Name: Condition for Resonance
Uploaded: 2019-07-19
Duration: 5 min 17 s
Description: Do you know you can shatter the glass goblet without touching it? Yes, it is possible when the sound of a specific frequency is produced. Let's find out the...

Question

A uniform rod is placed on two spinning wheels as shown in figure. The axes of the wheels are separated by a distance $l = 20 cm$ , the coefficient of friction between the rod and the wheels is, $k = 0.18$ . Demonstrate that in this case the rod performs harmonic oscillations. Find the period of these oscillations. Take $g = 9.8 m {s^{-}}^{2}$

Accepted Answer

In this question, the uniform rod is oscillating due to rotation of two wheels in the opposite direction of each other, and we have to prove that its harmonic oscillations, and then, we have to find out the time period of these oscillations.
In the equilibrium position, the centre of mass of the rod lies midway between the two rotating wheels. Let us displace the rod horizontally by some small distance and then release it. Let us depict the forces acting on the rod when its centre of mass is at a distance $x$ from its equilibrium position (as shown in the figure).

Since there is no net vertical force acting on the rod, Newton's second law gives,

$N_{1} + N_{2} = m g . . . . . (1)$

where, $N_{1} & N_{2}$ are the normal reactions at the contact of two wheels and $m$ is the total mass of the rod.
Now, considering the friction forces as shown in the diagram,
$\Rightarrow k N_{1} - k N_{2} = m a \Rightarrow k (N_{1} - m g + N_{1}) = m a$

(By using equation $1$ )

$\Rightarrow k (2 N_{1} - m g) = m a . . . . . . . . (2)$

where, $a$ is the total acceleration of the rod.

As the rod experiences no net torque about an axis perpendicular to the plane of the figure and passing through the centre of mass of the rod. That is, the torque provided due to both the wheels is balanced with each other. Hence, about the centre of mass,
$\Rightarrow N_{1} (\frac{l}{2} + x) = N_{2} (\frac{l}{2} - x) \Rightarrow N_{1} (\frac{l}{2} + x) = (m g - N_{1}) (\frac{l}{2} - x) \Rightarrow \frac{N_{1} l}{2} + N_{1} x = \frac{m g l}{2} - m g x - \frac{N_{1} l}{2} + N_{1} x \Rightarrow 2 N_{1} - m g = - \frac{2 m g x}{l} . . . . . . . . (3)$

Solving the equations $(1)$ , $(2)$ and $(3)$ simultaneously, we get,
$a = - k \frac{2 g}{l} x$

But, $a = - ω^{2} x$

Therefore,

$ω^{2} = \frac{2 k g}{l} \Rightarrow ω = \sqrt{\frac{2 k g}{l}}$

$\Rightarrow T = \frac{2 π}{ω} = 2 π \sqrt{\frac{l}{2 k g}} = π \sqrt{\frac{2 l}{k g}}$ .

Now, given that $l = 20 cm = 0.2 m, k = 0.18, g = 9.8 m s^{- 2}$

$\Rightarrow T = 3.14 \times \sqrt{\frac{2 \times 0.20}{0.18 \times 9.8}} = 1.5 s$

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