The angular speed of hour hand is:
360 degrees per 12 hours = 30 degrees per hour;
The angular speed of minute hand is:
360 degrees per hour;
The angular speed of second hand is:
360 degrees per minute = 360*60 degrees per hour;
Let’s set hour hand speed to ω = 30 degrees per hour, then minute hand speed is 12ω, second hand speed is 720ω
Suppose hour hand and minute hand...
The angular speed of hour hand is:
360 degrees per 12 hours = 30 degrees per hour;
The angular speed of minute hand is:
360 degrees per hour;
The angular speed of second hand is:
360 degrees per minute = 360*60 degrees per hour;
Let’s set hour hand speed to ω = 30 degrees per hour, then minute hand speed is 12ω, second hand speed is 720ω
Suppose hour hand and minute hand overlap at t hour. Whenever they overlap, the minute hand must have rotated n (n is a whole number) number of cycles more than the hour hand. Hence we have the following equation:
12ωt – ωt = 360n
11ωt = 360n
11*30t=360n
t = 12n/11 (0 < t <= 24, n is a whole number)
0 < 12n/11 <= 24, so the range of n is [1, 22]
Therefore, the hour hand and the minute hand overlap 22 times in a day. They overlap at 1 1/11h (n = 1), 2 2/11h (n = 2), ... , 22 10/11h (n=21), 24h (n = 22).
Likewise, whenever the second hand and the minute hand overlap, the second hand must have rotated N (N is a whole number) number of cycles more than the minute hand. Hence we have the following equation:
720ωt – 12ωt = 360N
708ωt = 360N
708*30t = 360N
Since t = 12n/11 and n is a whole number between 1 and 22 when hour hand and minute hand overlap,
708 *30 *12n/11 = 360N
N = 108n/11
Only when n = 11 or n = 22 is N a whole number.
Therefore, the hour hand, the minute hand and the second hand overlap only twice a day. They overlap at 12h (n=11) and 24h (n=22).
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