Exercises — Circular motion — centripetal acceleration derivation

We are given $v$ and $r$, so use the form built from those two: $a_c=\frac{v^2}{r}=\frac{6^2}{2}=\frac{36}{2}=18{\text{m/s}}^2.$ What we did: squared the speed, divided by radius. Why this form: $v$ and $r$ are exactly what's handed to us — no need to detour through $\omega$. Answer: $a_c=18{\text{m/s}}^2$, directed toward the centre.

Given $\omega$ and $r$, use $a_c={\omega }^{2}r$ directly: $a_c={\omega }^{2}r=(4{)}^2(0.25)=16\times 0.25=4{\text{m/s}}^2.$ Why this form: angular speed is given, so ${\omega }^{2}r$ avoids computing $v$ first. Answer: $a_c=4{\text{m/s}}^2$ toward the centre.

It points toward the centre of the circle, always perpendicular to the velocity. Because it is perpendicular to $\vec{v}$, it only turns the velocity's direction and never changes its magnitude (the speed). See the figure below — the red arrow is always at $90^{\circ }$ to the black velocity.

We are given the period, not $v$ or $\omega$. First convert: one full loop is $2\pi$ radians, done in time $T$, so $\omega =\frac{2\pi }{T}=\frac{2\pi }{0.5}=4\pi \text{rad/s}\approx 12.566\text{rad/s}.$ Why this step: $\omega$ is the bridge from timing to acceleration. Now: $a_c={\omega }^{2}r=(4\pi {)}^2(0.8)=16{\pi }^2\times 0.8=12.8{\pi }^2\approx 126.3{\text{m/s}}^2.$ Answer: $a_c\approx 126.3{\text{m/s}}^2$. (Equivalently $a_c=\frac{4{\pi }^{2}r}{T^2}$.)

Friction supplies the centripetal acceleration, so the car is safe as long as the required $a_c=v^2/r$ does not exceed what friction gives. At the limit: $\frac{v^2}{r}=6\Rightarrow v^2=6r=6\times 45=270\Rightarrow v=\sqrt{270}\approx 16.43\text{m/s}.$ What we did: rearranged $a_c=v^2/r$ to solve for $v$. Why: we know the acceleration and radius, want the speed. Answer: $v_{\max }\approx 16.43\text{m/s}$ (about $59\text{km/h}$).

"Revolutions per minute" must become radians per second. Each revolution is $2\pi$ rad, and one minute is $60$ s: $\omega =600\times \frac{2\pi }{60}=600\times \frac{2\pi }{60}=20\pi \text{rad/s}\approx 62.83\text{rad/s}.$ Why this step: the formula only accepts $\omega$ in rad/s, never rpm. $a_c={\omega }^{2}r=(20\pi {)}^2(0.3)=400{\pi }^2\times 0.3=120{\pi }^2\approx 1184.4{\text{m/s}}^2.$ Answer: $a_c\approx 1184.4{\text{m/s}}^2$ — over $120g$! That's why fast-spinning parts fly apart.

Centripetal force is $F_c=\frac{m{v}^2}{r}$. With $m$ and $r$ fixed, $F_c\propto v^2$. Tripling the speed: $\frac{F_{\text{new}}}{F_{\text{old}}}=\frac{(3v{)}^2}{v^2}=9.$ What we did: kept the constants fixed and looked only at the varying part $v^2$. Why: the ratio method cancels every constant, so we never need actual numbers. Answer: the force becomes $9\times$ larger.

Same turntable ⇒ same $\omega$. Use the $\omega$ form since $\omega$ is shared: $a_c={\omega }^{2}r\Rightarrow \frac{a_B}{a_A}=\frac{{\omega }^2(2r)}{{\omega }^{2}r}=2.$ Why the $\omega$ form: with $\omega$ common, $a_c$ is directly proportional to $r$, and the comparison is instant. Had we used $v^2/r$ we'd first have to note $v=\omega r$ differs for each — more work, same answer. Answer: particle B (outer) has twice the centripetal acceleration.

Now $v$ is shared, so use the $v$ form: $a_c=\frac{v^2}{r}\Rightarrow \frac{a_B}{a_A}=\frac{v^2/(2r)}{v^2/r}=\frac{1}{2}.$ Why the switch: the quantity held constant decides which form gives the cleanest comparison. With $v$ fixed, $a_c\propto 1/r$, so the larger radius gives the smaller acceleration — the opposite of Q3.2. Answer: particle A (inner) has twice the acceleration of B. Same objects, opposite conclusion — because what is held fixed differs.

(a) Given $v$ and $r$: $a_c=\frac{v^2}{r}=\frac{3^2}{0.6}=\frac{9}{0.6}=15{\text{m/s}}^2.$ (b) By Newton's Second Law, the net inward force is $F_c=m{a}_c$. Here the string tension is that inward force (it is not an extra force — it is the centripetal force): $T=m{a}_c=0.2\times 15=3\text{N}.$ Why: we identify the real force that supplies the inward pull, then set it equal to $m{v}^2/r$. Answer: $a_c=15{\text{m/s}}^2$, tension $T=3\text{N}$.

At the top of the hill, both gravity ($mg$, downward toward the centre) and the normal force ($N$, upward away from the centre) act along the vertical line to the centre. The net downward = inward force supplies the centripetal requirement: $mg-N=\frac{m{v}^2}{r}.$ The car "just loses contact" when the road can no longer push — that's $N=0$: $mg=\frac{m{v}^2}{r}\Rightarrow v^2=gr=9.8\times 40=392\Rightarrow v=\sqrt{392}\approx 19.8\text{m/s}.$ What we did: wrote Newton's 2nd law along the radial (vertical) direction, then imposed the lift-off condition $N=0$. Why $N=0$: contact is lost exactly when the surface stops pushing — mass $m$ cancels, so it doesn't matter how heavy the car is. Answer: $v\approx 19.8\text{m/s}$ (about $71\text{km/h}$).

Two real forces act: tension $T$ along the string, and gravity $mg$ down. Split tension into components (see figure):

• Vertical: $T\cos \theta =mg$ (bob doesn't rise or fall).
• Horizontal: $T\sin \theta =\frac{m{v}^2}{R}$ (this horizontal part is the centripetal force).

The circle's radius is $R=L\sin \theta$ (horizontal distance from axis to bob). Divide the horizontal equation by the vertical one to cancel $T$: $\tan \theta =\frac{v^2}{Rg}\Rightarrow v^2=Rg\tan \theta =L\sin \theta g\tan \theta .$ Why divide: we don't know $T$ and don't need it — dividing eliminates it, a classic trick. Plug in $L=1$, $\theta =30^{\circ }$ ($\sin 30^{\circ }=0.5$, $\tan 30^{\circ }=0.5774$), $g=9.8$: $v^2=1\times 0.5\times 9.8\times 0.5774\approx 2.829\Rightarrow v\approx 1.682\text{m/s}.$ Answer: $v\approx 1.68\text{m/s}$.

(a) Here gravity is the centripetal force — see Gravitation — orbital motion. So the gravitational acceleration at that height equals the centripetal acceleration: $g_{\text{orbit}}=\frac{v^2}{r}\Rightarrow v^2=g_{\text{orbit}}r=8.0\times 7.0\times 10^6=5.6\times 10^7.$ $v=\sqrt{5.6\times 10^7}\approx 7483\text{m/s}\approx 7.48\text{km/s}.$ Why equate them: nothing else pulls the satellite inward, so gravity alone must supply the whole $v^2/r$. Mass $m$ cancels — orbital speed doesn't depend on the satellite's mass.

(b) The period is the circumference divided by the speed (using Angular velocity and period): $T=\frac{2\pi r}{v}=\frac{2\pi (7.0\times 10^6)}{7483}\approx \frac{4.398\times 10^7}{7483}\approx 5877\text{s}.$ Convert to minutes: $5877/60\approx 97.9\text{min}$. Answer: $v\approx 7.48\text{km/s}$, period $T\approx 97.9\text{minutes}$ — a realistic low-Earth-orbit value (about 90–100 min).

At the bottom of the loop, the centre is directly above the bead. So "toward the centre" means upward. The normal force $N$ pushes up (inward); gravity $mg$ pulls down (outward). Newton's 2nd law along the radial (vertical) direction, taking inward as positive: $N-mg=\frac{m{v}^2}{r}.$ Solve for $N$: $N=mg+\frac{m{v}^2}{r}=0.1\times 9.8+\frac{0.1\times 4^2}{0.5}=0.98+\frac{1.6}{0.5}=0.98+3.2=4.18\text{N}.$ What we did: identified which way "inward" points at the bottom, then summed real forces along that line. Why $N>mg$: at the bottom the wire must both hold the bead up and bend its path upward, so it pushes harder than the weight alone. Answer: $N\approx 4.18\text{N}$, directed upward (toward the centre).

• Can you pick $v^2/r$ vs ${\omega }^{2}r$ based on what's given? (L1–L2)
• Can you convert rpm and period into rad/s? (L2)
• Do you know that $a_c\propto r$ (fixed $\omega$) but $a_c\propto 1/r$ (fixed $v$)? (L3)
• Do you set a real force equal to $m{v}^2/r$ instead of inventing a new one? (L4)
• Do you re-point "inward" for each position on a vertical loop? (L5)