Question

Have to develop question

Topic: Curvature

Definition of curvature

The curvature at a point tells us how fast the curve is turning at that point.

• Consider a car travelling along a winding road. The tighter the curve, the more difficult the driving is. The number curvature, describe this “ tightness”.

If the curvature is zero, then the curve looks like a line near this point.

If the curvature is a large  number, indicate that the curve has a sharp bend.

Thus, for a curve ,the curvature is a measure of how sharply the curve is bending at a point.

k= [ |dT| / |dt| ] / [ |dr| / |dt| ]

• Your example can be develop based on several set of questions. It must be original question and answer.

• The question must be based on Taxonomy Bloom Level (Please refer to the low order thinking skills taxonomy level i.e. REMEMBER (C1) UNDERSTAND (C2) APPLY (C3))

• The example must provide complete solution; which includes the derivation and step by step solution to the final answer.

Answer 1

consider a car on a tight bend of road with radius of curvature R = 30m at the tightest bend
now,
for a circular trac
let T be the tangent vector
then dT/dt has to be calculated
now, dT/dt is the rate at which tangent moves along the circumference of circle
now, dT/dt is because of the rotation about the circumference as well as rotation about the circumference of circle
so dT/dt = w . w = w^2 = v^2/r^2
dr/dt = w^2*r = v^2/r
[dT/dt ]/ [dr/dt ]= 1/r
hence for circular track bank, curvature is reciprocal of the radius of curvature

so , for the tightest radius r = 30 m, k = 1/30= 0.033
if friction coefficient is u
then
maximum speed at which the car can move is v
where v can be found for non banked road as under
from force balacne
mv^2/r = u*mg
v = sqrt(rug) = sqrt(u*g/k)
so maximum velocity at which a car can move along a tight bend of curvature k is sqrt(ug/k)
where u is coefficient of static friction between tyres of vehicle and road, and g is acceleration due to gravity on the said planet