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# Solved Examples on Linear Speeds and Angular Speeds

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE and Malta Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions
State the formulas and reasons for applicable steps
Negative angles are not allowed
Show all work
Except stated otherwise, use:

$d = diameter \\[3ex] r = radius \\[3ex] L = arc\:\:length \\[3ex] t = time \\[3ex] \pi = \dfrac{22}{7} \\[5ex] \theta = central\:\:angle \\[3ex] v = linear\:\: velocity ... in\:\:meters\:\:per\:\:second \\[3ex] \omega = angular\:\:velocity ... in\:\:radians\:\:per\:\:second \\[3ex] RAD = radians \\[3ex] ^\circ = DEG = degrees \\[3ex] L = \dfrac{2\pi r\theta}{360} ...\theta \:\:in\:\:DEG \\[5ex] L = r\theta ...\theta \:\:in\:\:RAD \\[5ex] Circumference\:\:of\:\:a\:\:circle = 2\pi r \\[3ex] v = r\omega \\[3ex] \omega = \dfrac{v}{r} \\[5ex] v = \dfrac{L}{t} \\[5ex] \omega = \dfrac{\theta}{t}$

(1.) Assume the flywheel of a $2019$ Maserati Levanti has a diameter of $10\:cm$, and is rotating at a rate of $13$ radians per second.
Determine the linear speed in centimeters per minute of a point on its rim.
Round to the nearest integer as needed.

$d = 10\:cm \\[3ex] r = \dfrac{10}{2} = 5\:cm \\[3ex] \omega = 13\:rad/sec \\[3ex] v = r\omega \\[3ex] v = 5 * 13 = 65\:cm/sec \\[3ex] 60\:sec = 1\:min \\[3ex] \therefore 1\:sec = \dfrac{1}{60}\:min \\[5ex] 65\:cm/sec = 65\:cm \div 1\:sec \\[5ex] = 65\:cm \div \dfrac{1}{60}\:min \\[5ex] = 65 * \dfrac{60}{1} \\[5ex] = 3900 \\[3ex] \rightarrow v = 3900\:cm/min$
(2.) The radius of the planet, Jupiter is $43000$ miles.
Jupiter makes a revolution every $10$ hours.
Calculate the linear speed of a point on its equator in miles per hour.
Round to the nearest integer as needed.

$r = 43000\:mi \\[3ex] \omega = 1\:rev \:\:per\:\:10\:hr \\[3ex] Convert\:\:to\:\:rev/hr \\[3ex] 1\:rev \rightarrow 10\:hr \\[3ex] 10\:hr \rightarrow 1\:rev \\[3ex] \therefore 1\:hr = \dfrac{1}{10}\:rev/hr \\[5ex] Convert\:\:to\:\:rad/hr \\[3ex] 1\:rev = 360\:DEG = 2\pi\:RAD \\[3ex] \therefore \dfrac{1}{10}\:rev = \dfrac{1}{10} * 2\pi\:rad \\[5ex] \dfrac{1}{10} * 2\pi\:rad = \dfrac{2\pi\:rad}{10} = \dfrac{\pi\:rad}{5} \\[5ex] \dfrac{\pi\:rad}{5} = \dfrac{22}{7} \div 5 \\[5ex] \dfrac{22}{7} \div 5 = \dfrac{22}{7} * \dfrac{1}{5} = \dfrac{22 * 1}{5 * 35} \\[5ex] \dfrac{22 * 1}{5 * 35} = \dfrac{22}{35} \\[5ex] \rightarrow \omega = \dfrac{22}{35}\:rad/hr \\[5ex] v = r\omega \\[3ex] v = 43000\:mi * \dfrac{22}{35}\:rad/hr \\[5ex] v = \dfrac{43000 * 22}{35} = \dfrac{946000}{35} \\[5ex] v = 27028.57143\:mi/hr \\[3ex] v \approx 27029\:mi/hr$
(3.) The diameter of a cylinder is $20.42\:in$
The linear speed of a point on the cylinder's surface is $18.09\:ft/sec$
Determine the angular speed of the cylinder in revolutions per hour.
Round to the nearest integer as needed.

The units should be consistent.
Convert ft to inches

$d = 20.42\:in \\[3ex] r = \dfrac{d}{2} = \dfrac{20.42}{2} = 10.21\:in \\[5ex] v = 18.09\:ft/sec \\[3ex] 1\:ft = 12\:in \\[3ex] \therefore 18.09\:ft = 18.09(12) = 217.08\:in \\[3ex] \rightarrow v = 217.08\:in/sec \\[3ex] \omega = \dfrac{v}{r} \\[5ex] \omega = \dfrac{217.08\:in/sec}{10.21\:in} \\[5ex] \omega = 21.26150833\:rad/sec \\[3ex] Convert\:\:to\:\:rev/hr \\[3ex] 1\:rev = 360\:DEG = 2\pi\:RAD \\[3ex] 2\pi rad = 1\:rev \\[3ex] \therefore 1\:rad = \dfrac{1}{2\pi}\:rev \\[5ex] 1\:hr = 3600\:sec \\[3ex] 3600\:sec = 1\:hr \\[3ex] \therefore 1\:sec = \dfrac{1}{3600}\:hr \\[5ex] \rightarrow 21.26150833\:rad/sec = 21.26150833\:rad \div 1\:sec \\[3ex] 21.26150833\:rad \div 1\:sec = 21.26150833 * \dfrac{1}{2\pi}\:rev \div \dfrac{1}{3600}\:hr \\[5ex] 21.26150833 * \dfrac{1}{2\pi} \div \dfrac{1}{3600} = 21.26150833 * \dfrac{1}{2\pi} * \dfrac{3600}{1} \\[5ex] 2\pi = 2 * \dfrac{22}{7} = \dfrac{2 * 22}{7} = \dfrac{44}{7} \\[5ex] \dfrac{1}{2\pi} = 1 \div \dfrac{44}{7} = 1 * \dfrac{7}{44} = \dfrac{1 * 7}{44} = \dfrac{7}{44} \\[5ex] 21.26150833 * \dfrac{1}{2\pi} * \dfrac{3600}{1} = 21.26150833 * \dfrac{7}{44} * 3600 \\[5ex] 21.26150833 * \dfrac{7}{44} * \dfrac{3600}{1} = \dfrac{21.26150833 * 7 * 3600}{44} \\[5ex] \dfrac{21.26150833 * 7 * 3600}{44} = \dfrac{535790.0099}{44} = 12177.04568 \\[5ex] \rightarrow \omega = 12177.04568\:rev/hr \\[3ex] \therefore \omega \approx 12177\:rev/hr$
(4.)

$\tan\left(\cos^{-1}\left(-\dfrac{1}{2}\right)\right) \\[5ex] \cos^{-1}\left(-\dfrac{1}{2}\right) = 120^\circ ... Unit\:\: Circle\:\: Trigonometry \\[5ex] = \tan120 \\[3ex] = -\sqrt{3} ... Unit\:\: Circle\:\: Trigonometry$
(5.)

$L = \dfrac{2\pi r\theta}{360} ...\theta \:\:in\:\:DEG \\[5ex] \pi = \dfrac{22}{7} \\[5ex] r = 6400 km \\[3ex] \theta = 12^\circ \\[3ex] L = \left(2 * \dfrac{22}{7} * 6400 * 12\right) \div 360 \\[5ex] L = 2 * \dfrac{22}{7} * 6400 * 12 * \dfrac{1}{360} \\[5ex] L = \dfrac{2 * 22 * 6400 * 12}{7 * 360} \\[5ex] L = \dfrac{3379200}{2520} \\[5ex] L = 1340.952381 \\[3ex] L \approx 1341 \:km$
(6.)

$From\:\:12:40\:PM \:\:to\:\: 1:25\:PM = 45\:\:minutes \\[3ex] 1\:\:complete\:\:rotation = 1\:\:hour = 60\:\:minutes \\[3ex] L = 2\pi r \:\:for\:\: 1\:\:hour \\[3ex] \rightarrow 60\:\:minutes \rightarrow 2\pi r \\[3ex] \therefore 45\:\:minutes = \dfrac{45 * 2\pi r}{60} = \dfrac{3\pi r}{2} \\[5ex] \rightarrow L = \dfrac{3\pi r}{2} \\[5ex] r = r ...not\:\:given \\[3ex] \theta = \dfrac{L}{r} \\[5ex] \theta = \dfrac{3\pi r}{2} \div r \\[5ex] \theta = \dfrac{3\pi r}{2} * \dfrac{1}{r} \\[5ex] \theta = \dfrac{3\pi}{2}$