written 8.5 years ago by | modified 3.0 years ago by |
Mumbai University > MECH > Sem 5 > Theory Of Machines 2
Marks: 8 M
Year: May 2015
written 8.5 years ago by | modified 3.0 years ago by |
Mumbai University > MECH > Sem 5 > Theory Of Machines 2
Marks: 8 M
Year: May 2015
written 8.5 years ago by |
Let W be the weight acting at C.G at height h. Let θ be the maximum angle of tilt with vertical and C be the corresponding couple. Fc is the centrifugal force.
For stability
$∑M = 0 \hspace{3cm} \text{(from base point)} \\ W \times h sinθ – Fc cos θ – C = 0 \\ Mgh \sinθ – mv^2/r \cosθ – (2Iw + G x Ie)w_ew_p \cosθ = 0 \\ Mgh \sinθ – mv2/r \cosθ – (2Iw + G x Ie)V/r x V/R cosθ = 0$
$\tan \theta=(mh+(2/w+G/e)/r) \times v^2 / mghR$
Where
Θ = angle of heel
R = radius of the curve
r = radius of the wheel
G = engine to wheel speed ratio
V= velocity of the two wheeler
M = mass
g = accn due to gravity
H = height of CG above ground in vertical position