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A vehicle is moving with a velocity v on a curved road of width b and radius of curvature R. For counteracting the centrifugal force on the vehicle, the difference in elevation required in between the outer and inner edges of the road is?

a) \(\begin{array}{l}\frac{bv}{Rg}\end{array} \) b) \(\begin{array}{l}\frac{vb^{2}}{Rg}\end{array} \) c) \(\begin{array}{l}\frac{bv^{2}}{Rg}\end{array} \) d) \(\begin{array}{l}\frac{Rg^{2}}{bv}\end{array} \) Solution: c) \(\begin{array}{l}\frac{bv^{2}}{Rg}\end{array} \)

A wheel whose moment of inertia is 12 kg m^2 has an initial angular velocity of 40 rad/s. A constant torque of 20 Nm acts on the wheel. The time in which the wheel is accelerated to 100 rad/s is

Solution: \(\begin{array}{l}T=I\alpha\end{array} \) \(\begin{array}{l}20=12*\alpha\end{array} \) \(\begin{array}{l}\alpha =\frac{20}{12}\end{array} \) \(\begin{array}{l}\omega_{0}=40\end{array} \) \(\begin{array}{l}\omega =\omega_{0}+\alpha t\end{array} \) \(\begin{array}{l}100 =40+\frac{20}{12}t\end{array} \) \(\begin{array}{l}\frac{60*12}{20}=t\end{array} \) t=36... View Article