Answer :
The acceleration experienced by the particle is given by
[tex]a=113000 g=113000 \cdot 9.81 m/s^2[/tex]
This corresponds to the centripetal acceleration of the motion, which is related to the angular speed [tex]\omega[/tex] of the particle and its distance r from the axis by the relationship
[tex]a= \omega ^2 r [/tex]
In our problem, [tex]r=6 cm=0.06 m[/tex], so we can solve for [tex]\omega[/tex]:
[tex]\omega = \sqrt{ \frac{a}{r} } = \sqrt{ \frac{113000 \cdot 9.81 m/s^2}{0.06 m} }=4298 rad/s [/tex]
However, we must convert it into rpm (revolution per minute).
We know that 1 rad corresponds to [tex]( \frac{1}{2 \pi} )[/tex] revolutions, while [tex]1 s = \frac{1}{60} min[/tex]. So we the conversion is[tex]\omega = 4298 rad/s \cdot ( \frac{1}{2\pi} rev/rad )( 60 s/min)=41067 rpm[/tex]
[tex]a=113000 g=113000 \cdot 9.81 m/s^2[/tex]
This corresponds to the centripetal acceleration of the motion, which is related to the angular speed [tex]\omega[/tex] of the particle and its distance r from the axis by the relationship
[tex]a= \omega ^2 r [/tex]
In our problem, [tex]r=6 cm=0.06 m[/tex], so we can solve for [tex]\omega[/tex]:
[tex]\omega = \sqrt{ \frac{a}{r} } = \sqrt{ \frac{113000 \cdot 9.81 m/s^2}{0.06 m} }=4298 rad/s [/tex]
However, we must convert it into rpm (revolution per minute).
We know that 1 rad corresponds to [tex]( \frac{1}{2 \pi} )[/tex] revolutions, while [tex]1 s = \frac{1}{60} min[/tex]. So we the conversion is[tex]\omega = 4298 rad/s \cdot ( \frac{1}{2\pi} rev/rad )( 60 s/min)=41067 rpm[/tex]