Exercises — Vertical circular motion — minimum speed conditions

Minimum at the top is when tension $T=0$ and gravity alone gives the centripetal pull: $mg=\frac{m{v}_{\text{top}}^2}{r}\Rightarrow v_{\text{top}}=\sqrt{gr}.$ $v_{\text{top}}=\sqrt{9.8\times 2.0}=\sqrt{19.6}=4.43\text{m/s}.$

$v_{\text{bot}}=\sqrt{5gr}=\sqrt{5\times 9.8\times 2.0}=\sqrt{98}=9.90\text{m/s}.$

For the minimum-speed loop the bottom tension is $6mg$: $T_{\text{bot}}=6mg=6\times 0.50\times 9.8=29.4\text{N}.$

(a) $v_{\text{top}}=\sqrt{gr}=\sqrt{9.8\times 1.2}=\sqrt{11.76}=3.43\text{m/s}.$ (b) Using the estimate $T=\frac{2\pi r}{v}=\frac{2\pi (1.2)}{3.43}=\frac{7.540}{3.43}=2.20\text{s}.$ (The real period is shorter because the ball is faster at the bottom — this is only a top-speed estimate.)

(a) $v_{\text{bot}}=\sqrt{5gr}=\sqrt{5\times 9.8\times 6.0}=\sqrt{294}=17.15\text{m/s}.$ (b) At the side (height $r$): $v_{\text{side}}=\sqrt{3gr}=\sqrt{3\times 9.8\times 6.0}=\sqrt{176.4}=13.28\text{m/s}.$

At the top both gravity and tension point down (toward center), so: $T+mg=\frac{m{v}^2}{r}\Rightarrow T=\frac{m{v}^2}{r}-mg.$ $T=\frac{0.30\times 6.0^2}{0.80}-0.30\times 9.8=\frac{10.8}{0.80}-2.94=13.5-2.94=10.56\text{N}.$

Speed at top by energy conservation, height gained $=2r=10\text{m}$: $v_{\text{top}}^2=v_{\text{bot}}^2-4gr=18^2-4(9.8)(5.0)=324-196=128,$ $v_{\text{top}}=\sqrt{128}=11.31\text{m/s}.$ Normal force at top (both $N$ and $mg$ point down toward center): $N+mg=\frac{m{v}_{\text{top}}^2}{r}\Rightarrow N=\frac{m{v}_{\text{top}}^2}{r}-mg.$ $N=\frac{600\times 128}{5.0}-600\times 9.8=15360-5880=9480\text{N}.$

Force equations: $T_{\text{bot}}-mg=\frac{m{v}_{\text{bot}}^2}{r},T_{\text{top}}+mg=\frac{m{v}_{\text{top}}^2}{r}.$ Subtract: $T_{\text{bot}}-T_{\text{top}}-2mg=\frac{m}{r}\left(v_{\text{bot}}^2-v_{\text{top}}^2\right).$ Energy conservation over height $2r$: $v_{\text{bot}}^2-v_{\text{top}}^2=4gr$. Substitute: $T_{\text{bot}}-T_{\text{top}}-2mg=\frac{m}{r}(4gr)=4mg\Rightarrow T_{\text{bot}}-T_{\text{top}}=6mg.$ This holds for every speed, not just the minimum — the $v$ dependence cancelled.

To complete the loop it needs $v_{\text{bot}}^2=5gr$ at the loop bottom. Energy from the release height $h$: $mgh=\frac{1}{2}m{v}_{\text{bot}}^2=\frac{1}{2}m(5gr)\Rightarrow h=\frac{5r}{2}=2.5r.$ A neat, mass- and $g$-free result: start at least two and a half radii high.

(a) A rod can push, so the top condition is only $v_{\text{top}}\ge 0$; minimum is $v_{\text{top}}=0$. Then: $v_{\text{bot}}^2=v_{\text{top}}^2+4gr=0+4gr=4gr,$ $v_{\text{bot}}=\sqrt{4gr}=2\sqrt{gr}=2\sqrt{9.8\times 1.5}=2\sqrt{14.7}=7.67\text{m/s}.$ (b) At the top with $v_{\text{top}}=0$, the centripetal requirement is $m{v}^2/r=0$. The force equation (taking down = toward center as positive) is $F_{\text{rod}}+mg=0$, so $F_{\text{rod}}=-mg$. The negative sign means the rod force points up — i.e. the rod pushes the ball outward (upward, away from center) to hold it up. $|F_{\text{rod}}|=mg=0.40\times 9.8=3.92\text{N}\text{(a push)}.$

On top of a hill, the center is below and the ground pushes up (away from center), gravity down (toward center). Centripetal is downward, so: $mg-N=\frac{m{v}^2}{r}.$ The skier leaves the ground when $N=0$: $mg=\frac{m{v}^2}{r}\Rightarrow v=\sqrt{gr}=\sqrt{9.8\times 20}=\sqrt{196}=14.0\text{m/s}.$ Same $\sqrt{gr}$ form — but now it's a maximum allowed speed to stay grounded, not a minimum. Geometry (center below) is identical; only which force is "extra" flips.

(a) $v_{\text{bot}}=\sqrt{5gr}=\sqrt{5\times 9.8\times 0.90}=\sqrt{44.1}=6.64\text{m/s}.$ (b) $v_{\text{side}}=\sqrt{3gr}=\sqrt{3\times 9.8\times 0.90}=\sqrt{26.46}=5.14\text{m/s}.$ (c) At the side the center is horizontal, so only the horizontal component matters. Tension is entirely horizontal (toward center) and gravity is entirely vertical (contributes nothing to the horizontal/centripetal direction): $T_{\text{side}}=\frac{m{v}_{\text{side}}^2}{r}=\frac{0.25\times 26.46}{0.90}=\frac{6.615}{0.90}=7.35\text{N}.$ (Notice $T_{\text{side}}=3mg=3\times 0.25\times 9.8=7.35\text{N}$ — matches, since $v_{\text{side}}^2=3gr$.)

Set up geometry. Let $\theta$ be the angle swept from the top. The ball's height below the top is $r(1-\cos \theta )$, so its height above the bottom is $r(1+\cos \theta )$. Energy from the top ($v_{\text{top}}^2=gr$): $v^2=v_{\text{top}}^2+2g[r(1-\cos \theta )]=gr+2gr(1-\cos \theta )=gr(3-2\cos \theta ).$

Radial equation at angle $\theta$. The component of gravity toward the center is $mg\cos \theta$ (when the string makes angle $\theta$ with the vertical-up direction). So: $T+mg\cos \theta =\frac{m{v}^2}{r}=mg(3-2\cos \theta ).$ $T=mg(3-2\cos \theta )-mg\cos \theta =mg(3-3\cos \theta )=3mg(1-\cos \theta ).$

Impose $T=mg$: $3mg(1-\cos \theta )=mg\Rightarrow 1-\cos \theta =\frac{1}{3}\Rightarrow \cos \theta =\frac{2}{3}.$ $\theta =\arccos \left(\frac{2}{3}\right)=48.19^{\circ }.$ Check the endpoints: at $\theta =0$ (top) $T=0$ ✓; at $\theta =180^{\circ }$ (bottom) $T=3mg(1-(-1))=6mg$ ✓. The formula reproduces both known cases.

String needs $v_{\text{bot}}^2=5gr$; rod needs $v_{\text{bot}}^2=4gr$. Kinetic energy $\frac{1}{2}m{v}^2$ (same $m$, same $r$): $\frac{K{E}_{\text{string}}}{K{E}_{\text{rod}}}=\frac{5gr}{4gr}=\frac{5}{4}=\mathrm{1.25.}$ The string version needs $25\%$ more launch energy, purely because it must keep the string taut all the way over the top.

A bead on a wire is like a rod: the wire can push both inward and outward on the bead (it wraps around the bead), so there is no "slack" condition. The only requirement to complete the loop is that the bead still moves at the top, i.e. $v_{\text{top}}\ge 0$; the minimum is $v_{\text{top}}=0$. $v_0^2=v_{\text{top}}^2+4gr=4gr\Rightarrow v_{0,\min }=2\sqrt{gr}.$ Comparison: same as the rod ($2\sqrt{gr}$), less than the string ($\sqrt{5gr}$). A string can only pull inward, so it needs enough speed that the required inward force never drops to zero — that extra demand is the whole reason $\sqrt{5gr}>2\sqrt{gr}$.

$\sqrt{5gr}$, and why more than the top? The bottom. It must arrive at the top with $\sqrt{gr}$, and climbing height $2r$ adds $4gr$ to $v^2$: $v_{\text{bot}}^2=gr+4gr=5gr$.

$r$? $h=2.5r$ — from $mgh=\frac{1}{2}m(5gr)$.

$\theta$ from the top (minimum-speed loop)? $T=3mg(1-\cos \theta )$: gives $0$ at top and $6mg$ at bottom.