Physics 2008 JAMB Past Questions - Question 13: A motorcycle of mass 100kg moves round in a circle of radius 10m with a velocity of 5ms.Find the coefficient of friction between the road and the tyres .

Correct Answer

D

Explanation

To find the coefficient of friction between the road and the tires of the motorcycle as it moves in a circular path, we can use the following principles:

1. Centripetal Force: The motorcycle is moving in a circle, which means there must be a centripetal force acting on it to keep it in that circular path. The centripetal force required for an object moving in a circle is given by the formula:

  F_c = (m * v^2) / r

  Where:
  - F_c is the centripetal force.
  - m is the mass of the motorcycle (100 kg).
  - v is the velocity of the motorcycle (5 m/s).
  - r is the radius of the circle (10 m).

2. Frictional Force: The centripetal force required is provided by the frictional force between the tires and the road. This frictional force can be calculated using the formula:

  F_friction = μ * N

  Where:
  - F_friction is the frictional force.
  - μ (mu) is the coefficient of friction.
  - N is the normal force, which is equal to the weight of the motorcycle (mg) since it's on a horizontal surface.

Now, let's calculate the normal force (N) and the coefficient of friction (μ):

1. Calculate the weight of the motorcycle:
  Weight (W) = m * g
  where g is the acceleration due to gravity (approximately 9.81 m/s²).

  W = 100 kg * 9.81 m/s² = 981 N

2. Calculate the centripetal force:
  F_c = (100 kg * (5 m/s)^2) / 10 m = 250 N

3. Equate the centripetal force to the frictional force:
  F_c = F_friction

4. Now, we can calculate the coefficient of friction (μ):
  250 N = μ * 981 N

  μ = 250 N / 981 N ≈ 0.255

So, the coefficient of friction between the road and the tires of the motorcycle is approximately 0.255.