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Three identical particles are joined together by a thread as shown in figure .all the 3 particles are moving in a horizontal plane .if the velocity of the outermost particle isV○ ,then the ratio of tension in the three section of the string
1)3:5:7
2)3:4:5
3)7:11:6
4)3:5:6

sheema , 8 Years ago
Grade 12th pass
anser 2 Answers
Ashish Kumar

Last Activity: 8 Years ago

The answer is 4) 3:5:6
Assume mass of each ball is m.
For the ball C, Vo =ω3L, from here ω = Vo / 3L.
For ball A force due to rotation on it = mω2L=mVo / 9L
For ball B force due to rotation on it = mω22L= 2mVo / 9L 
For ball C force due to rotation on it = mω23L= mVo / 3L
By drawing free body diagrams of each ball you will get-
Tension in the string joining the ball B and C = mVo / 3L =T1
Tension in the string joining the ball A and B =2mVo / 9L + mVo / 3L= 5mVo/9L = T2
Tension in the string joining the ball O and A = mVo / 9L + 2mVo / 9L + mVo / 3L= 2mVo/3L = T3
Now T: T2 : T3 = mVo / 3L :  5mVo/9L : 2mVo/3L = 1/3 : 5/9 : 2/3
Now in the first and third fraction multiply 3 to the numerator and denominator to make the denominator of each fraction = 9
=>T: T2 : T3 =1x3/3x3 : 5/9 : 2x3/3x3 = 3/9 : 5/9 : 6/9 , Now since the denominator of each fraction is same so it will be cancelled from all the three fractions so you will get T: T2 : T3 = 3 : 5 : 6 Answer
 
 
 

Rishi Sharma

Last Activity: 4 Years ago

Dear Student,
Please find below the solution to your problem.

According to the question we know that we have to find the ratio of tension in the three section of string.
so let's suppose angular speed of thread is equal to "W"
by each strings length.
in case of particle C= T3=mW²3l.
in case of particle B = T2- T3= mW²3l. so T2=mW²5l
in case of particle A= T1- T2 =mW²2l so T1=mW²6l.
Hence the ratio will 3:5:6. which is our answer

Thanks and Regards

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