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If a point moves along a circle with con...

If a point moves along a circle with constant speed, prove that its angular speed about any point on the circle is half of that about the centre.

A

`2omega_(c)=omega_(0)`

B

`omega_(c)=omega_(0)`

C

`omega_(c)=2omega_(0)`

D

`omega_(c)=(1/2)omega_(0)`

Text Solution

Verified by Experts

The correct Answer is:
C
Let, `O` be a point on a circle and `P` be the position of the particle at any time `T` , such that
`/_POA=theta` . Then, `/_PCA=20`
Here, `C` is the centre of the circle.
Angular velocity of `P` about `O` is `omega_(0)=(d theta)/(dt)`
and angular velocity of `P` about `C` is,
`omega_(c)=(d)/(dt)(2theta)=2(d theta)/(dt)`
or `omega_(c)=2omega_(0)`
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