# Conservation of energy in vertical circular motion

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?

• The centripetal force does no work because it always acts along the radius vector and the length of the radius vector never changes. – John Rennie Feb 5 '16 at 6:20
• mathematically, sir, please ? – Subhranil Sinha Feb 5 '16 at 6:21
• posted the idea I came up with. Please check if it is correct. – Subhranil Sinha Feb 5 '16 at 6:38
• Yes, that looks fine – 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 $$dW=0$$

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.