Name: Two Block System
Uploaded: 2019-07-19
Duration: 6 min 20 s
Description: Two blocks placed on a surface exerts a force between them and also with the surface. Let us discuss how these forces can be solved numerically in this video.

Question

A ring of mass $m$ can slide along a fixed rough vertical rod as shown in the figure. The ring is connected by a spring of spring constant $k = \frac{4 m g}{R}$ where $2 R$ is the natural length of spring. The other end of the spring is fixed to the ground at a point $A$ at a horizontal distance of $2 R$ from the base of the rod. If the ring is released from a height of $3 R / 2$ and it reaches the ground with a speed $\sqrt{3 g R}$ , find a co-efficient of friction between the rod and ring.

Accepted Answer

Given,

Force constant of a spring $k = \frac{4 m g}{R}$ ,

The natural length of the spring is $2 R$ ,

Mass of the ring is $m$ and

Speed of the ring when it reaches the ground $v = \sqrt{3 g R}$ .

When the spring makes an angle $θ$ with the horizontal as shown.

Elongation in the spring, $x = \frac{2 R}{\cos θ} - 2 R$

$\Rightarrow x = 2 R (\frac{1}{\cos θ} - 1)$ .

From the free-body diagram of the ring, along the vertical,

$m g + k x \sin θ - F_{r} = m a$

$\Rightarrow m g + k x \sin θ - μ N = m a$

$\Rightarrow m g + k x (\sin θ - μ \cos θ) = m a$

$\Rightarrow m g + k 2 R (\frac{1}{\cos θ} - 1) (\sin θ - μ \cos θ) = m \frac{d v}{d t}$

$\Rightarrow m g + k 2 R (\frac{\sqrt{4 R^{2} + y^{2}}}{2 R} - 1) (\frac{y}{\sqrt{4 R^{2} + y^{2}}} - μ \frac{2 R}{\sqrt{4 R^{2} + y^{2}}}) = - m \frac{d v}{d y} \frac{d y}{d t}$

$\Rightarrow \int_{\frac{3 R}{2}}^{0} (m g + k (\sqrt{4 R^{2} + y^{2}} - 2 R) (\frac{y - 2 μ R}{\sqrt{4 R^{2} + y^{2}}})) d y = - m \int_{0}^{\sqrt{3 g R}} v d v$

$\Rightarrow \int_{\frac{3 R}{2}}^{0} m g d y + k \int_{\frac{3 R}{2}}^{0} (y - 2 μ R) d y - 2 k R \int_{\frac{3 R}{2}}^{0} (\frac{y - 2 μ R}{\sqrt{4 R^{2} + y^{2}}}) d y = - \frac{3 m g R}{2}$

$\Rightarrow m g (- \frac{3 R}{2}) + k (- \frac{9 R^{2}}{8}) - 2 μ k R (- \frac{3 R}{2}) - 2 k R \int_{\frac{3 R}{2}}^{0} (\frac{y}{\sqrt{4 R^{2} + y^{2}}}) d y + 4 k μ R^{2} \int_{\frac{3 R}{2}}^{0} (\frac{1}{\sqrt{4 R^{2} + y^{2}}}) d y = - \frac{3 m g R}{2}$

$\Rightarrow m g (- \frac{3 R}{2}) + \frac{4 m g}{R} (- \frac{9 R^{2}}{8}) - 2 μ R (\frac{4 m g}{R}) (- \frac{3 R}{2}) - 2 R (\frac{4 m g}{R}) (- \frac{R}{2}) + 4 μ R^{2} (\frac{4 m g}{R}) (- \ln (2)) = - \frac{3 m g R}{2}$

$\Rightarrow μ (12 m g R - 16 m g R (\ln (2))) = \frac{m g R}{2}$

$\Rightarrow μ = \frac{m g R}{8 m g R (3 - 4 (\ln (2)))}$

$\Rightarrow μ = \frac{1}{8 (3 - 4 (\ln 2))}$ .

Important Questions on Friction

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