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: The number of rotations completed must be integer, and the alignment occurs when both complete a number of rotations differing by full cycles. The time until both complete whole rotations and are aligned again is $ \frac{360}{\mathrm{GCD}(48, 72)} $ minutes? No — correct formula: For two periodic events with periods $ T_1, T_2 $, time until alignment is $ \mathrm{LCM}(T_1, T_2) $, where $ T_1 = 1/48 $, $ T_2 = 1/72 $. But in terms of complete rotations: Let $ t $ be time. Then $ 48t $ rows per minute — better: Let angular speed be $ 48 \cdot \frac{360}{60} = 288^\circ/\text{sec} $? No — $ 48 $ rpm means 48 full rotations per minute → period per rotation: $ \frac{60}{48} = \frac{5}{4} = 1.25 $ seconds. Similarly, 72 rpm → period $ \frac{5}{12} $ minutes = 25 seconds. Find LCM of 1.25 and 25/12. Write as fractions: $ 1.25 = \frac{5}{4} $, $ \frac{25}{12} $. LCM of fractions: $ \mathrm{LCM}(\frac{a}{b}, \frac{c}{d}) = \frac{\mathrm{LCM}(a, c)}{\mathrm{GCD}(b, d)} $? No — standard: $ \mathrm{LCM}(\frac{m}{n}, \frac{p}{q}) = \frac{\mathrm{LCM}(m, p)}{\mathrm{GCD}(n, q)} $ only in specific cases. Better: time until alignment is $ \frac{\mathrm{LCM}(48, 72)}{48 \cdot 72 / \mathrm{GCD}(48,72)} $? No.

Tue, 05 May 2026 03:47:31 +0000

Content under review