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 2 of 5.
Ahmad said:
9 years ago
Can a car move without friction?
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.
A. Centripetal force
B. Friction
C. Both
D. None
Answer the question.
(1)
Ihsan ullah said:
9 years ago
Thanks to everyone. Really helpful explanation.
Josephine said:
9 years ago
Thanks for all your good work. It is really helpful.
Antim said:
9 years ago
Centrifugal force = Friction force,
mv^2/r = friction * mg.
mv^2/r = friction * mg.
Kailas said:
9 years ago
Thanks to everyone. Really helpful explanation.
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.
You will get 14 m/sec.
Akin said:
1 decade ago
Please which law states this @Bhiva?
(1)
Supreeth said:
1 decade ago
mv2/r = μmg.
v2/100 = 0.2*9.8.
v2 = 196.
v = 14 m/s.
v2/100 = 0.2*9.8.
v2 = 196.
v = 14 m/s.
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.
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