A 2 kg steel rod of length 0.6 m is clamped on a table vertically at its lower end and is free to rotate in vertical plane. The upper end is pushed so that the rod falls under gravity, ignoring the friction due to clamping at its lower end, the speed of the free end of rod when it passes through its lowest position is \underline\hspace2cm ms - 1. (Take g = 10 ms - 2)

Rotational Motion PYQ — JEE Main 2021 · 1 Sep · Shift 2 · Q73 with Solution

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Correct answer: 6

Given data

Mass of rod: $m = 2\,\text{kg}$
Length of rod: $L = 0.6\,\text{m}$
Acceleration due to gravity: $g = 10\,\text{m s}^{-2}$

The rod is hinged at its lower end and falls from the vertical upward position to the vertical downward position.

Use conservation of mechanical energy

As the rod rotates without friction about the lower end, loss in gravitational potential energy equals gain in rotational kinetic energy.

Change in height of center of mass

For a uniform rod, the center of mass is at its midpoint, i.e. at a distance $L/2$ from the hinge.

Initially, center of mass is at height $+L/2$ above the hinge.
Finally, when rod is at lowest position, center of mass is at height $-L/2$ below the hinge.

So total drop in height of center of mass is

\Delta h = \frac{L}{2} + \frac{L}{2} = L

Hence decrease in potential energy is

\Delta U = mgL

Rotational kinetic energy at lowest position

Moment of inertia of a uniform rod about one end is

I = \frac{1}{3}mL^2

Thus rotational kinetic energy is

K = \frac{1}{2}I\omega^2 = \frac{1}{2}\left(\frac{1}{3}mL^2\right)\omega^2

By energy conservation,

mgL = \frac{1}{2}\cdot \frac{1}{3}mL^2\omega^2

Cancel $m$ :

gL = \frac{1}{6}L^2\omega^2

\omega^2 = \frac{6g}{L}

Speed of the free end

The free end is at distance $L$ from the hinge, so

v = \omega L

Therefore,

v^2 = \omega^2 L^2 = \frac{6g}{L}L^2 = 6gL

Substitute $g=10$ and $L=0.6$ :

v^2 = 6 \times 10 \times 0.6 = 36

v = 6\,\text{m s}^{-1}

Final answer

The speed of the free end at the lowest position is

\boxed{6}

More from Rotational Motion

View written solutionFree

Correct answer: 6

Given data

Mass of rod: $m = 2\,\text{kg}$
Length of rod: $L = 0.6\,\text{m}$
Acceleration due to gravity: $g = 10\,\text{m s}^{-2}$

The rod is hinged at its lower end and falls from the vertical upward position to the vertical downward position.

Use conservation of mechanical energy

As the rod rotates without friction about the lower end, loss in gravitational potential energy equals gain in rotational kinetic energy.

Change in height of center of mass

For a uniform rod, the center of mass is at its midpoint, i.e. at a distance $L/2$ from the hinge.

Initially, center of mass is at height $+L/2$ above the hinge.
Finally, when rod is at lowest position, center of mass is at height $-L/2$ below the hinge.

So total drop in height of center of mass is

\Delta h = \frac{L}{2} + \frac{L}{2} = L

Hence decrease in potential energy is

\Delta U = mgL

Rotational kinetic energy at lowest position

Moment of inertia of a uniform rod about one end is

I = \frac{1}{3}mL^2

Thus rotational kinetic energy is

K = \frac{1}{2}I\omega^2 = \frac{1}{2}\left(\frac{1}{3}mL^2\right)\omega^2

By energy conservation,

mgL = \frac{1}{2}\cdot \frac{1}{3}mL^2\omega^2

Cancel $m$ :

gL = \frac{1}{6}L^2\omega^2

\omega^2 = \frac{6g}{L}

Speed of the free end

The free end is at distance $L$ from the hinge, so

v = \omega L

Therefore,

v^2 = \omega^2 L^2 = \frac{6g}{L}L^2 = 6gL

Substitute $g=10$ and $L=0.6$ :

v^2 = 6 \times 10 \times 0.6 = 36

v = 6\,\text{m s}^{-1}

Final answer

The speed of the free end at the lowest position is

\boxed{6}

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