The edge of a flying disc with a radius of 0.13 m spins with a tangential speed of 3.3 m/s.

What is the centripetal acceleration of the disc edge in m/s^2

Respuesta :

a = v²/r

v = 3.3 m/s
r = 0.13 m

a: centripetal acceleration

By definition, centripetal acceleration is given by:

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

Where,

v: tangential disk speed

r: disk radius

Substituting values in the given equation we have:

[tex] a =\frac{3.3^2}{0.13}

a = 83.76923077
[/tex]

Rounding the result we have:

[tex] a = 83.8 \frac{m}{s^2} [/tex]

Answer:

The centripetal acceleration of the disc edge in m/s^2 is:

[tex] a = 83.8 \frac{m}{s^2} [/tex]