Question

Four billion years ago, the day (as defined by Earth’s rotation about its own axis) was about 14 hours long.

a. What is the Earth’s average angular acceleration during the past four billion years? [Hint: you have Ti and Tf. From these, find ωi and ωf and then apply the rotational kinematic equations.]

b. This slowing is due to tides. What is the average torque applied to the Earth due to tides during the last four billion years? [Hint: assume that the torque is due to a tangential force applied to a spherical Earth.]

Answer 1

a)
calculate angular frequency using 2pi/T where T = 14 hrs (convert it into seconds)
calculate both: acceleration = (wf - wi)/4 billion yrs in seconds

b) Torque = Inertia x angular accn (use this to calculate inertia of solid sphere)
http://upload.wikimedia.org/wikipedia/en/math/2/5/4/254af3f927c4a1294f2d234fa57fe6f6.png

though i never think tides would do such a thing.

Answer 2

To solve this, you really need to recall the definitions of angular frequency \(\omega\), frequency \(f\), period \(T\), angular velocity \(\omega\), and angular acceleration \(\alpha\). Take note that in the special case of uniform circular motion, angular frequency \(\omega\) is quantitatively equivalent to angular velocity \(\omega\). It is unfortunate that we denote these two quantities with the same letter by standard, so please keep in mind that we can only interchange angular frequency and angular velocity in the case of circular motion.

By definition, the relationship between angular frequency and frequency is \(\omega = 2\pi f\). Additionally, by definition, the relationship between period and frequency is \(T = \frac{1}{f} \). Hence, we can express \(\omega = \frac{2\pi}{T} \). Now, we define average angular acceleration by \(\bar \alpha = \frac{\Delta \omega}{\Delta t}\) for some change in angular velocity \(\Delta \omega\) and some change in time \(\Delta t\). We can now expand this, recognizing that \(\Delta \omega = \omega_f = \omega_i\).\[\bar \alpha = \frac{\omega_f - \omega_i}{\Delta t} = \frac{\frac{2\pi}{T_f}-\frac{2\pi}{T_i}}{\Delta t}=\boxed{\displaystyle \frac{2\pi}{\Delta t}\left(\frac{1}{T_f}-\frac{1}{T_i}\right)}\] Now, you can simply just plug in values to get your answer.

By definition, the relationship between average torque, average angular acceleration, and an object's moment of inertia \(I\) is given by analogy to Newton's second law: \(\bar \tau = I \bar \alpha\). The moment of inertia of a solid sphere is given by \(I = \frac{2}{5}MR^2\), where \(M\) is the mass of the earth and \(R\) is the radius of the earth. Substituting everything gives the following expression.\[\bar \tau = \frac{2}{5}MR^2 \bar \alpha = \boxed{\displaystyle \frac{4\pi MR^2}{5\Delta t}\left(\frac{1}{T_f}-\frac{1}{T_i}\right)} \]Again, plug in values, and you'll have your final answer.