A car starts from rest in a circular flat road of radius R with an acceleration a. The friction coefficient between the road and the tyred is mu Find the distance car will travel before is start skidding.
Answers
Explanation:
There would be two accelerations: Centripetal Acceleration & Tangential Acceleration. So first of all direction wont be towards the center of the path (i.e not only radial direction)
Direction would be: at theta angle from radially outward tanθ=
v
2
/R
a
=
v
2
aR
Total Force due to these acceleration $$\sqrt { { ({ mv }^{ 2 }/R) }^{ 2 }+{ (ma) }^{ 2 } } >\quad { mv }^{ 2 }/R$$
For the car to not skid: Maximum Frictional Force that can act should be greater than or equal to
(mv
2
/R)
2
+(ma)
2
. Therefore the magnitude of the frictional force on the car is greater than mv
2
/R
Friction Force: f=μN=μmg
$$\Rightarrow \quad f\ge \sqrt { { ({ mv }^{ 2 }/R) }^{ 2 }+{ (ma) }^{ 2 } } \\ \Rightarrow \quad \mu mg>ma\\ \Rightarrow \quad \mu >a/g$$
Its not a tilted track so the friction coefficient between the ground and the car is not equal to { tan }^{ -1 }(\frac { { v }^{ 2 } }{ rg } ). This is for banked roads.
In our case: $$\Rightarrow \quad \mu g\ge \sqrt { { ({ v }^{ 2 }/R) }^{ 2 }+{ (a) }^{ 2 } } \\ \Rightarrow \quad \mu >\sqrt { { ({ v }^{ 2 }/R) }^{ 2 }+{ (a) }^{ 2 } } /g$$
Explanation:
v^2=u^2+2as
v^2=0+2as
v=√2as
ac=v^2/R=2as/R
=√at^2+ac^2=√a^2+(2as/R)^2
force=manet=m√a^2+(2as/R)^2
Car will skid when
m=√a^2+(2as/R)^2= umg
s=√u^2g^2-a^2(R^2/4a^2)