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.

Kunal kshirsagar said:   1 decade ago
Maximum velocity corresponds to the maximum inward static frictional force exerted on the tyre.

= > mv^2/R = μmg.
= > v (maximum).
= (μgR).

= (0.20*9.81* 00).
= 14 m/s.

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.

Sara said:   1 decade ago
Force of friction is equal to required centripetal force to move the body in circular path. Equate both forces as,

mu m g = m v*2 /r.
Cancelling mass on both sides we get the formula of velocity by simplifying as,

V= sq root mu r g.

Amit bole said:   1 decade ago
But the distance covered by the car is the circumference of a circle.

125 said:   1 decade ago
(co-efficient of friction X radius X acceleration due to gravity).

Can I know how this formula has been derieved.

Can you please explain this formula?

Sanjay said:   1 decade ago
Can I know how this formula has been derived?

Mukuuuuu champ physics said:   1 decade ago
tan$=v2/rg
tan$=miu
sqrtrgmiu=velocity

Rishabh said:   1 decade ago
9. 8 where it came from?

Bhavy said:   1 decade ago
9.8 comes from the g, which is gravitational constant.


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