Let v and a be the instantaneous velocity and acceleration of a particle moving in a circle. Then what is the rate of change of speed dv/dt of the particle?
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In a circle the particle moves with a uniform speed | v |.
a = instantaneous acceleration = dv / dt
Let the center of circle be O. Let at t = t, the particle be at A (r , 0), where r = radius of the circle. At time t = t +dt the particle moves to B by dθ radians. Let dv be the change in the velocity from v to v+dv. The angle between these two vectors is dθ.
If we construct a triangle with vectors v and v+dv, then dv is the third side. That is equal to v dθ = length of the arc in the infinitesimally small triangle. Magnitude of the vector v+dv = | v | = constant speed.
dv = v dθ
dv/dt = v dθ/dt = v ω = r ω² = v²/r = a , as v = r ω
a = instantaneous acceleration = dv / dt
Let the center of circle be O. Let at t = t, the particle be at A (r , 0), where r = radius of the circle. At time t = t +dt the particle moves to B by dθ radians. Let dv be the change in the velocity from v to v+dv. The angle between these two vectors is dθ.
If we construct a triangle with vectors v and v+dv, then dv is the third side. That is equal to v dθ = length of the arc in the infinitesimally small triangle. Magnitude of the vector v+dv = | v | = constant speed.
dv = v dθ
dv/dt = v dθ/dt = v ω = r ω² = v²/r = a , as v = r ω
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