Question

A curve of radius 28 m is banked so that a 990 kg car traveling at 41.1 km/h can round it even if the road is so icy that the coefficient of static friction is approximately zero. You are commissioned to tell the local police the range of speeds at which a car can travel around this curve without skidding. Neglect the effects of air drag and rolling friction. If the coefficient of static friction between the road and the tires is 0.300, what is the range of speeds you tell them?

Answer 1

For no friction condition there is no range of speed only on possible speed is 41.1 km/h

while for the case of 0.300 friction coefficient the range of speed is from 6.5 m/s to 15.75 m/s

Step-by-step explanation:

When there is no friction on the turn of road then the centripetal force is due to the component of Normal force only

So we will have

`Ncos\theta = mg`

`Nsin\theta = (mv^2)/(R)`

so we have

`tan\theta = (v^2)/(Rg)`

`\theta = tan^(-1)(v^2)/(Rg)`

v = 41.1 km/h = 11.42 m/s

`\theta = tan^(-1)(11.42^2)/(28(9.8))`

`\theta = 25.42^o`

so here in this case there is no possibility of range of speed and only one safe speed is possible to take turn

Now in next case if the coefficient of static friction is 0.300

then in this case we have

`v_(max) = \sqrt{((\mu + tan\theta)/(1 - \mu tan\theta))Rg}`

`v_(max) = \sqrt{((0.3 + 0.475)/(1 - (0.3)(0.475)))(28 * 9.8)}`

`v_(max) = 15.75 m/s`

Similarly for minimum speed we have

`v_(min) = \sqrt{((\mu - tan\theta)/(1 + \mu tan\theta))Rg}`

`v_(min) = \sqrt{((-0.3 + 0.475)/(1 + (0.3)(0.475)))(28 * 9.8)}`

`v_(min) = 6.5 m/s`

So the range of the speed is from 6.5 m/s to 15.75 m/s