The mass distribution of a rugby ball can be approximated by an ellipsoid (figur

Question

The mass distribution of a rugby ball can be approximated by an ellipsoid (figure obtainedby rotating an ellipse around its axis of symmetry, here the major axis) with the mass concentrated near the surface with uniform mass density. The length of the ball is h while its orthogonal cross-section is a circle of diameter d The ball is empty inside (assume vacuum) Assuming that the ball is rotationally symmetric along its length h, identify the principal axes of its moment of inertia. (a) Find the moment of inertia of the ball along the h axis, given that the mass of the ball is M the wall thicness is w and the wall material is umiform. Given: (b) where ?be is the volume of ellipsoid with half-axes a, b, c, while h, is the moment of inertia of solid ellipsoid with uniform mass density, for rotations around one of its half-axes (here a) (c) One of the basic skills in rugby is called spin pass. In that move the ball is thrown in the direction of the intended target with a linear velocity v and significant spinnins motion around the major axis of the ellipsoid, set by the passing player close to parallel to . If the goal was to pass the ball as quickly and precisely as possible in vacuum, would this be the most energy efficient strategy? Clearly explain your reasoning (d) In open air the spin pass seems to be the fastest and the most precise way of passing the ball. Briefly discuss the reasons for that by treating the rugby ball as a rigid body subject to gravitational pull and air drag

Explanation / Answer

for the rugby ball, its an ellipsoid, i.e. rotaiton of an ellipse about major axis

major axis = 2a = h

minor axis = 2b = d

mass density concentragted near surface and is constat = lambda

a. assuming the ball is rotatinoally symmetric along its length

the principal axis of moment of inertia from symmetry are

the z axis along the length of the ball

and x and y axis perpendicular to the ball's length, in the same plane

b. wall thickness = w

mass of ball = M

Vabc = 4*pi abc/3

Ia = 0.5M(b^2 + c^2)

hence

Ia = moment of inertia of solid ellipsoid - moment of inertia of solid removed part of the ellipsoid

mass density = lambda

lanbda = 3M/4pi(hd^2/8 - (h - w)(d/2 - w)^2/2) = 6M/pi(hd^2 - (h - 2w)(d - 2w)^2) = 6M/pi(hd^2 - (h - 2w)(d^2 + 4w^2 - 4dw)) = 6M/pi(hd^2 - hd^2 - 4hw^2 + 4dhw + 2d^2w + 8w^3 - 8dw^2)

lambda = 3M/pi*w( - 2hw + 2dh + d^2 + 4w^2 - 4dw)

hence

I = lambda*pi*hd^4/3*8 - lambda*pi*(h- 2w)(d - 2w)^4/3*8 = lambda*pi(hd^4 - (h - 2w)(d - 2w)^4)/24

I = M*(hd^4 - (h - 2w)(d - 2w)^4)/8w( - 2hw + 2dh + d^2 + 4w^2 - 4dw)

c. in vaccum, the energy spent in rotaitng the ball is a waste because that does not contribute to the range of the projectile

hence for tha ball to move most dquakly and precisely, yhe most efficient way, energy wise would be to throw a ball without any spin along any axis directly at the location with center of mass of the ball traversing a parabolic trajectroy

d. in presence of air, the differential air drag will cause change in aur pressure on the upper and lower surface of the ball, and specific rotations can help in decrease pressureon upper surface and increase it on lower surface to provide a lift force to the ball that will increase its range and is energetically more feasible then throwint he ball without any spin

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