Question

It takes 23 hours 56 minutes and 4 seconds for the earth to make one revolution (mean sidereal day). What is the angular speed of the earth?

angular speed = 7.29x10^-5 rad/s
Assume the earth is spherical. Relative to someone on the rotation axis, what is the linear speed of an object on the surface if the radius vector from the center of the earth to the object makes an angle of 49.0° with the axis of rotation. The radius of the earth is 6.37×103 km.
linear speed = 3.51 x 10^2 m/s
* What is the acceleration of the object on the surface of the earth in the previous problem?

Answer 1

Answer: See below

Step-by-step explanation:

Part (a):

The angular speed of the Earth can be calculated as,

`\omega &= \frac{{2\pi }}{T}\\ \omega &= \frac{{2\pi }}{{\left( {23 * 3600} \right)\left( {56 * 60} \right) + 4}}\\ \omega &= 7.29 * {10^( - 5)}\;{\rm{rad/s}}`

Part (b):

The linear velocity of the object can be calculated as,

`v &= \omega r\sin \theta \\ v &= \left( {7.29 * {{10}^( - 5)}} \right)\left( {6.37 * {{10}^6}} \right)\sin 47^\circ \\ v &= 339.62\;{\rm{m/s}}`

Part (c):

The acceleration of the object can be calculated as,

`a &= \frac{{{v^2}}}{{r\sin \theta }}\\ a &= \frac{{{{339.62}^2}}}{{6.37 * {{10}^6}\sin 47^\circ }}\\ a &= 0.024\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}`