6.3.2 Loop-the-loop

A ball of mass m is tied to a string, and swung in a vertical circle of radius r.

PCOE

Because of gravity, the speed of the ball decreases on the way up, and increases on the way down. In fact, assuming negligible losses, the speed at the top and bottom of the circle are related by PCOE:

\displaystyle \begin{aligned}(KE+GPE)\text{ at the top}&=(KE+GPE)\text{ at the bottom}\\\frac{1}{2}m{{v}_{t}}^{2}+mg(2r)&=\frac{1}{2}m{{v}_{b}}^{2}+0\quad \cdots (1)\end{aligned}

Required Centripetal Force

This would also mean that the (magnitude of the) required centripetal force is also not constant. It is largest at the bottom \displaystyle (m\frac{{{{v}_{b}}^{2}}}{r}) and smallest at the top \displaystyle (m\frac{{{{v}_{t}}^{2}}}{r}).

Maximum and Minimum Tension

Let’s now apply N2L to figure out the tension T in the string.

At the bottom, the centripetal direction is up. So

\displaystyle \begin{aligned}({{F}_{{net}}}&=ma)\\T-mg&=m\frac{{{{v}_{b}}^{2}}}{r}\\T&=m\frac{{{{v}_{b}}^{2}}}{r}+mg\quad \cdots (2)\end{aligned}

At the top, the centripetal direction is down. So

\displaystyle \begin{aligned}({{F}_{{net}}}&=ma)\\T+mg&=m\frac{{{{v}_{t}}^{2}}}{r}\\T&=m\frac{{{{v}_{t}}^{2}}}{r}-mg\quad \cdots (3)\end{aligned}

The maximum value of tension occurs at the bottom because (1) the required centripetal force \displaystyle ({{F}_{c}}=m\frac{{{{v}_{b}}^{2}}}{r})is largest at this position and (2) mg is completely opposite to the centripetal direction.

Likewise, the minimum value of tension occurs at the top because (1) the required centripetal force \displaystyle ({{F}_{c}}=m\frac{{{{v}_{t}}^{2}}}{r})is smallest at this position and (2) mg is completely in the centripetal direction.

Critical speed

From equation (3), it looks like if vt is too small, T must become negative. This means that mathematically speaking, the string must provide an upward tensional force on the ball, if it were to keep the ball in circular motion. Physically, this is impossible since the string can only pull inward. It is not possible for the string to push the ball upward when the ball is at the top.

This suggests that there is a critical speed for the ball when it is at the top, below which the ball will not be able to complete the circular motion. This critical speed Vt (at the top) can be derived by putting T=0 in equation (3).

\displaystyle \begin{aligned}T&=m\frac{{{{V}_{t}}^{2}}}{r}-mg\\0&=m\frac{{{{V}_{t}}^{2}}}{r}-mg\\{{V}_{t}}&=\sqrt{{rg}}\end{aligned}

Remember that the ball slows down on its way up because of gravity. To derive Vb, the minimum speed the ball must start from (at the bottom) in order to complete the loop, we go back to PCOE by using equation (1).

\displaystyle \begin{aligned}\frac{1}{2}m{{V}_{t}}^{2}+mg(2r)&=\frac{1}{2}m{{V}_{b}}^{2}+0\\\frac{1}{2}(rg)+2rg&=\frac{1}{2}{{V}_{b}}^{2}\\{{V}_{b}}&=\sqrt{{5rg}}\end{aligned}

Why there is a critical speed

Basically, Vt is the speed at which the required centripetal force is equal to mg. Any speed higher than this means that mg alone is insufficient to keep the ball in circular motion. The ball’s inertia (and gravity) would have carried it along a parabolic path (drawn in blue) that cuts outside the circular motion (shown by the dashed line). But the ball is tied to the string which has a fixed length. So string would naturally step in to provide the amount of tension required to keep the ball in circular motion.

On the other hand, if the ball is travelling at a speed lower than Vt, its inertia (and gravity) would have carried it along a parabolic path (drawn in red) that cuts inside the circular motion (shown by the dashed line). The string may be of fixed length but it is not rigid. It simply goes limp and allows the ball to do the projectile motion.

Video Explanation

Ball Swung in Vertical Circle (with demo)

Water Pail in Vertical Circle (with demo)

Roller Coaster (with demo)

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