Minimum Speed at the Top of a Vertical Loop

At the top of a vertical circle, gravity itself can provide the centripetal force. If the ball is moving fast enough, it needs more than just gravity, so the string also pulls inward (downward at the top) to make up the difference. If it's moving exactly at the minimum speed, gravity alone is enough — and the string tension drops to zero.

If the ball is moving any slower than the minimum, gravity provides more centripetal force than needed, and the ball "falls inward" — the string goes slack and the ball departs from the circular path.

The Minimum-Speed Condition

At the top of the loop, with the string just barely taut (tension $T = 0$):

$\dfrac{m\,v_{\min}^{2}}{r} = mg$

Solving for $v_{\min}$:

$v_{\min} = \sqrt{g\,r}$

Notice: mass cancels — the minimum speed depends only on the loop radius and gravity, not on how heavy the ball is.

Step-by-Step Solution

Given: $r = 1\;\text{m}$, $g = 9.81\;\text{m/s}^{2}$.

Find: The minimum speed at the top of the loop.

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Step 1 — Apply Newton's Second Law at the top.

At the very top, both gravity and any string tension point downward (toward the center of the loop). Setting the sum equal to the centripetal force:

$mg + T = \dfrac{mv^{2}}{r}$

Step 2 — Set $T = 0$ for the minimum-speed case.

$mg = \dfrac{mv_{\min}^{2}}{r}$

Step 3 — Cancel mass and solve.

$g = \dfrac{v_{\min}^{2}}{r}$

$v_{\min}^{2} = gr$

$v_{\min} = \sqrt{gr} = \sqrt{(9.81)(1)} = \sqrt{9.81} \approx 3.132\;\text{m/s}$

Step 4 — Compute the corresponding angular velocity.

$\omega_{\min} = \dfrac{v_{\min}}{r} = \dfrac{3.132}{1} = 3.132\;\text{rad/s}$

Step 5 — Find the bottom-of-loop speed using energy conservation.

The ball must descend from the top (height $2r$ above the bottom) while losing PE and gaining KE:

$\tfrac{1}{2}mv_{\text{bot}}^{2} = \tfrac{1}{2}mv_{\min}^{2} + mg(2r)$

$v_{\text{bot}}^{2} = v_{\min}^{2} + 4gr = gr + 4gr = 5gr$

$v_{\text{bot}} = \sqrt{5gr} = \sqrt{5 \times 9.81 \times 1} = \sqrt{49.05} \approx 7.004\;\text{m/s}$

So at the bottom of the loop, the ball must be moving at least $\sqrt{5gr} \approx 7.0\;\text{m/s}$ to make it around without going slack at the top.

Step 6 — String tension at the bottom of the loop.

At the bottom, the centripetal force points upward (toward the center). Tension is up, gravity is down:

$T_{\text{bot}} - mg = \dfrac{mv_{\text{bot}}^{2}}{r} = m \cdot 5g$

$T_{\text{bot}} = mg + 5mg = 6mg$

The tension at the bottom is six times the weight of the ball — that's the famous result for a minimum-speed loop. A 1 kg ball would have $T = 58.86\;\text{N}$ at the bottom.

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Answer:

$\boxed{\;v_{\min}^{\text{top}} = \sqrt{gr} \approx 3.132\;\text{m/s}\;}$

To make it around a 1-meter vertical loop without the string going slack, the ball needs at least 3.13 m/s at the top — equivalent to 7.00 m/s at the bottom, with the string tension reaching $6mg$ there.

Try It

• Adjust the loop radius and the bottom speed with the sliders.
• If the bottom speed is too low, the ball fails to make it over the top — the visualization shows the trajectory departing the circle.
• The HUD shows whether the current speed is sufficient.
• Bigger loops require more speed: $v_{\min} \propto \sqrt{r}$.