Q6.4 Watermelon Seed

A record turntable is rotating with a constant angular velocity of 15 rad s-1.

A watermelon seed is placed on the turntable 6cm from the axis of rotation.

  1. Calculate the magnitude of the acceleration of the seed assuming that it does not slip.
  2. What is the minimum value of the coefficient of static friction between the seed and the turntable if the seed is not to slip.
Hide

Hint

Identify all the forces acting on the seed.
 

Solution

Reveal
Hide

Solution

Graphic - WaterMelonSeedFigure
  1. Reaction force is given by Newton’s second law

    FN=mg

    Newton’s second in the horizontal direction gives

    \[ F_F = ma = \frac{mv^2}{r} = mr\omega^2 \]

    Therefore acceleration is given by

    a=rω2

    Plugging in the numbers

    a=0.06×(15)2=13.5ms-2

  2. At the point of slipping the frictional force is given by

    FF=μSFR

    Therefore

    \[ \mu_Smg = mr\omega^2 \hspace{0.5cm}\mbox{\rm so}\hspace{0.5cm} \mu_S = \frac{r\omega^2}{g} \]

    \[ \mu_S = \frac{0.06 \times (15 )^2}{9.81} = 1.4 \hspace{0.5cm}\mbox{\rm to 2sf}\hspace{0.5cm} \]