-2
$\begingroup$

I am trying to calculate the force a body exerts on a vertical circular track with no friction at its highest point I know the centripetal force at that point is mg and the given data is the velocity the mass of the body the gravitational constant and the radius of the track what do I do now?

$\endgroup$
  • $\begingroup$ "I know the centripetal force at that point is mg" Are you sure that the only centripetal force is $mg$? $\endgroup$ – lucas May 29 '16 at 20:34
  • $\begingroup$ The formula for the centripetal force is F=(mv^2)/r and the speed given is the square root of g*r from that I arrived at mg $\endgroup$ – Lendion May 29 '16 at 20:38
  • $\begingroup$ Let me exapnd of @lucas's comment: it is not, in general, true that the centripetal acceleration at the top of the loop is $g$. That is: you can't take this as a given and have to prove it for any particular case you wish to use it in. $\endgroup$ – dmckee May 29 '16 at 21:51
  • $\begingroup$ It is a particular teoretical case where the speed at the highest point is given to be the square root of m*g $\endgroup$ – Lendion May 30 '16 at 5:58
1
$\begingroup$

If you know velocity, you also know acceleration.

If you know acceleration, you can set up Newton's second law and all is known except the push on the track. This push is missing in your statement about the centripetal force, as @lucas comments.

$\endgroup$
  • $\begingroup$ That is what I'm trying to calculate is it possible to find using the given data? $\endgroup$ – Lendion May 29 '16 at 20:44
  • $\begingroup$ What part are you confused about? The acceleration expression or Newton's law or which forces there are at that point? $\endgroup$ – Steeven May 29 '16 at 20:56
  • $\begingroup$ Newton's law F=ma and a=v^2/r so F=(m(v^2))/r then I know the speed at the highest point is the squer root of g*r so the centripetal force is equal to mg and it is in the same direction as mg how do I calculate the normal force? $\endgroup$ – Lendion May 29 '16 at 21:18
  • $\begingroup$ Newton's law is not exactly $F=ma$ but rather $\sum F=ma$. The $\sum F$ must include all forces in the vertical-direction. You mention gravity $w=mg$, yes, but as @lucas refers to in the comments above, this is not the only force in the vertical direction. $\endgroup$ – Steeven May 29 '16 at 21:49

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.