Laws Of Motion Question 144
Question: A circular road of radius R is banked for a speed v = 40 km/hr. A car of mass m attempts to go on the circular road, the friction co-efficient between the tyre and road is negligible:
Options:
A)the car cannot make a turn without skidding
B)if the car runs at a speed less than 40 km/hr, it will slip up the slope
C)if the car runs at the correct speed of 40 km/hr, the force by the road on the car is equal to $ mv^{2}/r $
D)if the car runs 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 $ mv^{2}/r $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Car can make a turn without skidding at 40 km/hr. The car will slip down the slope if it runs at a speed less than 40 km/hr. $ N\sin \theta =\frac{mv^{2}}{r}\Rightarrow N=\frac{1}{\sin \theta }( \frac{mv^{2}}{r} ) $ and $ n\cos \theta =mg\Rightarrow N=\frac{mg}{\cos \theta } $
$ \because \sin \theta $ and $ \cos \theta <1 $ So $ N>mg $ and $ N>\frac{mv^{2}}{r} $
BETA
