two blocks of mass m1 & m2 are connected to each other bya massiess inextensible string of length 0.3m are placed along a diameter of turn table.the coefficient of friction b/w table &m1 is 0.5 while there is no friction between m2 &table.the table rotating with angular velocity of 10rad/sec.about vertical axis passing through its center.the masses are placed along diameter of table on either side of center o such that m1 is at a distance of 0.124m from ground.the masses are observed to be at rest as observed from turn table.m1=10kg & m2=5kg. 1)calculate friction force on m1?

2)what shoud be minimum angular speed of table so that masses will slip?

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. Let's break this down step by step to find the friction force on m1 and the minimum angular speed required for the masses to slip.

Understanding the Forces on m1

First, let's focus on mass m1, which is placed on the turntable with a coefficient of friction (μ) of 0.5. The gravitational force acting on m1 can be calculated using the formula:

• Weight of m1 (W1): W1 = m1 * g

Here, g is the acceleration due to gravity, approximately 9.81 m/s². For m1 = 10 kg:

• W1 = 10 kg * 9.81 m/s² = 98.1 N

Calculating the Normal Force

The normal force (N) acting on m1 is equal to its weight since it is resting on a horizontal surface:

• N = W1 = 98.1 N

Friction Force on m1

The maximum static friction force (F_friction) can be calculated using the formula:

• F_friction = μ * N

Substituting the values:

• F_friction = 0.5 * 98.1 N = 49.05 N

This is the maximum friction force that can act on m1 before it starts to slip. Since the problem states that the masses are at rest relative to the turntable, the actual friction force acting on m1 is equal to the centripetal force required to keep it moving in a circular path.

Finding the Centripetal Force

The centripetal force (F_c) required to keep m1 moving in a circle can be calculated using the formula:

• F_c = m1 * ω² * r

Where:

• ω = angular velocity = 10 rad/s
• r = distance from the center to m1 = 0.124 m

Substituting the values:

• F_c = 10 kg * (10 rad/s)² * 0.124 m = 124 N

Since the centripetal force (124 N) exceeds the maximum friction force (49.05 N), m1 would indeed start to slip if the angular velocity remains at 10 rad/s.

Determining Minimum Angular Speed for Slipping

To find the minimum angular speed (ω_min) at which m1 will start to slip, we set the centripetal force equal to the maximum friction force:

• F_friction = m1 * ω_min² * r

Rearranging this gives us:

• ω_min² = F_friction / (m1 * r)

Substituting the known values:

• ω_min² = 49.05 N / (10 kg * 0.124 m)
• ω_min² = 49.05 N / 1.24 kg = 39.6 (rad/s)²

Taking the square root to find ω_min:

• ω_min = √39.6 ≈ 6.29 rad/s

Summary of Results

1) The friction force acting on m1 is 49.05 N.

2) The minimum angular speed of the table required for the masses to start slipping is approximately 6.29 rad/s.

This analysis shows how the forces interact on a rotating system and how friction plays a crucial role in maintaining equilibrium. If you have any further questions or need clarification on any part, feel free to ask!