Question

A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/srev/s . The wheel can be considered a uniform disk of mass 4.5 kgkg and diameter 0.30 mm . The potter then throws a 2.8-kgkg chunk of clay, approximately shaped as a flat disk of radius 8.0 cmcm , onto the center of the rotating wheel. Part A What is the frequency of the wheel after the clay sticks to it

Answer 1

Answer:1.7 rev/s

Step-by-step explanation:

Given

Frequency of wheel
`N_1=2\ rev/s`

angular speed
`\omega_1=2\pi N_1=4\pi\ rad/s`

mass of wheel
`m_1=4.5\ kg`

diameter of wheel
`d_1=0.30\ m=30\ cm`

radius of wheel
`r_1=(d_1)/(2)=(30)/(2)=15\ cm`

mass of clay
`m_2=2.8\ kg`

the radius of the chunk of clay
`r_2=8\ cm`

Moment of inertia of Wheel

`I_1=(m_1r_1^2)/(2)=(4.5* 15^2)/(2)\ kg-cm^2`

Combined moment of inertia of wheel and clay chunk

`I_2=(m_1r_1^2)/(2)+(m_2r_2^2)/(2)=(4.5* 15^2)/(2)+(2.8* 8^2)/(2)\ kg-cm^2`

Conserving angular momentum

`\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow (4.5* 15^2)/(2)\cdot 4\pi=((4.5* 15^2)/(2)+(2.8* 8^2)/(2))\omega_2\\\\\Rightarrow \omega _2=(4\pi )/(1+(2.8)/(4.5)* ((8)/(15))^2)=(4\pi)/(1+0.1769)=0.849* 4\pi`

Common frequency of wheel and chunk of clay is

`\Rightarrow N_2=(4\pi * 0.849)/(2\pi)=1.698\approx 1.7\ rev/s`