Sunday
March 26, 2017

Post a New Question

Posted by on .

A 5.0 gram coin is placed 5.0 cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a rate of 36 RPM is reached and the coin slides off.

A.) What is the coefficient of static friction between the coin and the turntable?

B.)What is the acceleration at that moment?

C.) What is the centripetal force then?

  • College Physics - ,

    At 36 rpm, the angular velocity is
    w = 36*(2 pi)/60 = 6pi/5 = 3.770 radians/sec.
    A) mus*M g = M r w^2
    mus = rw^2/g = (0.05)(3.77)^2/9.81 = 0.072

    B)Before it starts slipping, the acceleration is
    rw^2 = (0.05)(3.77)^2 = 0.71 m/s^2

    After it starts slipping, it depends upon the kinetic coefficient of friction. More information is needed to get that.

    c)centripetal force = M r w^2

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question