A small block placed on the surface of a horizontal disc at a point 6 cm from the centre of the disc .The block is on the point of slipping when the disc rotates at 1 1/2 rev/s .Find the coefficient of friction between the block and the surface of the disc.

Answer 1

To find the coefficient of friction between the block and the surface of the disc, we need to use the condition for slipping.

When the block is on the point of slipping, the maximum static friction force between the block and the surface of the disc is equal to the centripetal force acting on the block.

The maximum static friction force, Fmax, can be calculated as:

Fmax = μN

where μ is the coefficient of friction and N is the normal force.

The normal force, N, can be calculated as:

N = mg

where m is the mass of the block and g is the acceleration due to gravity.

The centripetal force, Fc, can be calculated as:

Fc = mrω²

where r is the distance of the block from the center of the disc, ω is the angular velocity of the disc, and m is the mass of the block.

We are given that the distance of the block from the center of the disc, r, is 6 cm. We also know that the angular velocity of the disc, ω, is 1 1/2 rev/s, or (3/2) × 2π rad/s.

Now, let's substitute these values into the equations to find the coefficient of friction, μ.

N = mg = m × 9.8 m/s²

Fc = mrω² = m × (6/100) m × (3/2) × 2π rad/s²

Since Fc = Fmax, we can equate these equations:

m × 9.8 m/s² = m × (6/100) m × (3/2) × 2π rad/s²

Simplifying the equation:

9.8 m/s² = (6/100) × (3/2) × 2π rad/s²

m cancels out from both sides, and we are left with:

9.8 = (6/100) × (3/2) × 2π rad/s²

Simplifying further:

9.8 = (6/100) × (3/2) × 2π rad/s²

9.8 = (6/100) × (3/2) × 2π rad/s²

9.8 = (18/100) × π rad/s²

Dividing both sides by π:

9.8/π = 18/100 rad/s²

Simplifying:

0.3125 = 18/100 rad/s²

Now, we can isolate the coefficient of friction, μ:

Fmax = μN

μ = Fmax/N

Substituting the values, we get:

μ = (9.8/π)/(m × 9.8 m/s²)

μ = 1/π

Therefore, the coefficient of friction between the block and the surface of the disc is 1/π, or approximately 0.318.