Two blocks of mass m1 = 10 Kg and m2 = 5 Kg connected to each other by a massless inextensible string of length 0.3 m are placed along a diameter of turn table. The coefficient of friction between the table and m1 is 0.5 while there is no friction between m2 and the table. The table is rotating with an angular velocity of 10 rad/s about a vertical axis passing through its centre O. The masses are placed along the diameter of the table on either side of the centre O such that the mass m1 is at a distant of 0.124 m from O. The masses are observed to be at rest with respect to an observer on the turn table.(i) Calculate the frictional force on m1.(ii) What should be the minimum angular speed of the turn tale so that the masses will slip from this position.(iii) How should the masses be placed with the string remaining taut so that there is no frictional force acting on the mass m1.

To tackle this problem, we need to analyze the forces acting on the two masses, m1 and m2, as they are placed on a rotating turntable. The key concepts involved here include centripetal force, frictional force, and the conditions for slipping. Let's break this down step by step.

Understanding the Forces at Play

When the turntable rotates, the mass m1 experiences a centripetal force that keeps it moving in a circular path. This force is provided by the frictional force between m1 and the table. The mass m2, being on the other side of the string, does not experience any friction since it is not in contact with the table.

Calculating the Frictional Force on m1

The centripetal force required to keep m1 moving in a circle is given by the formula:

F_c = m1 * ω² * r

Where:

• m1 = mass of block 1 = 10 kg
• ω = angular velocity = 10 rad/s
• r = distance from the center O to m1 = 0.124 m

Substituting the values:

F_c = 10 kg * (10 rad/s)² * 0.124 m

F_c = 10 * 100 * 0.124 = 124 N

This centripetal force must be provided by the frictional force, which can be calculated using the formula:

F_friction = μ * N

Where:

• μ = coefficient of friction = 0.5
• N = normal force = m1 * g = 10 kg * 9.81 m/s² = 98.1 N

Now, calculating the frictional force:

F_friction = 0.5 * 98.1 N = 49.05 N

Since the required centripetal force (124 N) is greater than the maximum frictional force (49.05 N), the frictional force acting on m1 is limited to its maximum value:

F_friction = 49.05 N

Determining Minimum Angular Speed for Slipping

To find the minimum angular speed at which m1 will start to slip, we set the required centripetal force equal to the maximum frictional force:

m1 * ω² * r = μ * N

Substituting the known values:

10 kg * ω² * 0.124 m = 49.05 N

Solving for ω²:

ω² = 49.05 N / (10 kg * 0.124 m)

ω² = 49.05 / 1.24 = 39.6

Thus, taking the square root:

ω = √39.6 ≈ 6.29 rad/s

This means that if the angular speed exceeds approximately 6.29 rad/s, m1 will start to slip.

Arranging the Masses to Avoid Frictional Force

To ensure that there is no frictional force acting on m1, we need to adjust the position of m1 and m2 such that the tension in the string provides the necessary centripetal force without relying on friction. This can be achieved by placing m1 at the edge of the turntable, where the radius is maximized.

Let’s denote the distance from the center to m1 as r1 and the distance to m2 as r2. The total length of the string is 0.3 m, so:

r1 + r2 = 0.3 m

To eliminate friction, we want the tension in the string to equal the required centripetal force:

T = m1 * ω² * r1

Assuming m2 is placed at the maximum distance (0.3 m) from the center, we can set r2 = 0.3 m and r1 = 0.0 m. However, this is impractical since m1 cannot be at the center. Instead, we can find a balance:

Let’s say we place m1 at 0.15 m from the center, then m2 must be at 0.15 m as well:

0.15 + 0.15 = 0.3 m

In this configuration, the tension in the string will provide the necessary centripetal force for m1 without any frictional force acting on it.

In summary, we calculated the frictional force acting on m1, determined the minimum angular speed for slipping, and discussed how to arrange the masses to avoid friction. Each of these steps is crucial for understanding the dynamics of rotating systems.