Round and about

A small bead of mass M M is given an initial velocity of magnitude 3 m/s 3 \text{ m/s} on a horizontal circular track of radius 1 m 1 \text{ m} . If the coefficient of of kinetic friction is μ k = 0.5 \mu_{k}=0.5 , what is the distance covered by the bead before coming to rest?

Assumptions and Details

  • Take acceleration due to gravity as 9.8 m/s 2 9.8 \text{ m/s}^{2} .
  • The track goes through a small hole in the bead. The surface of the hole is uniform so that the coefficient of friction is the same in all directions.


The answer is 0.822457.

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1 solution

Abhishek Sinha
Sep 8, 2015

The working differential equation is m v d v d s = μ k ( m g ) 2 + ( m v 2 R ) 2 mv \frac{dv}{ds}= -\mu_k \sqrt{(mg)^2+(\frac{mv^2}{R})^2} which can be directly solved to get the answer.

I used circular motion and got a trigonometric equation in some and cosine which was much simpler than solving differential equations

Anubhav Tyagi - 5 years, 3 months ago

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