help@askiitians.com

+91-735-322-1155

General Physics> A disc of radius R has a light pole fixed...

question mark

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....

Vaibhavi , 9 Years ago

Grade 11

3 Answers

Shaswata Biswas

Last Activity: 9 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.sin45^০ = $R[\frac{1}{ \sqrt{2}} + 1]$

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, $\tan 30^{o} = \frac{m \omega ^{2} l}{mg} = \frac{\omega^{2}l}{g}$

Or, $\omega^{2} = \frac{gtan30^{o}}{l} = gtan30^{o} \div R[ \frac{1}{ \sqrt{2}} + 1]$

So the required angular velocity, $\omega = \sqrt{gtan30^{o} \div R[ \frac{1}{ \sqrt{2}} + 1]}$

THANKS

Vaibhavi

Last Activity: 9 Years ago

The answer is given [2g/3√3R]^1/2 ..

Shaswata Biswas

Last Activity: 9 Years ago

Your answer is correct.

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.sin30০ = $\frac{R}{ 2} + R = \frac{3R}{2}$

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, $\tan 30^{o} = \frac{m \omega ^{2} l}{mg} = \frac{\omega^{2}l}{g}$

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

LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

Full Live Access

Study Material

Live Doubts Solving

Daily Class Assignments

In viewing through a magnifying glass, one usually positions one's eyes very close to the lens. Does angular magnification change if the eye is moved back?

general physics

1 Answer Available

Last Activity: 2 Months ago

moment of inertia of a equilateral triangle about an axis passing through centre and perppendicular to the plane

general physics

1 Answer Available

Last Activity: 3 Years ago

. Because of the rotation of the earth, a plumb bob may not hang exactly along the direction of
the earth’s gravitational pull (its weight) but deviates slightly from this direction. Calculate the
deviation (a) at 40 latitude, (b) at the poles, and (c) at the equator

general physics

1 Answer Available

Last Activity: 3 Years ago

The length of a massless rod is measured using vernier callipers having 10 vernier scale divisions and least count of 0.1 mm. Two masses 1.00 ± 0.01 kg and 4.00 ± 0.02 kg are attached to its ends. If reading of vernier scale is 5 divisions coinciding with 1.3 cm, what is the % error in locating centre of mass from lighter mass

general physics

2 Answers Available

Last Activity: 3 Years ago

A block is kept on x-y plane at a point (2m,3m). It is pushed with a force of magnitude 10 Ntowards point (5m,7m). The force can be represented as?

general physics

1 Answer Available

Last Activity: 3 Years ago

View all Questions