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A disc of radius R has a light pole fixed perpendicular to the disc at the circumference which in turn has a pendulum of length R attached to its other end as shown in figure. The disc is rotated with constant angular velocity omega . The string is making an angle 30° with the rod . Then the angular velocity omega of disc is....

Shaswata Biswas
132 Points
4 years ago
The ball suspended with a thread is scted upon by two forces : its weight mg, and tension in the string T. The resultant of these two forces produces the necessary centrepetal force along the horizontal so as to make the ball move along a cirular path of radius l (say).
Then, l = R + R.sin45 = $R[\frac{1}{ \sqrt{2}} + 1]$
And, on the ball :
• Weight of the ball mg acts vertically downwards
• Centrepetal force mw2l acts radially inward
• Tension acts along the string of pendulum.
Now, $\dpi{120} \tan 30^{o} = \frac{m \omega ^{2} l}{mg} = \frac{\omega^{2}l}{g}$
Or, $\dpi{120} \omega^{2} = \frac{gtan30^{o}}{l} = gtan30^{o} \div R[ \frac{1}{ \sqrt{2}} + 1]$
So the required angular velocity, $\dpi{120} \omega = \sqrt{gtan30^{o} \div R[ \frac{1}{ \sqrt{2}} + 1]}$
THANKS
Vaibhavi
12 Points
4 years ago
The answer is given [2g/3√3R]^1/2 ..
Shaswata Biswas
132 Points
4 years ago

The ball suspended with a thread is scted upon by two forces : its weight mg, and tension in the string T. The resultant of these two forces produces the necessary centrepetal force along the horizontal so as to make the ball move along a cirular path of radius l (say).
Then, l = R + R.sin30 = $\frac{R}{ 2} + R = \frac{3R}{2}$
And, on the ball :
• Weight of the ball mg acts vertically downwards
• Centrepetal force mw2l acts radially inward
• Tension acts along the string of pendulum.
Now, $\dpi{120} \tan 30^{o} = \frac{m \omega ^{2} l}{mg} = \frac{\omega^{2}l}{g}$
Or, $\omega^{2} = \frac{gtan30^{o}}{l} = gtan30^{o} \div \frac{3R}{ 2} = \frac{2g}{3 \sqrt{3} R}$
So the required angular velocity,
$\omega = \sqrt{ \frac{2g}{3 \sqrt{3} R}}$
Identify the mistake I've done before and forgive me for that.
THANKS