a machine has 3 wheels each of radius 14 meter revolving at a speed of 5.5m/s, 11m/s, and 22m/s respectively. whenever 3 wheels comes at initial position a bell rings.
now the question is when will the bell ring for the 12th time if it first rings at 12 o'clock.
2007-02-17 03:56:05 · 2 answers · asked by nikki 2 in Science & Mathematics ➔ Mathematics
A wheel will go a full rotation in a time equal to (circle length)/wheel speed
As circle length is equal to 2*PI*radius, we will have
t1=2*PI*radius/5.5
t2=2*PI*radius/11
t3=2*PI*radius/22
the wheels will arrive for the first time simultaneously after
t1 seconds (which will mean 1 rotation for the first wheel, 2 for the second and 4 for the third).
the bell will ring or the 12th time at t1*11s=2*(wheel length)seconds
2007-02-17 04:25:32 · answer #1 · answered by Monica 1 · 0⤊ 0⤋
The time will be 12h 2m 55.929... s. Here's how I found this time :
The wheels rotate at multiples of (1, 2, and 4) times 5.5 m/s, the slowest one. So, in the basic rotation time of that slowest one, the 3 wheels will respectively rotate (4, 2, and 1) time(s).
Since it first rings at 12 o'clock the second ring is one basic time unit later, etc. ... so that the 12th time occurs 11 basic times later than 12 o'clock.
O.K., so what is that basic time? It's the circumference of the wheel, 2*14Ï m, divided by the speed at which that rotates, 5.5 m/s. So, the time following 12 o'clock is:
11*28Ï/5.5 s = 56Ï s = 175.929... 2m 55.929... s. QED
Live long and prosper.
2007-02-17 13:28:38 · answer #2 · answered by Dr Spock 6 · 0⤊ 0⤋