F16–5. a wheel has an angular acceleration of a = (0.5 u) rad> s 2 , where u is in radians. determine the magnitude of the velocity and acceleration of a point p located on its rim after the wheel has rotated 2 revolutions. the wheel has a radius of 0.2 m and starts at v0 = 2 rad> s.

The magnitude of the velocity of a point P located on its rim after 2 revolutions is V= 0.814 m/s.

The magnitude of the acceleration of a point P located on its rim after 2 revolutions is a= 3.31 m/s².

Explanation:

α= 0.5 rad/s²

r= 0.2 m

ωi= 2 rad/s

θf= 4π rad

a=?

Vt=?

θf=ωi * t + 1/2 * α * t²

clearing t:  (time to rotate 2 revolutions)

t= 4.14 s

ωf= ωi + α* t

ωf= 4.07 rad/s

Vt= ωf * r

Vt= 0.814 m/s

ac= ωf² * r

ac= 3.31m/s²

at= α * r

at= 0.1 m/s²

a= √(ac² + at²)

a= 3.31 m/s²

higher potential

explanation:

because it is in a high speed so it must move into a of high potential