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 small identical discs, each of mass

m

, lie on a smooth horizontal plane. The discs are interconnected by a light non-deformed spring of length

l_{0}

and stiffness

χ

. At a certain moment one of the discs is set in motion in a horizontal direction perpendicular to the spring with velocity

v_{0}

. Find the maximum elongation of the spring in the process of motion, if it is known to be considerably less than unity.

Accepted Answer

We know that the linear momentum of the system in the centre of the mass frame is zero. Hence, for the given system, we have,

${\vec{p}}_{1} + {\vec{p}}_{2} = \vec{0} \Rightarrow {\vec{p}}_{1} = - {\vec{p}}_{2}$

Clearly, $|{\vec{p}}_{1}| = |{\vec{p}}_{2}| = μ v_{rel} . . . (1)$

The kinetic energy of the system in the centre of the mass frame is given by,

$K = \frac{1}{2} μ v_{rel}^{2} . . . (2)$

where $μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}} =$ reduced mass

and $v_{rel} = |\vec{v_{1}} - \vec{v_{2}}|$ .

Here, $μ = \frac{m m}{m + m} = \frac{m}{2} . . . (3)$

The initial kinetic energy in the centre of the mass frame is,

$K_{1} = \frac{1}{2} μ v_{rel}^{2} = \frac{1}{2} (\frac{m}{2}) {(v_{0} - 0)}^{2} \Rightarrow K_{1} = \frac{m}{4} v_{0}^{2} . . . (4)$

Initial potential energy $U_{1} = 0$ .

Elongation in the spring is maximum, when both the discs move with same projected velocity, i.e., their projected relative velocity is zero. Let the maximum elongation be $x$ .

Since, the system is a conservative closed system, the mechanical energy in the centre of the mass frame remains constant. So,

$K_{1} + U_{1} = K_{2} + U_{2} \Rightarrow \frac{m}{4} v_{0}^{2} = \frac{1}{2} χ x^{2} + \frac{1}{2} (\frac{m}{2}) v_{rel}^{2} \Rightarrow \frac{m v_{0}^{2}}{2} = χ x^{2} + \frac{m}{2} v_{rel}^{2} . . . (5)$

As the net torque about the centre of mass of the system is zero, so we can apply the law of conservation of angular momentum about the centre of mass. Hence, we have,

$2 (\frac{l_{0}}{2}) (\frac{m}{2} v_{0}) = 2 (\frac{l_{0} + x}{2}) \frac{m}{2} v_{rel} \Rightarrow v_{rel} = \frac{v_{0} l_{0}}{(l_{0} + x)} \Rightarrow v_{rel} = v_{0} {(1 + \frac{x}{l_{0}})}^{- 1} \Rightarrow v_{rel} = v_{0} (1 - \frac{x}{l_{0}}) . . . (6)$

[using binomial expansion as $x ≪ l_{0}$ .]

Putting $(6)$ into $(1)$ , we get,

$\frac{m v_{0}^{2}}{2} = χ x^{2} + \frac{m}{2} {[v_{0} (1 - \frac{x}{l_{0}})]}^{2}$

$\Rightarrow \frac{1}{2} m v_{0}^{2} [1 - {(1 - \frac{x}{l_{0}})}^{2}] = χ x^{2}$

$\Rightarrow \frac{1}{2} m v_{0}^{2} [1 - (1 - \frac{2 x}{l_{0}})] = χ x^{2}$

[using binomial expansion as $x ≪ l_{0}$ .]

$\Rightarrow m v_{0}^{2} \frac{x}{l_{0}} = χ x^{2}$

Since the maximum elongation $x$ cannot be zero, so,

$x = \frac{m v_{0}^{2}}{χ l_{0}}$ .

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