Lesson 6, Topic 5
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Angular Speed and Velocity

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When a stone is tied to the end of a string or rope and whirled around, the stone moves in a circular path as shown in the diagram below.

Suppose that as the stone is being whirled around, it moves from point M to N, in t seconds, so that the radius OM sweeps through an angle Î¸ at the same time.

As the stone moves around the circular path and sweeps through angle Î¸, the stone moves with angular velocity, $$\scriptsize \omega$$

Angle $$\scriptsize \hat{MON} = \theta ,$$ the angular velocityof motion, $$\scriptsize \omega$$ can be defined as;

$$\scriptsize \omega = \normalsize \frac{\theta}{t}$$ ……………..(1)

We can say that the angular velocity,$$\scriptsize \omega ,$$ is the angle turned through, with respect to time.

Recall that linear velocity, v is given by the formula:

$$\scriptsize v = \normalsize \frac{s}{t}$$ …………….(2)

where s is the length of the arc MN.

Comparing equations 1 and 2, instead of using linear displacement in 1, we used angular displacementÎ¸.

The image illustrates the relationship between the radius and the central angleÎ¸ in radians.

We define the rotation angle Î¸ to be the ratio of the arc length to the radius of curvature:

The formula is $$\scriptsize \theta = \normalsize \frac{s}{r}$$

Therefore,

s = rÎ¸  …………………..(3)

where s represents the arc length,

Î¸ represents the central angle in radians.

r is the length of the radius.

Now we substitute equation (3) (s = rÎ¸) into equation (2), $$\scriptsize v = \normalsize \frac{s}{t}$$

we have,

$$\scriptsize v = \normalsize \frac{s}{t} \\ = \frac{r\theta}{t} \\ \scriptsize = r. \normalsize \frac{\theta}{t} \\ \scriptsize but \: \normalsize \frac{\theta}{t} \scriptsize = \omega\\ \scriptsize \therefore v = r \omega$$

$$\scriptsize v = r \omega$$ ………….(4)

Units:

r is in metres, m

s is in metres, m

v is in metre per second, $$\scriptsize m/s \: or \: ms^{-1}$$

w is in radians per second, $$\scriptsize rad/s \: or \: rads^{-1}$$

Also, 1 revolution = one complete turn around a circle

The distance around a circle =  Circumference

The circumference of a circle is 2Ï€r.

So the distance travelled, by the stone in one full revolution is;

s =$$\scriptsize 2 \pi r$$

Thus for one complete revolution, the rotation angle is

$$\scriptsize \theta = \normalsize \frac{2 \pi r}{r}$$ (Remember: $$\scriptsize \theta = \normalsize \frac{s}{r}$$)

$$\scriptsize \therefore\: \theta = 2 \pi$$

1 complete revolution:Î¸ = 360Â° or $$\scriptsize 2\pi \: radians$$

Let us substitute s (the distance for 1 complete revolution) into equation (2)

So $$\scriptsize v = \normalsize \frac{2 \pi r}{t}$$ ……….(5)

But we know that,

$$\scriptsize v = r \omega$$ from equation (4)

Substitute v in equation (4) into equation (5)

$$\scriptsize r \omega = \normalsize \frac{2 \pi r}{t}$$

makeÏ‰ the subject of the formula

$$\scriptsize \omega = \normalsize \frac{2 \pi r}{t r}$$

Therefore,

$$\scriptsize \omega = \normalsize \frac{2 \pi}{t}$$

And,

$$\scriptsize t = \normalsize \frac{2 \pi}{w}$$

Also,

Frequency, f = $$\frac{1}{t}$$

Therefore,

$$\scriptsize \omega = 2 \pi f$$

Since v =Ï‰r

Then we can also say,

v = $$\scriptsize 2 \pi f r$$

Example 1

Calculate the angular speed of the earth which rotates on its axis every 24 hours.

Solution

In 1 rotation Î¸ = 2Ï€

The time it takes to rotate is 24 hours, so t = 24 hr.

Convert t to seconds

t = 24 hr Ã— 60 min/hr Ã— 60 sec/min

= 86400 sec

We will use the formula for angular speed which is;

$$\scriptsize \omega = \normalsize \frac{\theta}{t}$$

Substitute the values ofÎ¸ and t into the equation;

$$\scriptsize \omega = \normalsize \frac{2 \pi}{86400}$$

$$\scriptsize \omega = 0.0000726 \: radians \: per \: second$$

$$\scriptsize \omega = 7.26 \: \times \: 10^{-5} \: rad/s$$

Therefore, the angular speed of the earth is 7.26 Ã— 10-5 rad/s

Example 2

A stone of mass 0.8kg is tied to one end of a rope and spun so that it moves in a circular path of radius 0.15m with a velocity of 15m/s for 2 minutes.

(i) calculate the angular velocity of the object.

(ii) angle sweep through.

(iii) distance traveled.

Solution:

Data from the question.

Mass = 0.8kg

Velocity = 15m/s

radius of the circular path = 0.15m

time = 2 minutes.

Convert Time to seconds

time = 2 x 60 sec = 120 seconds

Linear Velocity = $$\scriptsize \omega \: \times \: r$$

Substitute the data values into the equation

Therefore, you will get,

15 m/s = $$\scriptsize \omega \: \times \: 0.15$$

make Ï‰ the subject of the formula

Ï‰ = $$\frac{15}{0.15}$$

(ii) What we are looking for isÎ¸. We know the value forÏ‰ and t, so we can use the formula,

$$\scriptsize \omega = \normalsize \frac{\theta}{t}$$

$$\scriptsize 100 rads^{-1} = \normalsize \frac{\theta}{120}$$

MakeÎ¸ the subject of the formula,

$$\scriptsize \theta = 100 rads^{-1} \: \times \: 120 seconds$$

$$\scriptsize \theta = 12000 radians$$

Now we have to convert 2000 radians to degrees.

360 degrees = 2 x 3.142

Now we will calculate how many degrees make 1 radian.

1 rad = $$\frac{360}{6.284} \\ \scriptsize = 57.29^o$$

Now we will convert12000 radians to degrees by multiplying 12000 by 55.29Â°.

Therefore, 12000 rad = 12000 x 57.29Â° = 687480Â°.

The angle Î¸ which the object swept through = 687480Â°.

We can change this 687480Â° to numbers of revolutions by dividing it by 360Â° because 360Â° makes 1 revolution.

Therefore,

Number of revolutions = $$\frac{687480}{360}$$

= 1909.67 revolutions

(iii) To calculate the distance travelled, we will use the formula,

$$\scriptsize v = \normalsize \frac{s}{t}$$

make distance, s the subject of the formula

$$\scriptsize s = vt$$

$$\scriptsize s = 15 \: \times \: 120$$

$$\scriptsize s = 1800m$$

Example 3

A stone whirled at the end of a rope 30 cm long makes 10 complete revolutions in 2 seconds. Find

(i) the angular velocity in radian per second.

(ii) the linear speed.

(iii) the distance covered in 5 seconds.

Solution

(i) The angle converted in 1 revolution is given by360Â° or 2Ï€ radians

Angle covered in 10 revolutions

Time for 10 revolutions is 2 seconds

Angular velocity $$\scriptsize \omega = \normalsize \frac{\theta}{t} \\ = \frac{20 \pi}{2} \\ \scriptsize = 10\pi \: radians \: per \: second \\ \scriptsize 10 \: \times \: 3.14 \\ \scriptsize 31.4 rad.s^{-1}$$

(ii) Linear speed is given by

$$\scriptsize v = r \omega \\ \scriptsize = 30cm \: \times \: 31.4 \: rad \: s^{-1} \\ \scriptsize = 942 \: cms^{-1} \\ \scriptsize = 0.942 ms^{-1}$$

Note:

Radians has no dimension since it is a ratio between two lengths, remember $$\scriptsize \theta = \normalsize \frac{s}{r}$$

So $$\scriptsize cm \: \times \: rads^{-1} = cms^{-1}$$

(iii) distance covered in 5 seconds is given by

$$\scriptsize s = vt \\ \scriptsize = 0.942 \: \times \: 5 \\ \scriptsize = 4.71m$$

1. thank you