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.
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)
Ashish said:
6 years ago
mv^2/r = u(friction)mg.
By using above formula "v" can be calculated.
By using above formula "v" can be calculated.
(2)
Nuclei said:
7 years ago
It is based on the Newton 2nd law of Motion.
(1)
Deriba Fufa said:
7 years ago
Please explain the formula.
(1)
Praveen said:
8 years ago
Mv2/r = co.eff.friction x m x g.
(3)
Vvn reddy said:
8 years ago
.2*100*9.8=196.
Root applied by 196=14 so the answer is 14.
Root applied by 196=14 so the answer is 14.
(2)
Jasbir Ranga said:
8 years ago
Yes, I agree @Sayyad Touffiq Ali.
Kriti said:
8 years ago
Yes, I agree @M. Sathvik.
M.sathvik said:
8 years ago
Option C is the correct answer.
(1)
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)
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