Dynamics_Part73

Dynamics_Part73 - Problem Set 9: Problem 7. Problem: A...

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Unformatted text preview: Problem Set 9: Problem 7. Problem: A projectile of mass m has a radius of gyration R about its axis of symmetry (x axis) and a radius of gyration 4R about the transverse axis (y axis). Its angular velocity, , can be resolved into two components. The first is the rate of spin, s , which is directed along the axis of symmetry. The second is the rate of precession, p , which is parallel to the projectile's velocity vector (axis GD). For the projectile shown, the angular momentum vector relative to its center of mass is HG = H(i-0.032 j), where H is a constant. (a) Determine as a function of H, m and R. (b) Determine s and p as functions of H, m, R and the angle between the symmetry axis and the velocity vector. HINT: Let = s i + p p, where p is a unit vector parallel to the velocity vector. (c) If the projectile has R = 5 cm, m = 30 kg, H = 0.75 kgm2 /sec and = 5o , compute the projectile's angular velocity vector, = x i + y j, and its spin and precession rates, s and p , respectively. Solution: (a) The moment of inertia tensor for the center-of-mass based principal axis system is mR2 0 0 0 16mR2 [I] = 0 0 0 16mR2 where we use the fact that Iy and Iz are equal for this axisymmetric body. So, in general, the angularmomentum vector will be HG = [I] = mR2 x i + 16mR2 y j + 16mR2 z k For the given HG , we have mR2 x = H, 16mR2 y = -0.032H, 16mR2 z = 0 Solving for x , y and z , the angular-velocity vector, , is = H (i - 0.002 j) mR2 (b) First, note that a unit vector p tangent to the projectile's velocity vector is related to the unit vectors i and j according to p = i cos - j sin = j = i cot - p csc ...
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