How is the velocity vector (2.6) found? Can you show a step by step process? By

Question

How is the velocity vector (2.6) found? Can you show a step by step process?

By definition, the position vector OQ is (Fig. 2.2) OQ = h(- tan gamma j + k) This expression can he differentiated with respect to time to obtain V_Q - V_O = h(- tan gamma j + k) + h(omega_z tan gamma i - gamma/cos^2 gamma j) = h omega_z tan gamma i - (h tan gamma + h gamma/cos^2 gamma)j + hk since dj/dt = -omega_z i. Even in steady-state conditions, that is to h = gamma = 0, we have V_Q = V_O + h omega_z tan gamma i and hence the two velocities are not exactly the same, unless also gamma = 0. The camber angle gamma is usually very small in cars, but may be quite large in motorcycles. The velocity of point O has, in general, longitudinal and lateral components V_0 = V_O = V_ox i + V_oy j

Explanation / Answer

OQ = h ( -tan g j + k)

From the 3rd figure

i = sin (wz * t)

j = cos (wz * t)

dj/dt = [-sin (wz * t) * wz]

= -wz i

Vq - Vo = d/dt (OQ)

= dh/dt ( -tan g j + k) + h d(-tan g j + k )/dt d(AB)/dt = BdA/dt + A dB/dt

= dh/dt ( -tan g j + k) + h (wz tan g i - (sec^2 g )(dg/dt)j ) d(tan g)/dt = (sec^2 g) dg/dt and dj/dt = -wz i

= (h * wz * tan g) i - (dh/dt * tan g + h * dg/dt * sec^2 g) j + dh/dt k

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