Question

a grinding wheel of radius 7.6 cm is rotating at 1750 rpm.what is the speed of a point on the outer edge of the wheel?what is the centripetal accleration of the point?​

Answer 1

Given that,

Radius = 7.6 cm

frequency = 1750 rpm

We need to calculate the speed of a point on the outer edge

Using formula of speed

$v = r\omega$

Put the value into the formula

$v= 7.6\times10^{-2}\times\dfrac{2\pi\times1750}{60}$

$v=13.93\ m/s$

We need to calculate the centripetal acceleration of the point

Using formula of acceleration

$a_{c}=\dfrac{v^2}{r}$

Put the value into the formula

$a_{c}=\dfrac{13.92^2}{7.6\times10^{-2}}$

$a_{c}=2553.2\ m/s^2$

Hence, The speed of a point on the outer edge is 13.93 m/s.

The centripetal acceleration of the point is 2553.2 m/s²

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Topic : centripetal acceleration

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