Exercises — Centripetal force — what provides it in various situations

Answer: the tension in the string. The string can only pull along its own length, toward your hand at the centre. That inward pull is exactly the job the centripetal requirement demands. Nothing else has an inward horizontal component (gravity points straight down, air drag is tiny). So $T=\frac{m{v}^2}{r}$.

Answer: gravity — the Earth's gravitational pull on the Moon. It points from the Moon toward the Earth (the centre of the orbit). We equate it to the requirement: $\frac{GMm}{r^2}=\frac{m{v}^2}{r}$.

Answer: static friction between the tyres and the road, pointing horizontally inward (toward the centre of the roundabout). It is static friction, not sliding friction, because the tyre's contact patch is not skidding sideways — it grips. That sideways grip is the only horizontal force available, so it supplies $\frac{m{v}^2}{r}$.

Only tension acts inward, so $T=\frac{m{v}^2}{r}$. $T=\frac{0.20\times 3.0^2}{0.50}=\frac{0.20\times 9.0}{0.50}=\frac{1.8}{0.50}=3.6\text{N}$ Tension $=3.6\text{N}$, directed from the ball toward the hand.

Friction can supply at most ${\mu }_{s}mg$. Setting that equal to the requirement: ${\mu }_{s}mg=\frac{m{v}_{\max }^2}{r}\Rightarrow v_{\max }=\sqrt{{\mu }_{s}gr}$ The mass $m$ cancels — heavy and light cars have the same limit. $v_{\max }=\sqrt{0.80\times 9.8\times 50}=\sqrt{392}\approx 19.8\text{m/s}$ $v_{\max }\approx 19.8\text{m/s}$ (about $71\text{km/h}$).

The magnetic force $qvB$ is always perpendicular to velocity — it never speeds the electron up, it only bends it, which is exactly what a centripetal force does. Set $qvB=\frac{m{v}^2}{r}$ and solve: $r=\frac{mv}{qB}=\frac{9.11\times 10^{-31}\times 2.0\times 10^6}{1.60\times 10^{-19}\times 0.010}$ $r=\frac{1.822\times 10^{-24}}{1.60\times 10^{-21}}\approx 1.14\times 10^{-3}\text{m}\approx 1.1\text{mm}$

Look at the figure: the road surface tilts, so the normal force $N$ (perpendicular to the surface) also tilts. It splits into a vertical slice $N\cos \theta$ and a horizontal slice $N\sin \theta$ pointing toward the centre.

• Vertical (no up/down acceleration): $N\cos \theta =mg$.
• Horizontal (this is the centripetal requirement): $N\sin \theta =\frac{m{v}^2}{r}$.

Divide the second by the first — $N$ and $m$ both cancel, leaving the clean relation: $\tan \theta =\frac{v^2}{rg}\Rightarrow v=\sqrt{rg\tan \theta }$ $v=\sqrt{80\times 9.8\times \tan 20^{\circ }}=\sqrt{80\times 9.8\times 0.3640}\approx \sqrt{285.3}\approx 16.9\text{m/s}$ $v\approx 16.9\text{m/s}$.

The string tension $T$ tilts, just like the normal force above:

• Vertical: $T\cos \theta =mg$, so $T=\frac{mg}{\cos \theta }=\frac{0.30\times 9.8}{\cos 30^{\circ }}=\frac{2.94}{0.8660}\approx 3.39\text{N}$.
• Horizontal: $T\sin \theta =\frac{m{v}^2}{r}$, and the circle's radius is $r=L\sin \theta =1.2\times 0.5=0.60\text{m}$.

Solve for $v$: $v=\sqrt{\frac{T\sin \theta \cdot r}{m}}=\sqrt{\frac{3.39\times 0.5\times 0.60}{0.30}}=\sqrt{3.39}\approx 1.84\text{m/s}$ $T\approx 3.39\text{N}$, $v\approx 1.84\text{m/s}$.

(a) Top. Look at the top of the figure: both gravity ($mg$, down) and tension ($T$, also down toward the centre) point inward. So $T+mg=\frac{m{v}_{\text{top}}^2}{r}.$ The string can pull but never push, so the least it can do is $T=0$. Then gravity alone must supply the whole requirement: $mg=\frac{m{v}_{\min }^2}{r}\Rightarrow v_{\min }=\sqrt{gr}=\sqrt{9.8\times 1.0}\approx 3.13\text{m/s}.$

(b) Bottom. Use energy conservation. The bottom is $2r$ lower than the top, so the ball gains kinetic energy equal to $mg(2r)$: $\frac{1}{2}m{v}_{\text{bot}}^2=\frac{1}{2}m{v}_{\text{top}}^2+mg(2r)$ $v_{\text{bot}}^2=v_{\text{top}}^2+4gr=gr+4gr=5gr=5\times 9.8\times 1.0=49$ $v_{\text{bot}}=7.0\text{m/s}.$

(c) Tension at bottom. At the bottom, tension points up (inward) but gravity points down (outward). Net inward is $T-mg$: $T-mg=\frac{m{v}_{\text{bot}}^2}{r}\Rightarrow T=mg+\frac{m{v}_{\text{bot}}^2}{r}$ $T=0.50\times 9.8+\frac{0.50\times 49}{1.0}=4.9+24.5=29.4\text{N}.$ $v_{\min }\approx 3.13$ m/s, $v_{\text{bot}}=7.0$ m/s, $T_{\text{bot}}=29.4$ N.

Gravity provides the centripetal force: $\frac{GMm}{r^2}=\frac{m{v}^2}{r}$. Cancel $m$ and one $r$: $v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{3.986\times 10^{14}}{7.0\times 10^6}}=\sqrt{5.694\times 10^7}\approx 7546\text{m/s}.$ The period is the circumference divided by speed: $T=\frac{2\pi r}{v}=\frac{2\pi \times 7.0\times 10^6}{7546}\approx 5828\text{s}\approx 97\text{min}.$

At maximum speed the car tends to slide outward/up the slope, so friction $f={\mu }_{s}N$ points down the slope (opposing the impending slide). Resolve $N$ and $f$ into vertical and horizontal, as in the figure.

Vertical (no vertical acceleration): $N\cos \theta =mg+f\sin \theta =mg+{\mu }_{s}N\sin \theta .$ Horizontal (centripetal): both the normal's horizontal slice and friction's horizontal slice point inward: $N\sin \theta +{\mu }_{s}N\cos \theta =\frac{m{v}^2}{r}.$ Solve the first for $N$: $N(\cos \theta -{\mu }_s\sin \theta )=mg$. Substitute into the second and cancel $m$: $v_{\max }=\sqrt{rg\frac{\sin \theta +{\mu }_s\cos \theta }{\cos \theta -{\mu }_s\sin \theta }}.$ Plug in ($\sin 15^{\circ }=0.2588,\cos 15^{\circ }=0.9659$): $v_{\max }=\sqrt{100\times 9.8\times \frac{0.2588+0.40\times 0.9659}{0.9659-0.40\times 0.2588}}=\sqrt{980\times \frac{0.6452}{0.8624}}$ $=\sqrt{980\times 0.7481}\approx \sqrt{733.1}\approx 27.1\text{m/s}.$ $v_{\max }\approx 27.1\text{m/s}$ (about $98\text{km/h}$).

Geometry first. Each string is the hypotenuse ($1.25\text{m}$); the horizontal reach is $r=1.0\text{m}$, so the vertical reach is $\sqrt{1.25^2-1.0^2}=\sqrt{0.5625}=0.75\text{m}$ — consistent with the $1.5\text{m}$ separation ($2\times 0.75$). Thus $\cos \theta =r/L=1.0/1.25=0.80$ (horizontal) and $\sin \theta =0.75/1.25=0.60$ (vertical). The upper string pulls inward and up; the lower pulls inward and down.

Vertical balance (up positive): upper string's up-part minus lower string's down-part minus weight $=0$: $T_1(0.60)-T_2(0.60)-mg=0\Rightarrow T_1-T_2=\frac{mg}{0.60}=\frac{2.0\times 9.8}{0.60}=32.67\text{N}.(\star )$ Horizontal (centripetal): both inward horizontal parts add: $T_1(0.80)+T_2(0.80)=\frac{m{v}^2}{r}=\frac{2.0\times 36}{1.0}=72\text{N}\Rightarrow T_1+T_2=\frac{72}{0.80}=90\text{N}.(\star \star )$ Add $(\star )$ and $(\star \star )$: $2T_1=122.67\Rightarrow T_1\approx 61.3\text{N}$. Then $T_2=90-61.3\approx 28.7\text{N}$. $T_1\approx 61.3\text{N}$ (upper), $T_2\approx 28.7\text{N}$ (lower). Both positive → both strings genuinely taut. ✓

L1 taught you to name the provider ::: tension / gravity / friction / normal / electromagnetic. L2 taught you $F_{\text{provider}}=m{v}^2/r$ ::: solve for one unknown, and mass often cancels. L3 taught you to resolve a tilted force ::: vertical part balances weight, horizontal part is centripetal. L4 taught you position matters ::: at the top gravity adds inward, at the bottom it subtracts. L5 taught you what breaks first / which way it slides ::: friction direction and taut-string checks decide the answer.