A cyclist bend
on
a
Circular ara
20 cm when
he turns if the cofficient of static friction
between the type
of
his cycle
and road
is
0.8what velocity should he
has while turning?
given g=10 m/s2
•
Answers
Here,r=3m, μ=0.1,g=10 ms
−2
The maximum speed with which a cyclist can take a turn without skidding is
=
μrg= .01×(3m)(10ms −2 )
= 3ms−1
.
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Some teachers say: centripetal force must equal friction force, so
[math]m*V^2/R = f*m*g[/math]
m is eliminated and you are left with
[math]V^2 = f*g*R[/math]
on Earth g = 9,81 m/s^2 and then this equation relates 3 variables : f friction, R radius and V speed. So problems will give you two and ask for the third.
Here, R=30 m, f = 0.4 and then V is calculated as 10.85 m/s (39 km/hr or 24,4 mph)
There is just one thing that can produce an external force up on our car: the road.
Because air friction is too complicated and we will deal with it in a second course in college, and gravitation is countered by the road’s vertical reaction. This last sentence is in math language
gravitation (m*g) = N ( normal reaction )
Now they say Friction = f* N = f*m*g
But this is yet another conceptual mistake.
Consider a 10 kg block on a table with f=0.5. now a mouse is pushing it to the right with a force of 10 Newtons ( 2 pounds). Will it move? Let’s see.
In our car friction = m*V^2/R
if V is low friction will be equal to m*V^2/R, and the mass is important, we can continue increasing V and friction will counter it by increasing right to the new value of m*V^2/R and friction will increase until it reaches its maximum value of f times Normal. This maximum is
FrictMax = f*m*g
and will be reached only when V be equal to Vmax
FrictMax = f*m*g = m*Vmax^2/R
This leads to the equation [math]m*V^2/R = f*m*g[/math]
that we started with.
.
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