To understand why the term mR² is replaced with mR in the context of a block rotating in a hemispherical bowl, we need to delve into the dynamics of the system and the forces acting on the block. Let's break this down step by step.

Understanding the Forces at Play

When the bowl rotates, the block experiences several forces: gravitational force, normal force from the bowl's surface, and frictional force. The gravitational force acts downward, while the normal force acts perpendicular to the surface of the bowl. The frictional force is what prevents the block from slipping as it rotates with the bowl.

Analyzing the Motion

As the bowl rotates, the block moves in a circular path. The centripetal force required to keep the block moving in this circular motion is provided by the horizontal component of the normal force and the frictional force. The key here is to analyze the forces in terms of their components.

• The gravitational force acting on the block is given by mg, where m is the mass of the block and g is the acceleration due to gravity.
• The normal force N acts perpendicular to the surface of the bowl.
• The frictional force f acts parallel to the surface, preventing slipping.

Components of Forces

When the block is at an angle θ with the vertical, we can resolve the gravitational force into two components:

• The component acting along the direction of the normal force: mg cos(θ)
• The component acting perpendicular to the normal force: mg sin(θ)

Applying Newton's Second Law

For the block to rotate without slipping, the frictional force must be sufficient to provide the necessary centripetal force. The centripetal force required for circular motion is given by:

F_c = mω²R,

where ω is the angular speed and R is the radius of the circular path at angle θ.

Frictional Force and Normal Force Relationship

The maximum frictional force that can act on the block is given by:

f_max = μN,

where μ is the coefficient of friction. The normal force can be expressed as:

N = mg cos(θ).

Setting Up the Equation

For the block to not slip, the centripetal force must be less than or equal to the maximum frictional force:

mω²R ≤ μmg cos(θ).

Why mR² Becomes mR

Now, when we analyze the effective radius of the circular motion, we consider the radius of the circular path at angle θ, which is R sin(θ). This means that the centripetal force can be rewritten in terms of the effective radius:

F_c = mω²(R sin(θ)).

However, when we rearrange the terms and simplify, we find that the angular speed ω is related to the radius R and the angle θ, leading to the simplification where mR² is effectively replaced with mR in the context of balancing forces.

Conclusion on Angular Speed

Thus, the range of angular speed for which the block will not slip can be derived from the inequality:

ω² ≤ (μg cos(θ)) / R.

This gives us the maximum allowable angular speed based on the frictional force and the geometry of the bowl. By understanding the relationship between the forces and the geometry, we can see how the terms simplify and why mR² transitions to mR in this context.