General Knowledge - Physics - Discussion

Discussion Forum : Physics - Section 1 (Q.No. 23)
23.
Find the maximum velocity for the overturn of a car moving on a circular track of radius 100 m. The co-efficient of friction between the road and tyre is 0.2
0.14 m/s
140 m/s
1.4 km/s
14 m/s
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
46 comments Page 3 of 5.

Pratiksha Pimple said:   1 decade ago
By formula:

V = Under root(mu.r.g).
V = Under root(0.2*100*9.8).
V = Under root(20*9.8).
V = Under root(196).

Therefore V = 14m/s.

Ashok kumar said:   1 decade ago
Traction force = centripetal force.

umg = mv^2/r.

u = coefficient of friction.
m = mass.
g = 9.8.

r = radius.
v = max velocity at turning.
So v= under root ugr.

v = under root 0.2*9.8*100.
v =14 m/sec.

Arav said:   1 decade ago
Dear @Bhavna will you please check units of your formula, you will find physically impossible.

Gerald said:   1 decade ago
Can someone who knows this make it clear to us please.

Pablo said:   1 decade ago
But max speed can be 140m/sec. Because even it will result in the overturn.

Deep said:   1 decade ago
Where this formula is given?

Joshi said:   1 decade ago
v=mu*r*g is the correct formula.

Bhavana said:   1 decade ago
Maximum velocity of car = coefficient of friction X radius X acceleration due to gravity.
= 0.2X 100X 9.8.
=14m/s.

Amisha said:   1 decade ago
r = 100
v = ?
coefficient of friction = ?

As v = underroot(coefficient of friction X radius X acceleration due to gravity).
v = underroot (0.2 X 100 X 9.8).
v = 14m/s.

Nikhil said:   1 decade ago
As the S.I unit of acceleration is m/s.


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