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Fawz Naim Grade:
        

To one end of a light inextensible string oflength L is attached a particleof mass m. The particle is on a smooth horizintal table. The string passes through a hole in the table and to its other end is attached a small particle of equal mass m. The system is set in motion with the first particle describing a circle on the table with constant angular velocity omega1 and the second particle moving in a hrizintla circle as a conical pendulum with constant angular velocity omrga2.


(a) then the length of the poritond of string on either side of the hole are in the ratio omega2^2:omega1^2


(b) omega1 and omega2 satisfy the relation


1/omega1^2 +1/omega2^2 > L/g


(c) tension in the two parts of the string are equal


(d) omega1 and omega2 satisfy the relation


1/omega1^2 +1/omega2^2 < L/g

6 years ago

Answers : (1)

vikas askiitian expert
510 Points
										

let the length of thread over the table is L1 & below the table is L2 ...


L1+L2 = L        ...........1


tension in upper part as wel as lower part will be same because the thread is massless  , let it beT ....


mass m is circulating with angular velocity W1 so


 T = mW12L1         ........2                  (L1 is the radius of circle)


(tension in the thread provides necessary centripital force )


now ,


          another mass is moving in conical pendulam , let it is making an angle @ with verticle then


 Tcos@ = mg         ..........3


 Tsin@ = mW22 R            ........4               (R is the radius of circular path)


 R = L2sin@


putting value of R in eq4 we get


 T = mW22L2            ........5


equating 2 & 5


 L1/L2 = W22/W12                                 ans  (a)


 1/W12 + 1/W22 = mL1/T + mL2/T                                  (from eq 2 & 5 )


                        =m(L1+L2)/T


                        =mL/T                                                  (L1+L2=L)


                       =Lcos@/g                                               ( substituting T from eq 3)


 cos@ = max =1 so


 1/W12+1/W22 < L/g                          ans (d )


thus option a,c,d are correct


 

6 years ago
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