Question

A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/s. The wheel can be considered a uniform disk of mass 4.9 kg and diameter 0.40m. The potter then throws a 2.7 kg chucnk of clay, approximately shaped as a flat disk of radius 8.0 cm, onto the center of the rotating wheel.
What is the frequency of the wheel after the clay sticks to it? Ignore friction/.

Answer 1

Answer:

$\omega_f=12.2954 \,rad.s^{-1}$         i.e.

$Frequency=1.9569\,rev.s^{-1}$

Explanation:

Given:

angular speed,$\omega_i =2\times 2\pi \,rad.s^{-1}$

mass of the disk, $M=4.9\,kg$

radius of the disk, $R=0.2\,m$

mass of the chunk, $m=2.7\,kg$

radius of the chunk, $r=0.04\,m$

We know that the angular momentum is given by:

$L=I.\omega$

and moment of inertia for a disc:

$I=\frac{1}{2} m.r^2$

According to the conservation of angular momentum, the final angular momentum is equal to the initial angular momentum.

$\frac{1}{2} \times M.R^2\times \omega= \frac{1}{2} \times (M.R^2+m.r^2) \omega_f$

$4.9\times 0.2^2\times (2\times 2\pi)=(4.9\times 0.2^2+2.7\times 0.04^2 )\tiems \omega_f$

$\omega_f=12.2954 \,rad.s^{-1}$

$Frequency=1.9569\,rev.s^{-1}$