Books usually use conservation of energy to solve problems in vertical circular motion. But, principle of conservation of energy for one particle is,

If all the $n$ forces $\vec F_i (i=1,2,3,..,n)$ acting on a particle are conservative, each with its corresponding potential energy $U_i(\vec r)$, the total mechanical energy, defined as $$E=\frac 12 mv^2 + \Sigma_i U_i(\vec r)$$ is constant in time.

but why are we not considering the centripetal force ( Tension, Normal reaction whatever that may be)? I don't understand this. Please help. How is this centripetal force taken into account?

  • 2
    $\begingroup$ The centripetal force does no work because it always acts along the radius vector and the length of the radius vector never changes. $\endgroup$ – John Rennie Feb 5 '16 at 6:20
  • $\begingroup$ mathematically, sir, please ? $\endgroup$ – Subhranil Sinha Feb 5 '16 at 6:21
  • $\begingroup$ posted the idea I came up with. Please check if it is correct. $\endgroup$ – Subhranil Sinha Feb 5 '16 at 6:38
  • $\begingroup$ Yes, that looks fine $\endgroup$ – John Rennie Feb 5 '16 at 6:38

I came up with a solution seeing John Rennie's comment.

The centripetal force, $\vec F= -F \hat r $ so infinitesimal work done by centripetal force, $$dW=\vec F.d \vec r= -F \hat r.d\vec r$$ but, $\hat r⊥d \vec r$ so $$dW=0$$ the image

is this correct ?


You know that the centripetal force is given by $\vec F_z = m\omega^2r \, \vec e_r$ ,where $\vec e_r = \cos \theta \, \vec e_x + \sin \theta \, \vec e_y$ is the unit vector in radial direction. We want to calculate the work given by the line integral $$ \int_C \vec F_z \cdot \mathrm d \vec r $$

where the position of the point mass $\vec r = r\, \vec e_r$ is parametrized by the angle $\theta$

$$ \mathrm d \vec r = -r \sin \theta\, \mathrm d \theta \, \, \vec e_x +r \cos \theta\, \mathrm d \theta \, \, \vec e_y \implies \int_C \vec F_z \cdot \mathrm d \vec r = \int_{\theta = \theta_0}^{\theta_1} \vec F_z \cdot \frac{\mathrm d \vec r}{\mathrm d \theta} \, \mathrm d \theta $$

$$ \implies W = \int_{\theta = \theta_0}^{\theta_1} m \omega^2 r^2 \cdot ( \sin\theta \cos \theta - \cos \theta \sin \theta) \, \mathrm d \theta = 0 $$

Thus the zentripetal force doesn't do any work on the point mass.


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