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A car of mass m travelling at speed v moves on a horizontal track so that the centre of mass describes a circle of radius r . Show that the limiting speed beyond which the car will overturn is given by v^(2) = ("gra")/(h) where 2a is the separation of the inner and outer wheels and h is the height of the centre of mass above the ground . |
Answer» Solution :We consider the rear view of the CAR , centripetal force `(mv^(2))/(r)` to the left is provided by frictional force acting on the tyres. From the force diagram ,  Sum of forces in vertical direction , `R_(1) + R_(2) = "mg" ........... (i)` Sum of forces in horizontaldirection , `F_(1) + F_(2) = (mv^(2))/(r)` ........ (ii) Taking moments about the centre of mass G, `(F_(1) + F_(2) )h + R_(1)a = R_(2)a` `therefore "" F_(1) + F_(2) = (R_(2) - R_(1)) (a)/(h) "" ...... (iii)` Combining this with Eqn. (ii) to eliminate `F_(1) + F_(2)` gives `R_(2) - R_(1) = (hmv^(2))/(ar) "" ...... (iv)` We now have two simultaneous EQNS.(i) and (iv) , for `R_(1)` and `R_(2)` . Solving these by adding and subtracting , we find that `2R_(1) = mg - (hmv^(2))/(ar)` and `2R_(2) = mg + (hmv^(2))/(ar)` From these expressions it is clear that the inner wheels will leave the ground when `R_(1)` will become zero and the car begins to overturn , i .e.,`""mg = (hmv^(2))/(ar)` So , the LIMITING speed is given by `v^(2) = ("GRA")/(h)` as required Alternative solution : We choose the frame of refernce attached to the car , i.e., a FICTITIOUS centrifugal force of magnitude `mv^(2)//r` , acts through the centre of mass G and directed away from the centre of the circle . At the instant when the car begins to overturn , the force `R_(1)` and `F_(1)` are zero , so the only forces acting on the car are `R_(2) , F_(2)` the centrifugal force and the weight mg . If the resultant P of the centrifugal force and the weight passes outside the point of contact with the ground (the outer wheels) as shown in Fig . 7.66 (b) , the couple on the car will be clockwise , causing it to overturn . The critical condition , when overturning is just about to occur , happens when the resultant P passes through the contact point . By using similar triangles , we can see that the condition for overturning is thus `(mv^(2))/(r) + mg gt (a) /(h)` which gives `v^(2) gt ("gra")/(h)` as before . |
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