Math Problem Statement

A rough inverted cone rotates at a constant speed about its vertical axis slowly enough so a block inside does not slide down. Given the block’s free-body diagram, which of the following components contribute to the block’s centripetal acceleration?

Solution

To solve this problem, we need to determine which components contribute to the block's centripetal acceleration in the rotating inverted cone.

Analyzing the diagram:

  • The forces acting on the block are:
    • ww is the gravitational force (weight).
    • nn is the normal force from the cone's surface.
    • ff is the frictional force preventing the block from sliding down.

The block undergoes centripetal acceleration due to the rotation of the cone, and this acceleration points toward the axis of rotation. For centripetal acceleration, we are concerned with the horizontal components of the forces, as they contribute to the inward radial force.

Components of Forces:

  1. Normal Force (nn): The normal force nn is directed perpendicular to the cone's surface. We can resolve it into two components:

    • One along the horizontal (radial) direction, which is nsinθn \sin \theta.
    • One along the vertical direction, which is ncosθn \cos \theta.
  2. Friction Force (ff): The frictional force ff is along the surface of the cone and also has two components:

    • One along the horizontal (radial) direction, which is fcosθf \cos \theta.
    • One along the vertical direction, which is fsinθf \sin \theta.

Since centripetal acceleration is directed horizontally, the components of the forces that contribute to it will be the horizontal components:

  • The horizontal component of the normal force is nsinθn \sin \theta.
  • The horizontal component of the friction force is fcosθf \cos \theta.

Answer:

Thus, the total force contributing to the centripetal acceleration is the difference of these two components: nsinθfcosθ.n \sin \theta - f \cos \theta.

This matches the option: nsinθfcosθ.\boxed{n \sin \theta - f \cos \theta}.


Do you want further details or clarifications on any part of this solution?

Here are some related questions you might want to explore:

  1. How do we derive the expression for centripetal force in a rotating system?
  2. What role does friction play in preventing the block from sliding?
  3. How does the angle θ\theta affect the magnitude of the centripetal force?
  4. Can the block slide down if the rotation speed is increased?
  5. What happens if the surface of the cone is smooth (i.e., no friction)?

Tip: When analyzing forces in rotating systems, always separate the components of forces into radial (toward the center of rotation) and vertical (parallel to gravity) directions.

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Math Problem Analysis

Mathematical Concepts

Centripetal Force
Vector Components
Trigonometry

Formulas

n sin(θ) - f cos(θ)
Centripetal Force = m * v^2 / r

Theorems

Newton's Second Law
Frictional Force
Normal Force

Suitable Grade Level

Grade 10-12