Centripetal Acceleration metres per second squared

[Chica1975]

Homework Statement

What is the magnitude of the centripetal acceleration of a point on the circumference of a 0.32m radius bicycle wheel rotating at 22 rad/s ?

Give your answer in metres per second squared

Homework Equations

To be honest I am stumped I have no idea

The Attempt at a Solution

I am lost.

Please help!

 

[issacnewton]

centripetal acceleration is given by

[tex]a_c=\frac{v^2}{r}[/tex]

where r is the radius which you can find since circumference is given. v is tangential velocity
you have been given angular velocity [tex]\omega[/tex]. the relation between
tangential velocity and angular velocity is

[tex]v=r\omega[/tex]

 

[Chica1975]

Thank you!

 

[Chica1975]

thanks so much for explaining why this is so clearly!

 

[rwisz]

I can help you with this problem. The centripetal acceleration, a, of a point on the circumference of a rotating object is given by the equation a = v^2/r, where v is the linear velocity of the point and r is the radius of the rotation. In this case, the linear velocity, v, can be calculated by multiplying the angular velocity, ω, by the radius, r. So, v = ωr = (22 rad/s)(0.32 m) = 7.04 m/s. Plugging this value into the equation for centripetal acceleration, we get a = (7.04 m/s)^2 / 0.32 m = 155.52 m/s^2. Therefore, the magnitude of the centripetal acceleration is 155.52 m/s^2. I hope this helps!