2
$\begingroup$

Suppose we have on a horizontal timetable

$m\textbf{a}_{rot}=\textbf{F}-\textbf{F}_{corialis}-\textbf{F}_{centrifugal}$

where $\textbf{a}_{rot}$ is the acceleration in a rotating frame. Suppose an ant on the turntable walks from the centre at constant speed. Suppose an ant slips an experiences a horizontal force exceeding a value $F$.

Explain why $F^2=|\textbf{F}_{corialis}|^2+|\textbf{F}_{centrifugal}|^2$

I have no idea why the above holds. May someone elaborate

$\endgroup$
1
  • $\begingroup$ This is a gentle reminder to consider accepting an answer if your question has been resolved. $\endgroup$ – Sal Feb 27 at 13:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.