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