Question

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?

Answer 1

centripetal acceleration will be equal to 4
`m/sec^2`

Step-by-step explanation:

We have given radius r = 1 m

Constant angular speed
`\omega =2rad/sec`

We have to find the centripetal acceleration

Linear velocity id equal to
`v=\omega r=2* 1=2m/sec`

Centripetal acceleration will be equal to
`a=(v^2)/(r)=(2^2)/(1)=4m/sec^2`

So centripetal acceleration will be equal to 4
`m/sec^2`