Two turns of a smooth road are banked with the same angle. What is the ratio of maximum velocities for the two turns if the ratio of radii of curvature is 1:5?
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Class 11
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Circular Motion and Centripetal Force
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EXAMPLE
Cars moving on smooth circular banked roads
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 coefficient between the tyre and the road is negligible. Then comment on the speed of car when it turns.
Solution:
Applying Newton's laws in horizontal and vertical directions. we get
Nsinθ=
r
mv
2
... (I) and Ncosθ=mg ... (II)
from these two equations, we get
tanθ=
rg
v
2
v=
rgtanθ
This is the speed at which the car doesn't slide down even if there is no
friction. Since the car is in horizontal and vertical equilibrium. So the option 'A' is wrong.
If the car's speed is less then the banking speed then It will slip down to reduce the r.
If the car turns at correct speed of 40 m/s. Then the force by the road on the car is given by
N=
rsinθ
mv
2
=
cosθ
mg
....(III) [from (I) and (II)]
By looking at equation (III). we can say that, the force by the road on
the carN is equal to
rsinθ
mv
2
, not
r
mv
2
,
Since θ<
2
π
;
⇒sinθ<1&cosθ<1
⇒
sinθ
1
>1&
cosθ
1
>1
⇒
rsinθ
mv
2
>
r
mv
2
&
cosθ
mg
>mg
⇒N>
r
mv
2
&N>mg