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
Discussion:
46 comments Page 4 of 5.
Prashant Khandai said:
1 decade ago
We Know that, force F=m.a; mu.m.g = m.a
That implied, a = mu.g
Again, v2 - u2 = 2.a.s
v2 = 2. mu. g. s
v2 = sqrt(2. mu. g. s)
Where, m=mass, a= acceleration, mu=co-efficient of friction, g=acceleration due to gravity, s=distance, v=final speed and u=initial speed.
That implied, a = mu.g
Again, v2 - u2 = 2.a.s
v2 = 2. mu. g. s
v2 = sqrt(2. mu. g. s)
Where, m=mass, a= acceleration, mu=co-efficient of friction, g=acceleration due to gravity, s=distance, v=final speed and u=initial speed.
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.
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.
Tamanna said:
1 decade ago
Can you please explain how this formula came?
Pawan said:
1 decade ago
If we substitute the given values in (co-efficient of friction X radius X acceleration due to gravity) , we will get the answer as 196 but not 14. I think there is formula related to friction from the above problem.
Abhi said:
1 decade ago
@Pawan.
Can't you see there is a underroot written in the formula? underroot 196 = 14.
Can't you see there is a underroot written in the formula? underroot 196 = 14.
Sayyed Toufique Ali said:
1 decade ago
The maximum speed for the overturn of a car moving on a circular track is given by,
v=underroot(co-efficient of friction X radius X acceleration due to gravity)
v=underroot(0.2 X 100 X 9.8)
v=14m/s.
v=underroot(co-efficient of friction X radius X acceleration due to gravity)
v=underroot(0.2 X 100 X 9.8)
v=14m/s.
Prajakta Patil said:
1 decade ago
Because there is one formula for max. velocity i.e.
V.V = urg.
Where u=coefficient of friction of that road.
r=radius of road.
g=acceleration due to gravity,which is on the earth equal to 9.8 m/s.
V.V = urg.
Where u=coefficient of friction of that road.
r=radius of road.
g=acceleration due to gravity,which is on the earth equal to 9.8 m/s.
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.
mg = mv(sq)/r.
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.
v2 = {0.2 x 100 x 9.8}.
= 196.
v = 14m/sec.
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.
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.
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