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 1 of 5.
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.
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.
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.
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.
Chinmaya said:
5 years ago
Overturning is counterbalanced by lateral friction in order to prevent from skid.
Overturning (centrifugal force) = frictional force.
Mv^2/r = UN=Umg (m is cancelled out both side).
V^2/100 = .2*10,
V = √(2*100),
V = 14.142 m/s..
Overturning (centrifugal force) = frictional force.
Mv^2/r = UN=Umg (m is cancelled out both side).
V^2/100 = .2*10,
V = √(2*100),
V = 14.142 m/s..
(5)
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.
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.
Sai manoj said:
8 years ago
It's very simple sum because
Velocity(V) = ?.
Radius(R) = 100m,
Co-efficient of friction(f) = 0.2,
Gravitational force(g) = 9.8.
V=F * V * G.
V^2=0.2 * 100 * 9.8 = 196underroot,
v = root 196,
V = 14m/s.
Velocity(V) = ?.
Radius(R) = 100m,
Co-efficient of friction(f) = 0.2,
Gravitational force(g) = 9.8.
V=F * V * G.
V^2=0.2 * 100 * 9.8 = 196underroot,
v = root 196,
V = 14m/s.
(3)
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.
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.
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.
= > mv^2/R = μmg.
= > v (maximum).
= (μgR).
= (0.20*9.81* 00).
= 14 m/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.
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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