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Physics Centripetal force and Centripetal Acceleration

Centripetal force

Centripetal force is the force required to move a body uniformly in circle, this force act along the radius towards the center of circle.
If the body of mass m is moving in horizontal circle of radius of r with constant speed v the equation for centripetal force is given by
Fc = mv2/r
Fc= mr2ω2/r
Fc= mrω 2

#Derivation Of Centripetal Acceleration And Force

Derivation Of Centripetal Acceleration And Force
Consider a body of mass m is moving in horizontal circle with a constant speed v and angular velocity ω at time t.
The particle is at point p were radius vector op= r after the time t +ΔT particle reached at point Q and angle made by the radius vector at the center of the circle is Δ θ
Angular velocity ω = Δ θ / Δ T
Δ θ = ω Δ T --------------------(2)
Let V1 and V2 be the velocity vector at location P and Q respectively.
In vector of the body is always directed towards the tangent.
In circular motion the speed of the particle remains constant.
|PA| = |QB| = |V|
|V1| = |V2| = |V|  ----------------(3)
Centripetal forceTo calculate the change in velocity of particle in moving from P to Q in time Δ T.
For the calculation we take on external point P1 and we draw P1A1 and P1B1 parallel to PA and PB respectively.
 ∴ ∠ A1 P1 B1 = Δθ
From triangle law of vector
 P1A1 + P1B1 = P1B1
A1 B1 = P1B1 - P1A1
A1 B1 = V2 – V1 = ΔV --------------------------(4)
Centripetal accelerationIf ΔT = 0 then A1 and B1 comes closer to each other and the point can be contact by an arc of circle.
∴ Angle Δθ = A1 B1 / P1 A1  = ΔV/|V|
Putting the value of Δθ form equation (2)
ω.Δt = ΔV/|V|
ω.|V| = ΔV/| Δt |
ωrw = rω2 ---------------(5)
If  LT ΔV/Δt  the instantaneous Δt = 0 acceleration at particle at point P which called centripetal acceleration.
∴ ac = 2  ----------------------(6)
According to Newton 2nd law of the magnitude
Fc = mac = mrω2  ------------------------(7)

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