question_answer

A toy car travels in a horizontal circle of radius 2a, kept on the track by a radial elastic string of unstretched length a. The period of rotation is T. Now the car is speeded up until it is moving in a circle of radius 3a. Assuming that the string obeys Hookes law then the new period will be

A)  \[\sqrt{\frac{4}{3}}T\]                  

B)  \[\frac{{{3}^{2}}}{{{4}^{2}}}T\]

C)  \[\frac{\sqrt{3}}{2}T\]                  

D)  \[\frac{3}{4}T\]

Correct Answer: C

Solution :

                 For a body executing circular motion, centripetal force is given by                                 \[F=mr{{\omega }^{2}}=mr{{\left( \frac{2\pi }{T} \right)}^{2}}\]                 ... (i) If k is force constant of string, then elastic force                 F = ka                                                    ... (ii) From Eqs. (i) and (ii), we get                 \[ka+m\,(2a)\,{{\left( \frac{2\pi }{T} \right)}^{2}}\]                          ... (iii) In second case New length of string                                 = new radius of circle = 3d Stretching of string = 3a - a = 2 a Hence,  elastic force = k . 2 a So,          \[k\,.\,2a=m\,(3\,a)\,{{\left( \frac{2\,\pi }{T} \right)}^{2}}\]                         ... (iv) Dividing Eq. (iv) by Eq. (iii), we get                                 \[2=\frac{3}{2}{{\left( \frac{T}{T} \right)}^{2}}\] \[\Rightarrow \]               \[{{\left( \frac{T}{T} \right)}^{2}}=\frac{4}{3}\] \[\Rightarrow \]               \[T=\frac{\sqrt{3}}{2}\,T\]