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In the solution of above question, they solved it by moving from ground frame to the frame of centre of mass. But we know that, to move from one frame to another ,we have to subtract the mass-acceleration product. But in the above case, pseudoforce is applied in two opposite direction. Is their any other approach to solve it from ground frame?

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  • $\begingroup$ Sure, just set up Newton's 2nd law on the block M and then on the block m, and then remember that their accelerations are equal. $\endgroup$ – Steeven Jul 12 '18 at 20:36
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    $\begingroup$ FYI, your title doesn't really make sense. The pseudo force does not act in the opposite direction when in the ground frame (inertial frame) compared to in the centre-of-mass frame (non-inertial frame), because there is no pseudo force in the ground frame. That pseudo force is an addition for non-inertial frames only, because Newton's laws otherwise wouldn't hold true in non-inertial frames. $\endgroup$ – Steeven Jul 12 '18 at 20:39
  • $\begingroup$ There is no need to consider conceptually changing frame 0f references to solve this, you can just try to calculate the difference between F and f when m reaches the same speed as M. Although actually it's the same calculation. $\endgroup$ – JMLCarter Jul 12 '18 at 20:46
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From the ground frame, let $X$ represent the displacement of the mass M to the right (from its original location), and let $x$ represent the displacement of the mass m to the right (from its original location). Then, from free body diagrams on each of the two masses, the force balances are: $$M\frac{d^2X}{dt^2}=F-k(X-x)$$and$$m\frac{d^2x}{dt^2}=+k(X-x)$$

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  • $\begingroup$ And how do I find acceleration of particular body? $\endgroup$ – The ReBel Jul 13 '18 at 4:45

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