Question

Why do we say that a particle moving in a circle with a uniform speed still has an acceleration. Derive an expression for this accele ration

Answer 1

Answer:

That's because rate of change in velocity is defined as acceleration and because of continuous change in direction the velocity in circular motion is said to be changing over time.

The acceleration is given by,

$\frac{mv^{2} }{r}$

Where m is mass, v is speed of the body and r is the distance from center i.e radius.

Explanation:

Consider the attachment,

We have distance(r) split into two components, i.e r sinθ + r cosθ,

Now,

$\frac{d(rcosθ +r sinθ)}{dt} \\=\frac{d(r cosθ)}{dt} +\frac{d(r sinθ)}{dt} \\ \\Solving this using chain reaction,=r(-sinθ)\frac{dθ}{dt} + rcosθ \frac{dθ}{dt} \\=-ωr(sinθ -cosθ)\\which is equal to v (velocity).$

Differentiating the terms again to get acceleration,

$\frac{dv}{dt}= \frac{d(-ωr (sinθ-cosθ))}{dt} \\\\= -ω^{2} r(cosθ +sinθ)\\=-ω^{2} r$

Now we have, ω=$\frac{v}{r}$

Therefore (considering the magnitude only),

$\frac{v^{2} }{r^{2} } (r)\\ =\frac{v^{2} }{r}$

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