Math Problem Statement

Dudukan terdiri dari alas bermassa M yang dihubungkan ke tiang tak bermassa. Massa m dipasang ke ujung tiang melalui tali panjang tak bermassa l dan pada awalnya tidak bergerak pada tingkat horizontal yang sama dengan ujung kutub. Massa m kemudian dilepaskan. Tentukan koefisien minimum gesekan statis μ antara dudukan dan meja jika dudukan tidak terjemahkan sementara massa m mengalami gerakan melingkar di sekitar ujung tiang.

Solution

The image describes a physical scenario where a mass mm is attached to the end of a massless pole and connected to a platform with mass MM. The setup involves a frictionless table, and the question asks for the minimum static friction coefficient μ\mu between the platform and the table so that the platform remains stationary while the mass mm rotates around the pole.

Key elements to extract:

  1. System setup:

    • A mass mm is attached to a massless pole of length ll and is released from rest.
    • The platform has mass MM, and the goal is to prevent it from moving.
    • The mass mm rotates around the pole in a circular motion.
  2. Centripetal force:

    • When the mass mm moves in a circular path, the tension in the rope provides the centripetal force.
    • The formula for centripetal force is Fcentripetal=mv2lF_{\text{centripetal}} = \frac{m v^2}{l}, where vv is the tangential velocity of the mass mm, and ll is the length of the rope.
  3. Friction force on platform:

    • To prevent the platform from moving, the friction force between the platform (mass MM) and the table must balance the horizontal force exerted by the tension in the rope.
    • The friction force Ffriction=μMgF_{\text{friction}} = \mu M g, where μ\mu is the static friction coefficient, and gg is gravitational acceleration.
  4. Force equilibrium:

    • The tension in the rope causes a force on the platform, which is opposed by friction. Hence, T=FfrictionT = F_{\text{friction}}.

Solution Approach:

  • The tension in the rope can be expressed as T=mv2lT = \frac{m v^2}{l}.
  • The platform remains stationary if the friction force equals this tension: μMg=T\mu M g = T.
  • Substituting TT: μMg=mv2l\mu M g = \frac{m v^2}{l}
  • Rearranging for μ\mu: μ=mv2Mgl\mu = \frac{m v^2}{M g l}

However, since no specific velocity is given, the expression in this form represents the minimum static friction coefficient μ\mu that depends on how fast the mass mm is rotating.


Would you like to dive into more details on how we can compute the velocity vv, or do you have any other specific questions about this setup?

Here are five follow-up questions to expand on this concept:

  1. How would the coefficient of friction change if the mass MM of the platform increases?
  2. What role does the length of the rope ll play in determining the tension?
  3. Can the mass mm ever move fast enough that the platform moves, even with high friction?
  4. What assumptions are we making about the rotational motion of the mass mm?
  5. How can the velocity vv of the mass be calculated in this scenario?

Tip:

Always analyze the forces acting on each component of the system independently and identify which force provides the centripetal acceleration for circular motion.

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Math Problem Analysis

Mathematical Concepts

Circular Motion
Friction
Force Equilibrium

Formulas

Centripetal Force: F = (m * v^2) / l
Friction Force: F_friction = μ * M * g
Force Equilibrium: μ * M * g = (m * v^2) / l

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 10-12