Tag Archives: period

Problem 6-4: Period, angular velocity, and linear velocity of the Earth


(a) What is the period of rotation of Earth in seconds? (b) What is the angular velocity of Earth? (c) Given that Earth has a radius of 6.4×106 m at its equator, what is the linear velocity at Earth’s surface?


Solution:

Part A

The period of a rotating body is the time it takes for 1 full revolution. The Earth rotates about its axis, and complete 1 full revolution in 24 hours. Therefore, the period is

Period=24 hoursPeriod=24 hours×3600 seconds1 hourPeriod=86400 seconds  (Answer)\begin{align*} \text{Period} & = 24 \ \text{hours} \\ \\ \text{Period} & = 24 \ \text{hours} \times \frac{3600 \ \text{seconds}}{1 \ \text{hour}} \\ \\ \text{Period} & = 86400 \ \text{seconds} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The angular velocity ω\omega is the rate of change of an angle,

ω=ΔθΔt,\omega = \frac{\Delta \theta}{\Delta t},

where a rotation Δθ\Delta \theta takes place in a time Δt\Delta t.

From the given problem, we are given the following: Δθ=2πradian=1 revolution\Delta \theta = 2\pi \text{radian} = 1 \ \text{revolution}, and Δt=24 hours=1440 minutes=86400 seconds\Delta t =24\ \text{hours} = 1440 \ \text{minutes}= 86400 \ \text{seconds}. Therefore, the angular velocity is

ω=ΔθΔtω=1 revolution1440 minutesω=6.94×104 rpm  (Answer)\begin{align*} \omega & = \frac{\Delta\theta}{\Delta t} \\ \\ \omega & = \frac{1 \ \text{revolution}}{1440 \ \text{minutes}}\\ \\ \omega & = 6.94 \times 10^{-4}\ \text{rpm}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

We can also express the angular velocity in units of radians per second. That is

ω=ΔθΔtω=2π radian86400 secondsω=7.27×105 radians/second  (Answer)\begin{align*} \omega & = \frac{\Delta\theta}{\Delta t} \\ \\ \omega & = \frac{2\pi \ \text{radian}}{86400 \ \text{seconds}}\\ \\ \omega & = 7.27 \times 10^{-5}\ \text{radians/second}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The linear velocity vv, and the angular velocity ω\omega are related by the formula

v=rωv = r \omega

From the given problem, we are given the following values: r=6.4×106 metersr=6.4 \times 10^{6} \ \text{meters}, and ω=7.27×105 radians/second\omega = 7.27 \times 10^{-5}\ \text{radians/second}. Therefore, the linear velocity at the surface of the earth is

v=rωv=(6.4×106 meters)(7.27×105 radians/second)v=465.28 m/s  (Answer)\begin{align*} v & =r \omega \\ \\ v & = \left( 6.4 \times 10^{6} \ \text{meters} \right)\left( 7.27 \times 10^{-5}\ \text{radians/second} \right) \\ \\ v & = 465.28 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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