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 2 of 5.

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?

Rishikesh kumar suman said:   1 decade ago
@Bhavy.

9.8 is the acceleration due to gravity not the gravitational constant.
Gravitational constant=6. 67x10 to the power -11 N m*m/kg*kg.

Pankaj pundir said:   1 decade ago
If it travel in circular motion it affected by two forces centripetal force and gravitational force expression.

mg = mv(sq)/r.

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.

Abiha said:   1 decade ago
v2 = {co-efficient of friction X radius X acceleration due to gravity).

v2 = {0.2 x 100 x 9.8}.
= 196.

v = 14m/sec.

Ahmad said:   9 years ago
Forces required to move a car in a circular track?

A. Centripetal force
B. Friction
C. Both
D. None

Answer the question.
(1)

Ekhayemhe Eugene said:   10 years ago
Equating Centripetal force (F = mv^2/r) formula and that of coefficient of friction (U = F/R).

You will get 14 m/sec.

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.

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

Abhi said:   1 decade ago
@Pawan.

Can't you see there is a underroot written in the formula? underroot 196 = 14.


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