Minimum speed at top of vertical loop for ball to complete the loop

Question

A ball moves in a vertical circular loop of radius $r$ . Find the minimum speed at the top of the loop for the ball to maintain contact with the track throughout the loop.

Solution — Step by Step

At the top of the vertical circle, both gravity ( $mg$ , downward) and the normal force from the track ( $N$ , downward for inner surface) provide the centripetal force directed toward the centre (downward):

N + mg = \frac{mv^2}{r}

The ball maintains contact with the track as long as $N \geq 0$ .

The minimum speed at the top corresponds to the limiting case where contact is just barely maintained: $N = 0$ .

Setting $N = 0$ :

0 + mg = \frac{mv_{\min}^2}{r}

mg = \frac{mv_{\min}^2}{r}

v_{\min}^2 = gr

\boxed{v_{\min} = \sqrt{gr}}

Using conservation of energy between the bottom (speed $v_b$ ) and top (speed $v_{\min}$ ), with height difference = $2r$ :

\frac{1}{2}mv_b^2 = \frac{1}{2}mv_{\min}^2 + mg(2r)

v_b^2 = gr + 4gr = 5gr

\boxed{v_b = \sqrt{5gr}}

The ball needs a minimum speed of $\sqrt{5gr}$ at the bottom of the loop to complete the full circle.

At the minimum speed, gravity alone provides the centripetal force at the top — the track exerts zero normal force. If the speed is any less, gravity exceeds the centripetal requirement and the ball falls away from the track before completing the loop.

Why This Works

The critical point for a ball in a vertical loop is the top, not the bottom. At the top, both gravity and normal force point toward the centre (inward). As speed decreases, we need less centripetal force, but we can’t go below $N = 0$ — the track can only push, not pull.

At the bottom, both $N$ (upward) and $mg$ (downward) are present, but they act in opposite directions. The normal force here exceeds $mg$ even at minimum loop speed: $N = mg + \frac{mv^2}{r} = 6mg$ (at $v = \sqrt{5gr}$ ).

Common Mistake

Students sometimes set $N = 0$ at the bottom of the loop, not the top. This is wrong — at the bottom, the normal force is always positive (the track surface pushes upward). The critical constraint is at the top where the normal force could potentially go to zero. Setting $N = 0$ at the bottom would give zero-speed, which is nonsensical.

The ratio of speeds: $v_{\text{top}} : v_{\text{bottom}} = \sqrt{gr} : \sqrt{5gr} = 1 : \sqrt{5}$ . The three key speeds to remember: at top = $\sqrt{gr}$ , at bottom = $\sqrt{5gr}$ , at side (height = r) = $\sqrt{3gr}$ .

Question

Solution — Step by Step

Why This Works

Common Mistake

Try These Next