Advertisements
Advertisements
Question
If the horizontal force needed for the turn in the previous problem is to be supplied by the normal force by the road, what should be the proper angle of banking?
Solution
Given:
Speed of the scooter = v = 36 km/hr = 10 m/s
Radius of turn = r = 30 m
Let the angle of banking be \[\theta\]
We have :
\[\tan\theta = \frac{\text{v}^2}{\text{rg}}\]
\[\Rightarrow \tan\theta = \frac{100}{30 \times 10} = \frac{1}{3}\]
\[ \Rightarrow \theta = \tan^{- 1} \left( \frac{1}{3} \right)\]
APPEARS IN
RELATED QUESTIONS
You are driving a motorcycle on a horizontal road. It is moving with a uniform velocity. Is it possible to accelerate the motorcycle without putting higher petrol input rate into the engine?
Is it necessary to express all angles in radian while using the equation ω = ω0 + at ?
A car driver going at some speed v suddenly finds a wide wall at a distance r. Should he apply brakes or turn the car in a circle of radius r to avoid hitting the wall?
A circular road of radius r is banked for a speed v = 40 km/hr. A car of mass of m attempts to go on the circular road. The friction coefficient between the tyre and the road is negligible.
(a) The are cannot make a turn without skidding.
(b) If the car turns at a speed less than 40 km/hr, it will slip down.
(c) If the car turns at the car is equal to \[\frac{\text{mv}^2}{\text{r}}\]
(d) If the car turns at the correct speed of 40 km/hr, the force by the road on the car is greater than mg as well as greater than \[\frac{\text{mv}^2}{\text{r}}\]
A park has a radius of 10 m. If a vehicle goes round it at an average speed of 18 km/hr, what should be the proper angle of banking?
If the road of the previous problem is horizontal (no banking), what should be the minimum friction coefficient so that scooter going at 18 km/hr does not skid?
A circular road of radius 50 m has the angle of banking equal to 30°. At what speed should a vehicle go on this road so that the friction is not used?
