If a cyclist moving with a speed of 4.9 m/s on a level road can take a sharp circular turn of radius 4 m,then what is the coefficient of friction between the cycle tyres and the road?

mu*mg=mv^2/r

mu=v^2/rg

0. 61

To determine the coefficient of friction between the cycle tires and the road, we can use the centripetal force equation.

The centripetal force required to keep the cyclist moving in a circular path is provided by the frictional force between the tires and the road. The equation for centripetal force is given by:

F = (mv^2) / r

where:
F is the centripetal force,
m is the mass of the cyclist,
v is the velocity of the cyclist, and
r is the radius of the circular path.

Given:
v = 4.9 m/s
r = 4 m

We need to find the coefficient of friction (μ) between the cycle tires and the road.

First, let's rearrange the equation to solve for the force (F):

F = (mv^2) / r

Next, we need to recall Newton's second law of motion, which states that the force acting on an object is equal to the product of its mass and acceleration (F = ma).

In this case, the acceleration is the centripetal acceleration, given by a = v^2 / r. Substituting this into the rearranged equation gives:

F = m(v^2 / r)

Now, we can substitute the known values:

F = m(4.9^2 / 4)

Let's assume the mass of the cyclist is 70 kg:

F = 70(4.9^2 / 4)

Calculating:

F ≈ 70(24.01 / 4)
≈ 70(6.0025)
≈ 420.175 N

The centripetal force (F) required to keep the cyclist moving in a circular path is approximately 420.175 N.

To find the coefficient of friction (μ), we can use the equation relating frictional force (Ff) to the normal force (Fn) and the coefficient of friction (μ):

Ff = μFn

Since the cyclist is on a level road, the normal force (Fn) is equal to the weight of the cyclist, which is given by:

Fn = mg

where:
m is the mass of the cyclist,
g is the acceleration due to gravity (approximately 9.8 m/s²).

Let's assume the mass of the cyclist is 70 kg:

Fn = 70 × 9.8
= 686 N

Substituting this into the equation for frictional force:

Ff = μFn

420.175 = μ × 686

Now we can solve for the coefficient of friction (μ):

μ ≈ 420.175 / 686
≈ 0.613

Therefore, the coefficient of friction between the cycle tires and the road is approximately 0.613.