Q.28
Obtain an expression for maximum safety speed with which a vehicle should move along a curved horizontal road.
Moment of intertia of a solid sphere about its diameter is 25 kg m\(^2\). Calculate its moment of inertia about a tangent.
Best answer
Answer: \(v_{\max} = \sqrt{\mu_s r g}\); moment of inertia about a tangent = 87.5 kg m\(^2\).
(i) Maximum safety speed on a flat (horizontal) curved road:
Let a vehicle of mass \(m\) move with speed \(v\) on a horizontal circular road of radius \(r\). The forces on it are its weight \(mg\) (down), the normal reaction \(N\) (up) and the static friction \(f_s\) between tyres and road, acting horizontally towards the centre. Since there is no vertical motion,
\[N = mg\]The necessary centripetal force is provided only by friction:
\[f_s = \frac{mv^2}{r}\]Friction cannot exceed its limiting value: \(f_s \le \mu_s N = \mu_s mg\). So
\[\frac{mv^2}{r} \le \mu_s mg \Rightarrow v^2 \le \mu_s r g\]The maximum safety speed is therefore
\[\boxed{v_{\max} = \sqrt{\mu_s r g}}\]where \(\mu_s\) is the coefficient of static friction. It is independent of the mass of the vehicle. Above this speed the vehicle skids outwards.
(ii) Numerical: \(I_d = \dfrac{2}{5}MR^2 = 25\) kg m\(^2\), so \(MR^2 = \dfrac{5}{2}\times25 = 62.5\) kg m\(^2\).
A tangent is parallel to a diameter at distance \(R\). By the theorem of parallel axes,
\[I_t = I_d + MR^2 = 25 + 62.5 = 87.5\ \text{kg m}^2\](Equivalently \(I_t = \dfrac{7}{5}MR^2 = \dfrac{7}{2}I_d = 87.5\) kg m\(^2\).)
Log in or create a free account to answer this question.
