Name: Inertial Frames of Reference
Uploaded: 2019-07-19
Duration: 7 min 10 s
Description: Imagine if you are on a bus or a train, can you say you are in an inertial frame of reference? If yes, how can you predict it? Let us explore more in this vi...

Question

Two bars connected by a weightless spring of stiffness $x$ and length (in the non-deformed state) $l_{0}$ rest on a horizontal plane. A constant horizontal force $F$ starts acting on one of the bars as shown in figure. Find the maximum and minimum distances between the bars during the subsequent motion of the system, if the masses of the bars are,

$(a)$ equal,

$(b)$ equal to $m_{1}$ and $m_{2}$ , and the force $F$ is applied to the bar of mass $m_{2}$ .

Accepted Answer

Here, let us consider both blocks and spring as the physical system. Since the external force $F$ is acting the system, the centre of mass of the system will move with acceleration $a = \frac{F}{m_{1} + m_{2}}$ towards the right. Let us observe the masses from the frame of the centre of mass of the spring-mass system. As this frame is a non-inertial frame (which is accelerated with respect to the ground), we have to apply a pseudo force $m_{1} a$ towards the left on the block $m_{1}$ and $m_{2} a$ towards the left on the block $m_{2}$ .

As the centre of mass is at rest in the COM-frame, the blocks move in the opposite directions and come to instantaneous rest at some instant. The elongation of the spring will be maximum or minimum at this instant. Assume that the block $m_{1}$ is displaced by the distance $x_{1}$ and the block $m_{2}$ through a distance $x_{2}$ from the initial positions. From the energy equation given in the COM-frame, the work-energy theorem for this system will be, the change in kinetic energy of blocks + change in potential energy of spring = Work done by an external force,

$Δ K + U = W_{ext} . . . (1)$ (where $W_{ext}$ also includes the work done by the pseudo forces.)

Here, $Δ K = 0$ , because initially and finally blocks are at rest.

and $U = \frac{1}{2} k {(x_{1} + x_{2})}^{2}$ ,

$W_{ext} =$ (External force on $m_{2} \times$ Displacement of $m_{2}$ ) $+$ (External force on $m_{1} \times$ Displacement of $m_{1}$ )
$W_{ext} = (F - \frac{m_{2} F}{m_{1} + m_{2}}) x_{2} + \frac{m_{1} F}{m_{1} + m_{2}} x_{1} = \frac{m_{1} F (x_{1} + x_{2})}{m_{1} + m_{2}}$ ,

Write down the equation $(1)$ with putting the values of $U$ and $W_{ext}$ .

$U = W_{ext} \Rightarrow \frac{1}{2} k {(x_{1} + x_{2})}^{2} = \frac{m_{1} (x_{1} + x_{2}) F}{m_{1} + m_{2}}$

So, $x_{1} + x_{2} = 0$ or $x_{1} + x_{2} = \frac{2 m_{1} F}{k (m_{1} + m_{2})}$

Hence, the maximum separation between the blocks = $l_{0} + \frac{2 m_{1} F}{k (m_{1} + m_{2})}$ .

$(a)$ When $m_{1} = m_{2}$
Maximum separation $= l_{0} + \frac{2 m_{1} F}{k (m_{1} + m_{1})} = l_{0} + \frac{F}{k}$

The minimum separation corresponds to zero elongation (initial elongation) and is equal to $l_{0}$ .

$(b)$ For masses values of $m_{1}$ and $m_{2}$ , the maximum separation $= l_{0} + \frac{2 m_{1} F}{k (m_{1} + m_{2})}$ . The minimum separation corresponds to zero elongation (initial elongation) and is equal to $l_{0}$ .

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