Question

A 2.44 gram penny is on a turn table. It starts from rest at a radius of 0.136 meters and the turn table speeds up until the penny flies off with a tangential speed of 0.574 m/s.a. What coefficient of static friction exists between the penny and turntable if the penny doesn't fly off until it reaches this speed? b. Suppose a 5.7-gram quarter has the same coefficient of static friction between it and the turntable as does the penny? In what way will the speed at which it les off differ from that of the penny?

Answer 1

Answer:

a)  μ = 0.247, b) 0.574 m/s

Explanation:

a) Let's use Newton's second law

      F = m a

As the table is spinning the acceleration is centripetal

      a = v² / r

The force that holds the penny is the force of rubbing

      F = fr

      fr = μ N

On the vertical axis

      N-W = 0

     N = mg

Let' replace

     μ mg = m v² / r

  μi g = v² / r

    Mou = v² / gr

Let's calculate

     μ = 0.574² / (9.8 0.136)

     μ = 0.247

b) for this new mass

     fr = m a

     μ mg = m v² / r

     μ g= v² / r

We can see that the speed does not depend on the mass, so the speed of the coin will be the same