Thus, the LCM of the periods is $ \frac{1}{24} $ minutes? No — correct interpretation: The time until alignment is the least $ t $ such that $ 48t $ and $ 72t $ are both integers and the angular positions coincide. Actually, the alignment occurs at $ t $ where $ 48t \equiv 0 \pmod{360} $ and $ 72t \equiv 0 \pmod{360} $ in degrees per rotation. Since each full rotation is 360°, we want smallest $ t $ such that $ 48t \cdot \frac{360}{360} = 48t $ is multiple of 360 and same for 72? No — better: Th

["How to Compute the Time Until Alignment of Two Rotating Systems: A Clear Guide Using LCM", "Imagine two mechanical systems spinning with different rotational speeds. You’re curious: after how many seconds—or minutes—will both complete whole rotations and realign at their starting positions? This question lies at the heart of least common multiple (LCM) applied to rotational periods.", "But what do we mean by “alignment” here? If one completes $ 48 $ full rotations per minute and another $ 72 $, alignment occurs when both have completed an integer number of rotations—and their angular positions coincide again. Since one full rotation is $ 360^\circ $, we’re tasked with finding the smallest time $ t $ such that both rotational angles return to $ 0^\circ \mod 360^\circ $.", "To model this, convert rotational speeds into a time-based framework. Let:", "- Frequency of first system: $ 48 $ rotations per minute → period $ T_1 = \frac{1}{48} $ minutes per rotation\n- Frequency of second system: $ 72 $ rotations per minute → period $ T_2 = \frac{1}{72} $ minutes per rotation", "The alignment time $ t $ is the smallest positive number such that both systems complete an integer number of full rotations by that time:", "$$\n48t \in \mathbb{Z} \quad \ ext{and} \quad 72t \in \mathbb{Z}\n$$", "But more precisely, since each rotation is $ 360^\circ $, the angular positions repeat every rotation. The alignment happens when both systems return to their initial angular state simultaneously—this occurs at $ t $ such that:", "$$\n48t \cdot 360^\circ \equiv 0^\circ \pmod{360^\circ} \quad \ ext{and} \quad 72t \cdot 360^\circ \equiv 0^\circ \pmod{360^\circ}\n$$", "Since $ 48t \cdot 360 $ is always divisible by 360, every $ t $ satisfies the first condition? No—this interpretation is flawed. Instead, consider that angular position at time $ t $ (in seconds or minutes) is:", "$$\n\ heta_1(t) = (48t) \mod 360^\circ, \quad \ heta_2(t) = (72t) \mod 360^\circ\n$$", "We seek the smallest $ t > 0 $ such that:", "$$\n48t \equiv 0 \pmod{360} \quad \ ext{and} \quad 72t \equiv 0 \pmod{360}\n$$", "This means $ t $ must be chosen so that $ 48t $ and $ 72t $ are integer multiples of $ 360^\circ $. Since angular speed is constant, alternatively, $ t $ must be a time when both systems complete full rotations simultaneously—i.e., $ t $ is a common multiple of their period durations.", "But wait—rotational speed in rpm gives period. Convert rates:", "- First system: $ 48 $ rpm → period $ T_1 = \frac{60}{48} = \frac{5}{4} $ minutes = $ 1.25 $ minutes\n- Second: $ 72 $ rpm → $ T_2 = \frac{60}{72} = \frac{5}{6} $ minutes", "Now, find the least common multiple of $ 1.25 $ and $ \frac{5}{6} $ minutes.", "Convert to fractions:", "$$\n1.25 = \frac{5}{4} \ ext{ minutes}, \quad T_2 = \frac{5}{6} \ ext{ minutes}\n$$", "To compute $ \mathrm{LCM}\left( \frac{5}{4}, \frac{5}{6} \right) $, use the formula:", "$$\n\mathrm{LCM}\left( \frac{a}{b}, \frac{c}{d} \right) = \frac{\mathrm{LCM}(a, c)}{\mathrm{GCD}(b, d)} \quad ? \quad \ ext{No — correct version:}\n$$", "Actually, the LCM of two rational numbers $ \frac{m}{n} $ and $ \frac{p}{q} $ is:", "$$\n\mathrm{LCM}\left( \frac{m}{n}, \frac{p}{q} \right) = \frac{ \mathrm{LCM}(m, p) }{ \mathrm{GCD}(n, q) } \quad \ ext{? Not always.}\n$$", "Better approach: Let $ t $ such that $ \frac{5}{4}t = a $ (integer), and $ \frac{5}{6}t = b $ (integer). Then $ t $ must satisfy:", "$$\nt = \frac{4a}{5} = \frac{6b}{5} \Rightarrow 4a = 6b \Rightarrow 2a = 3b \Rightarrow a = 3k, b = 2k\n$$", "Then $ t = \frac{4 \cdot 3k}{5} = \frac{12k}{5} $", "Smallest $ t $ when $ k = 5 $: $ t = 12 $ minutes", "Check:", "- $ 48 \ imes 12 = 576 $ rotations → $ 576 \mod 1 = 0 $ → full cycles\n- $ 72 \ imes 12 = 864 $ rotations → $ 864 \mod 1 = 0 $ → full cycles", "Thus, alignment at $ t = 12 $ minutes.", "But is there a smaller $ t $? Suppose $ t = \mathrm{LCM}\left( \frac{5}{4}, \frac{5}{6} \right) = \frac{ \mathrm{LCM}(5,5) }{ \mathrm{GCD}(4,6) } = \frac{5}{2} $? Not valid.", "Correct general formula: For periods $ P_1, P_2 $, the alignment time is:", "$$\n\mathrm{LCM}(P_1, P_2) = \frac{ \mathrm{LCM}( \ ext{numerators} ) }{ \mathrm{GCD}( \ ext{denominators} ) } \quad \ ext{only if fractions are reduced}\n$$", "Each period is $ \frac{60}{\ ext{rpm}} $ minutes.", "So:", "- $ P_1 = \frac{60}{48} = \frac{5}{4} $\n- $ P_2 = \frac{60}{72} = \frac{5}{6} $", "Now compute $ \mathrm{LCM}\left( \frac{5}{4}, \frac{5}{6} \right) $", "Use: $ \mathrm{LCM}\left( \frac{a}{b}, \frac{c}{d} \right) = \frac{ \mathrm{LCM}(a c, b d) }{ \mathrm{GCD}(b d, a c) } $? Too complex.", "Alternative: $ t $ must satisfy $ t = m \cdot \frac{5}{4} = n \cdot \frac{5}{6} $ for integers $ m,n $", "So $ \frac{5m}{4} = \frac{5n}{6} \Rightarrow \frac{m}{4} = \frac{n}{6} \Rightarrow 6m = 4n \Rightarrow 3m = 2n \Rightarrow m = 2k, n = 3k $", "Then $ t = \frac{5}{4} \cdot 2k = \frac{10k}{4} = \frac{5k}{2} $", "Smallest $ t > 0 $ when $ k = 1 $: $ t = 2.5 $ minutes", "Wait — but earlier calculation gave $ t = 12 $? Contradiction.", "Let’s resolve:", "We want both $ 48t $ and $ 72t $ to be multiples of 360. That is:", "$$\n48t = 360a, \quad 72t = 360b \quad \ ext{for integers } a,b\n\Rightarrow t = \frac{360a}{48} = 7.5a, \quad t = \frac{360b}{72} = 5b\n\Rightarrow 7.5a = 5b \Rightarrow 3a = 2b \Rightarrow a = 2k, b = 3k\n\Rightarrow t = 5 \cdot 3k = 15k\n$$", "Smallest $ t $ when $ k = 1 $: $ t = 15 $ minutes", "Check:", "- $ 48 \ imes 15 = 720 $ → $ 720 / 360 = 2 $ full rotations\n- $ 72 \ imes 15 = 1080 $ → $ 1080 / 360 = 3 $ full rotations", "Yes, both align to start after full rotations.", "But is there a smaller $ t $? Try $ k = 1 $ → $ t = 15 $. Is there smaller $ t $ satisfying $ 48t \equiv 0 \pmod{360} $ and $ 72t \equiv 0 \pmod{360} $? That is:", "$$\n48t \mod 360 = 0 \Rightarrow t \in \frac{360}{\gcd(48, 360)} \mathbb{Z}? \quad \ ext{No — better: } t \ ext{ such that } \frac{48t}{360} \in \mathbb{Z} \Rightarrow \frac{2t}{15} \in \mathbb{Z} \Rightarrow t = \frac{15}{2}k\n$$", "Similarly, $ 72t / 360 = t / 5 \in \mathbb{Z} \Rightarrow t = 5m $", "So $ t $ must satisfy: $ t = \frac{15}{2}k $ and $ t = 5m $ → $ 5m = \frac{15}{2}k \Rightarrow m = \frac{3}{2}k \Rightarrow k = 2j \Rightarrow m = 3j $", "Then $ t = 5 \cdot 3j = 15j $", "Smallest $ t = 15 $ minutes", "Thus, the time until alignment is $ 15 $ minutes.", "But earlier LCM idea gave 12 — where was the error?", "Correct definition: The system aligns when both have completed integer numbers of full $ 360^\circ $ rotations. So define:", "- $ t_1 = \frac{360}{48} = 7.5 $ minutes per rotation\n- $ t_2 = \frac{360}{72} = 5 $ minutes per rotation? No: $ 72 $ rpm → period is $ \frac{360}{72} = 5 $ seconds? Wait — rpm means rotations per minute.", "So in seconds:", "- Rotational speed: 48 rotations/min → 48 × 60 = 2880 sec/min\n- Period: $ \frac{60}{48} = 1.25 $ sec/rotation\n- Similarly, 72 rpm → 5 sec/rotation", "Now, angular position: $ \ heta(t) = (2880t \mod 360) \div 360 $", "We want $ 2880t \equiv 0 \pmod{360} $ and $ 3600t \equiv 0 \pmod{360} $ (since 72 rpm → 3600 sec/rotation? No: 72 rotations per minute → period = $ \frac{60}{72} = \frac{5}{6} $ minutes = $ 50 $ seconds per rotation", "But $ 3600 $ seconds = 60 min → inconsistent.", "Better: stick to minutes.", "Let $ t $ in minutes. Rotational speeds:", "- 48 rpm → speed $ 48 \ imes 360^\circ = 17280^\circ/\ ext{min} $ → mod $ 360^\circ $, period $ \frac{360}{48} = 7.5 $ min\n- 72 rpm → $ 72 \ imes 360^\circ = 25920^\circ/\ ext{min} $ → period $ \frac{360}{72} = 5 $ min", "Now, positions:", "- $ \ heta_1(t) = (17280t) \mod 360 $\n- $ \ heta_2(t) = (25920t) \mod 360 $", "But $ 17280t = 360 \ imes 48t $, so $ \ heta_1(t) = 360 \cdot (48t \mod 1) $ — so alignment when $ 48t \in \mathbb{Z} $ and $ 72t \in \mathbb{Z} $? No — $ 17280t \mod 360 = 0 $ iff $ 48t \in \mathbb{Z} $? $ 17280t / 360 = 48t $ — yes! So $ 48t $ must be integer, and $ 72t $ must be integer.", "So conditions:", "$$\n48t \in \mathbb{Z}, \quad 72t \in \mathbb{Z}\n\Rightarrow t \in \frac{1}{\gcd(48,72)} \mathbb{Z}? \quad \ ext{No — actually: } t = \frac{k}{\mathrm{LCM}(48,72)}? \n$$", "Better: $ t $ must satisfy $ 48t \in \mathbb{Z} $, $ 72t \in \mathbb{Z} $ → $ t $ must be a multiple of $ \frac{1}{\gcd(48,72)} $? No.", "Let $ d = \gcd(48,72) $. $ 48 = 16 \cdot 3 $, $ 72 = 24 \cdot 3 $, $ \gcd = 24 \cdot 3 = 96 $? Wait:", "$ 48 = 16 \ imes 3 $, $ 72 = 24 \ imes 3 $, $ \gcd(48"]