Question

the acceleration of the point on the rim of flywheel 1 m in diameter, if it makes 1200 revolutions per minute is

Answer 1

using some concept of circular motion we can easily solve this problem..

acceleration in uniform circular motion is equal to w^2 r towards the centre..

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hope it will help u

Answer 2

Answer:

Required acceleration = $800\pi^2ms^{-2}$

Explanation:

Given : Diameter of the flywheel = 1m

Radius of the flywheel = 1/2 m = 0.5m

Angular velocity = 1200 revolutions/minute = $\frac{1200*2\pi }{60} rad/sec =40\pi rad/sec$

General formula of acceleration is given by,

$a=-w^2r$

Distance of rim from the center = radius = 0.5m

Acceleration of the point on the rim = $|a|=w^2r=(40\pi)^2(0.5)=800{\pi}^2$

Acceleration = $=800\pi ^2 ms^{-2}= 7895.68ms^{-2}$

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