As we saw in
Applications of Newton’s Laws , objects moving at constant speed in a circle have a centripetal acceleration directed toward the center of the circle, which means that there must be a net force directed toward the center of that circle. Since all objects on the surface of Earth move through a circle every 24 hours, there must be a net centripetal force on each object directed toward the center of that circle.
Let’s first consider an object of mass
m located at the equator, suspended from a scale (
[link] ). The scale exerts an upward force
${\overrightarrow{F}}_{\text{s}}$ away from Earth’s center. This is the reading on the scale, and hence it is the
apparent weight of the object. The weight (
mg ) points toward Earth’s center. If Earth were not rotating, the acceleration would be zero and, consequently, the net force would be zero, resulting in
${F}_{\text{s}}=mg$ . This would be the true reading of the weight.
With rotation, the sum of these forces must provide the centripetal acceleration,
${a}_{\text{c}}$ . Using Newton’s second law, we have
Note that
${a}_{\text{c}}$ points in the same direction as the weight; hence, it is negative. The tangential speed
v is the speed at the equator and
r is
${R}_{\text{E}}$ . We can calculate the speed simply by noting that objects on the equator travel the circumference of Earth in 24 hours. Instead, let’s use the alternative expression for
${a}_{\text{c}}$ from
Motion in Two and Three Dimensions . Recall that the tangential speed is related to the angular speed
$\left(\omega \right)$ by
$v=r\omega $ . Hence, we have
${a}_{c}=\text{\u2212}r{\omega}^{2}$ . By rearranging
[link] and substituting
$r={R}_{\text{E}}$ , the apparent weight at the equator is
Substituting for the values or
${R}_{\text{E}}$ and
$\omega $ , we have
${R}_{\text{E}}{\omega}^{2}=0.0337\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$ . This is only 0.34% of the value of gravity, so it is clearly a small correction.
Zero apparent weight
How fast would Earth need to spin for those at the equator to have zero apparent weight? How long would the length of the day be?
Strategy
Using
[link] , we can set the apparent weight (
${F}_{\text{s}}$ ) to zero and determine the centripetal acceleration required. From that, we can find the speed at the equator. The length of day is the time required for one complete rotation.
Solution
From
[link] , we have
$\sum F={F}_{\text{s}}}-mg=m{a}_{\text{c}$ , so setting
${F}_{\text{s}}=0$ , we get
$g={a}_{\text{c}}$ . Using the expression for
${a}_{\text{c}}$ , substituting for Earth’s radius and the standard value of gravity, we get
We will see later in this chapter that this speed and length of day would also be the orbital speed and period of a satellite in orbit at Earth’s surface. While such an orbit would not be possible near Earth’s surface due to air resistance, it certainly is possible only a few hundred miles above Earth.
A computer is reading from a CD-ROM that rotates at 780 revolutions per minute.What is the centripetal acceleration at a point that is 0.030m from the center of the disc?
the specific heat of hydrogen at constant pressure and temperature is 14.16kj|k.if 0.8kg of hydrogen is heated from 55 degree Celsius to 80 degree Celsius of a constant pressure. find the external work done .
Many amusement parks have rides that make vertical loops like the one shown below. For safety, the cars are attached to the rails in such a way that they cannot fall off. If the car goes over the top at just the right speed, gravity alone will supply the centripetal force. What other force acts and what is its direction if: (a) The car goes over the top at faster than this speed? (b) The car goes over the top at slower than this speed?