Question

A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 8.3 cm from the axis of rotation. (a) Calculate the acceleration of the seed, assuming that it does not slip. (b) What is the minimum value of the coefficient of static friction between the seed and the turntable if the seed is not to slip? (c) Suppose that the turntable achieves its angular speed by starting from rest and undergoing a constant angular acceleration for 0.17 s. Calculate the minimum coefficient of static friction required for the seed not to slip during the acceleration period

Answer 1

w = 33*2pi/60

=3.455 rad/s

a)acceleration = w^2*r

=3.455^2*0.083

=0.99 m/s^2

b)let it be x.so,

frictional force=xmg

so,

balancing forces,

frictional force = centrifugal force

xmg = mw^2r

or xg=w^2r

or x*9.81 = 0.99

or x=0.1

c)angular acceleration = 3.455/0.17

=20.32 rad/s^2

let it be x.

at the 0.17th second, the total acceleration will be maximum,

Atangential=(20.32*0.083)

=1.687 m/s^2

so,

Atotal = (1.687^2 + 0.99^2)^0.5

=1.956 m/s^2

so,

xmg = ma

or x*9.81=1.956

or x=0.1999

or x~0.2