A sculptor is sharpening a chisel on grindstone of radius 1.0 m that is spinning with a constant angular speed of 2.0 rad/s. What is the magnitude of the centripetal acceleration of a point on the rim of the grindstone?

Respuesta :

Answer:

centripetal acceleration will be equal to 4 [tex]m/sec^2[/tex]      

Explanation:

We have given radius r = 1 m

Constant angular speed [tex]\omega =2rad/sec[/tex]

We have to find the centripetal acceleration

Linear velocity id equal to [tex]v=\omega r=2\times 1=2m/sec[/tex]

Centripetal acceleration will be equal to [tex]a=\frac{v^2}{r}=\frac{2^2}{1}=4m/sec^2[/tex]

So centripetal acceleration will be equal to 4 [tex]m/sec^2[/tex]