The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the plane of the equator. assuming the earth is a sphere with a radius of 6.38 × 106 m, determine the speed and centripetal acceleration of a person situated (a) at the equator and (b) at a latitude of 30.0° north of the equator.

Answer

given,

Radius of earth = 6.38 x 10⁶ m

angular velocity  =

                            =

                            =

a) at equator

velocity v   = R ω  = 6.38 x 10⁶ x 7.27 x 10⁻⁵

             v = 463.8 m/s

centripetal acceleration = R ω² = 6.38 x 10⁶ x (7.27 x 10⁻⁵)²

                                        = 0.0337 m/s²

b) at a latitude of 30.0° north of the equator

r = R cos θ  =  6.38 x 10⁶ cos 30⁰ =  3.19 x  10⁶ m

velocity v   = R ω  = 3.19 x  10⁶ x 7.27 x 10⁻⁵

            v = 231.9 m/s

centripetal acceleration = R ω² = 3.19 x  10⁶  x (7.27 x 10⁻⁵)²

                                        = 0.0168 m/s²

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm