InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Two identical balls (A) and (B) each of mass (0.1 kg), are attached to two identical massless springs. The spring - mass system is constrained to move inside a riged smooth pipe bant in the form of a circle as shown in Fig. The pipe is fixed in a horizontal plane. The centres of the balls can move in a circle of radius (0.06 pi) meter. Each spring has a natural length `zof0.06 pi` meter and spring spring constant5 `0.1 N//m`. Initially, both the balls are displaced by an angle `theta = pi// 6` radian with respect to the diameter (pQ) of the circle (as shown in Fig.) and released from, rest. . (i) Calculate the frequency of oscillation of ball (B). (ii) Find the speed of ball (A) when (A) and (B) are at the two ends of the diameter (PQ). (iii) What is the total energy of the system. |
|
Answer» Correct Answer - A::B::C::D (i) As both the balls displaced by `thetapi//6` radian with espect to the diameter `PQ` of the circle and released from rest. It results into compression of spring in upper segment and an equal elongation of spring in lower segment. Let it x. Pband QA denote x in the figure. Compression =Rtheta= elongation =x` :. Force exerted by each spring on each ball=2 kx` :. Total force on each ball due to two spring `4kx` :. Restoring torque about origin `=O=-(4kx)R` :. tau=-4k (Rtheta) R, where theta =Anfular displacement or `tau=-4kR^(2)theta` Since torque (tau) is proprtional to theta, each ball executes angular SHM about the centreO. Again, `tau=-4kR^(2)theta` or `l alpha=-4kR^(2)theta` where alpha=angular acceleration` or `(mR^(2))alpha=-4kR^(2)theta or alpha=-((4k)/m)theta` :. Frequency `f=1/2_(pi)sqrtalpha/theta` :. `Frequency of each ball=1/2_(pi)sqrt(4k)/m` `=1/2_(pi)sqrt(4xx0. 1)/(0.1)=1/(pi)sec^(-1)` (ii) Let velocity at the mean position be v_(max)` Loss in elastic potential energy=Gain in kinetic energy` `[1/2K(2R(pi)/2)]` `2[1/1xx0.1(0.02pi)^(2)] `=3.95xx10^(-4) J`. |
|