First, we need to determine the angular velocity of the football in the position shown. The angular velocity of the football about its axis of symmetry is 80 rev/min, or 8.377 rad/s (since there are 2π radians in one revolution). The angular velocity of the axis of symmetry about axis A-A is 40 rev/min, or 4.188 rad/s. The angle between the two axes is constant at 15º.
Using the formula for the vector cross product, we can find the angular velocity of the football in the position shown:
w = w1 + w2 x r
where w1 is the angular velocity of the axis of symmetry about axis A-A, w2 is the angular velocity of the football about its axis of symmetry, and r is the vector from the axis of symmetry to the point where we want to find the angular velocity.
We can represent w1, w2, and r using their magnitudes and direction cosines. Let's assume that the football is centered at the origin, and that the x, y, and z axes are aligned with the principal axes of the football.
w1 = 4.188 [1 0 0]
w2 = 8.377 [cos(15º) 0 sin(15º)]
r = [0 0 -r]
where r is the radius of the football.
Taking the cross product of w2 and r, we get:
w2 x r = 8.377 [cos(15º) 0 sin(15º)] x [0 0 -r]
= 8.377r [sin(15º) 0 cos(15º)]
Adding w1 and w2 x r, we get:
w = 4.188 [1 0 0] + 8.377r [sin(15º) 0 cos(15º)]
The magnitude of w is:
|w| = sqrt((4.188)^2 + (8.377r sin(15º))^2 + (8.377r cos(15º))^2)
The angular acceleration of the football is given by the formula:
alpha = dw/dt
where w is the angular velocity of the football. We can find the time derivative of w by taking the partial derivatives of the components of w with respect to time:
dw/dt = [d/dt(4.188) d/dt(8.377r sin(15º)) d/dt(