Physics, asked by suni9346, 1 year ago

Derive the expression for total acceleration in the non uniform circular motion.

Answers

Answered by JinKazama1
20
Non-uniform Circular Motion :

Here, Speed changes with time.
=> In this case, we are throwing lights on when rare of change of speed is constant .

=> That is, Angular Acceleration is constant.
=> Net Acceratation lies inside the circle but does not passed through centre.
=> Net acceleration consists of two components - Radial and Transverse.


DERIVATION:
1) Let there be a point P at P(Rsin (a), R cos(a))
moving in a circle with angular acceleration α and angular velocity at that time be w


2) Position of particle :
OP = R( cos (a) , sin(a))

We will differentiate this to get velocity and then again to get acceleration.


3) At acceleration, we have two components.
Radial accerlation : Towards the centre
Tangential Acceleration : Along the tangent to circle.


4) a(net) = √(a(rad)^2 + a(tang)^2 )



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Answered by akashvibryd2003
5

Answer:

Explanation:

Non-uniform Circular Motion :

Here, Speed changes with time.

=> In this case, we are throwing lights on when rare of change of speed is constant .

=> That is, Angular Acceleration is constant.

=> Net Acceratation lies inside the circle but does not passed through centre.

=> Net acceleration consists of two components - Radial and Transverse.

DERIVATION:

1) Let there be a point P at P(Rsin (a), R cos(a))

moving in a circle with angular acceleration α and angular velocity at that time be w

2) Position of particle :

OP = R( cos (a) , sin(a))

We will differentiate this to get velocity and then again to get acceleration.

3) At acceleration, we have two components.

Radial accerlation : Towards the centre

Tangential Acceleration : Along the tangent to circle.

4) a(net) = √(a(rad)^2 + a(tang)^2 )

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