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Circular Motion

In the case of races and motions in straight line, we have observed that if the two bodies or persons are moving with different speeds in a straight line in one direction, then they will never meet. This is due to the fact that with the passage of time, the distance between them is increasing constantly.

Circular motion should be seen as a logical extension of Races where runners are running on a circular track. Since it an enclosed track (by virtue of it being circular), runners are bound to meet at some point or the other.

Case 1: When two or more than two persons are running around a circular track in the same direction.
 
Example-1
When will they meet for the first time anywhere on the track?
Solution
To understand the situation completely, let us assume that there are two persons A and B. Speed of A = 20 m/s, speed of B = 10 m/s, length of the track is 1000 m and they are running in the same direction. It can be seen in figure 1 that initially both of them are at the same point, i.e., the starting point.
 
They will be meeting for the first time only if the faster runner A has taken one more round of the track than the slower runner B. This can be interpreted as—A will have to cover 1000 m more than B.
 
It is understood with the figures given above that the distance will keep on increasing between them with the passage of time. And the moment distance between them becomes equal to 1000 m, they will be at the same point.
 
So, the time taken = distance/relative speed = Description: 3150.png
 
= 100 s
 
Or, this can be done by using unitary method also:
 
Distance of 10 m is created in 1 s
 
So, the distance of 1000 m will be created in 100 s
 
Now, let us assume that there are three persons A, B and C running with following speeds in the same direction:
 
Speed of A = 30 m/s
 
Speed of B = 20 m/s
 
Speed of C = 10 m/s
 
To calculate when will they meet for the first time, we are required to find the time taken by the fastest runner to take one round over the other runners.
 

 
Time taken by A to take one round over B = tA−B = 1000/10 = 100 s
 
Time taken by A to take one round over C = tA−c = 1000/20 = 50 s
 
Now, the LCM of these two values tA−B and tA−c will give us the time after which all of them will be meeting at the same place.
 
LCM = (100, 50) = 100 s
 
It can also be seen that they will be meeting after every 100 s.
 
 
Example-2
When will they meet for the first time at the starting point?
Solution
To calculate this, we will use the concept of LCM (Usage of LCM and HCF, chapter 2, case 2)
Find the time taken by each individual to take one round and then calculate LCM of these values.
 
Assume there are three persons A, B and C with a respective speed of 30 m/s, 20 m/s and 10 m/s running in the same direction. Length of the circular track is 1000 m.
 
Time taken by A to take one round = t1 = 1000/30
= 33.33 s
 
Time taken by B to take one round = t2 = 1000/20
= 50s
Time taken by C to take one round = t3 = 1000/10
= 100 s
 
LCM of t1, t2, t3 = 100 seconds.
 
 
Example-3
At how many different points of the track will they be meeting?
Solution
Let us assume that the speed of A = 25 m/s and the speed of B = 10 m/s and the length of the track = 1000 m
 
They will be meeting for the first time after a time-gap of 1000/15 = 66.66 s
 
Till this time, A has covered 1666.66 m and B has covered 666.66 m. This point is 666.66 m from the starting point. Now, this point can be assumed to be the starting point.
 
So, they will meet at a distance of 666.66 m from here. This is the second meeting point, at a distance of 333.33 m from the starting point. Next meeting point will be 666.66 m from here. This point will be nothing but the starting point again (Fig.1 and Fig.4 are same).
 
This can be seen through the figures given below:
 

 
So, there are a total of 3 distinct meeting points on the track.
 
In general, number of meeting points = difference of ratio of the speed the of A and B in its simplest form.
 
Ratio of speed of A and B = 5:2
 
So, the number of different meeting points = 5 − 2
= 3 points
 
 
Case 2: When two or more than two persons are running around a circular track in the opposite direction.
 
Here again there are two persons A and B with a speed of 20 m/s and 10 m/s respectively, and length of track is 1000 m.
 
Example-4
When will they meet for the first time anywhere on the track?
Solution
Since they are running in the opposite direction, relative speed = 10 + 20 = 30 m/s
 
So, time taken = distance/relative speed = 1000/30
= 33.33 seconds
 
 
Example-5
When will they meet for the first time at the starting point?
Solution
 
First,We calculate the time taken by each individual to take one round and then calculate the LCM of those values.
Time taken by A to take one round = t1 = 1000/20
= 50 seconds
 
Time taken by B to take one round = t2 = 1000/10
= 100 seconds
 
LCM of (t1, t2)= 100 s.
 

Example-6
At how many different points of the track will they be meeting?
Solution
They are meeting after 33.33 seconds for the first time. Till this time, A has covered 666.66 m and B has covered 333.33 m. So, obviously they are meeting at a distance of 666.66 m from starting point in the direction of A. Next point will be again 666.66 m ahead of this point. And, the next point will be another 666.66 m ahead of this point, which will be the starting point.
 
So, a total of 3 points will be there.
 
In general, number of distinct meeting points = addition of the ratio of the speed of A and B in its simplest form.
 
The ratio of speed of A and B = 2:1
 
So, the number of different meeting points = 2 + 1
= 3 points
 

Example-7
Anup and Shishir start running from the same point of a circular track at the same time. After how much time will Anup and Shishir, who are running with a speed of 35 m/s and 40 m/s, respectively, meet at diametrically opposite point?
Solution
The simplest ratio of speed of Anup and Shishir = 7:8
 
So, if they are running in the same direction, they will meet at 1 point and if they are running in the opposite direction, they will meet at 15 different points.
 
Now, for them to meet at a diametrically opposite point, there should be at least two meeting points or the number of meeting points should be a multiple of 2.
 
Since, they would meet either at 1 point or at 15 different points, depending on the direction of their movement, they will, therefore, not meet at a diametrically opposite point.
 





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