantes the coordiantes of the four points are

A = (0,0,R)

B = (0,0,-R)

C = (0,Rcos(theta0),Rsin(theta0))

D = (0,-Rcos(theta0),-Rsin(theta0))

hence net center of mass is at the cneter of the globe

a. at t= 0, moment of inertial = I

Ixx = mR^2 + mR^2 + mR^2cos^2(theta0) + m^2R^2*sin^2(theta0) + m^2R^2*cos^2(theta0) + m^2R^2*sin^2(theta0)

Ixx = 4mR^2

similiarly

Iyy = m(R^2 + R^2 + R^2sin^2(theta0)) = 2mR^2(sin^2(theta0) + 1)

Izz = m(2R^2cos^2(theta0)) = 2mR^2cos^2(theta0)

Ixy = Iyx = 0

Ixz = Izx = 0

Iyz = Izy = 2mR^2(cos(theta0)sin(theta0)) = mR^2sin(2*theta0)

b. at t = 0, w is in z direction

angular momentum = -Ixz*w i - Iyz*w j + Izz* w k

H = 2mR^2*cos^2(theta0)*w (k - j)

where k and ja re unit vectors along z and y axis

c. for w in x direction

H =(Ixx i - Iyx j - Izx k)w = (4mR^2 i )

d. w = w(sin(theta0/2)j + cos(theta0/2)k)

hence

at time t the coordiantes of the points are given by

A = (Rsin(w*sin(theta0/2)*t), 0, Rcos(w*sin(theta0/2)*t))

B = (-Rsin(w*sin(theta0/2)*t), 0, -Rcos(w*sin(theta0/2)*t))

C = (Rsin(theta0)sin(w*sin(theta0/2)) - Rcos(theta0)sin(wcos(theta0/2)), Rcos(theta0)cos(wcos(theta0/2)), Rsin(theta0)cos(w*sin(theta0/2)) )

D = (-Rsin(theta0)sin(w*sin(theta0/2)) + Rcos(theta0)sin(wcos(theta0/2)), -Rcos(theta0)cos(wcos(theta0/2)), -Rsin(theta0)cos(w*sin(theta0/2)) )

hence

Ixx = m[(Rcos(w*sin(theta0/2)*t))^2 + (Rcos(w*sin(theta0/2)*t))^2 + (Rcos(theta0)cos(wcos(theta0/2)))^2 + ( Rsin(theta0)cos(w*sin(theta0/2)) )^2 + (Rcos(theta0)cos(wcos(theta0/2)))^2 + (Rsin(theta0)cos(w*sin(theta0/2)))^2)

Ixx = mR^2[(cos^2(w*sin(theta0/2)*t) + cos^2(w*sin(theta0/2)*t) + cos^2(theta0)cos^2(wcos(theta0/2)) + sin^2(theta0)cos^2(w*sin(theta0/2)) + cos^2(theta0)cos^2(wcos(theta0/2)) + sin^2(theta0)cos^2(w*sin(theta0/2)))

Ixx = mR^2[(4cos^2(w*sin(theta0/2)*t))]

si8miliarly we can find

Iyy = m(xa^2 + za^2 + xb^2 + zb^2 + xc^2 + zc^2 + xd^2 + zd^2)

Izz = m(xa^2 + ya^2 + xb^2 + yb^2 + xc^2 + yc^2 + xd^2 + yd^2)

Ixy = m(xa*ya + xb*yb + xc*yc + xd*yd)

Iyz = m(ya*za + yb*zb + yc*zc + yd*zd)

Izx = m(za*xa + zb*xb + zc*xc + zd*xd)

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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