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 θ at the same time.

As the stone moves around the circular path and sweeps through angle θ, 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 displacementθ.

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

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

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

Therefore,

s = rθ  …………………..(3)

where s represents the arc length,

θ represents the central angle in radians.

r is the length of the radius.

Now we substitute equation (3) (s = rθ) 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 2πr.

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:θ = 360° 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)

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

makeω 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 =ωr

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 θ = 2π

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

Convert t to seconds

t = 24 hr × 60 min/hr × 60 sec/min

= 86400 sec

We will use the formula for angular speed which is;

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

Substitute the values ofθ 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 × 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 ω the subject of the formula

ω = $$\frac{15}{0.15}$$

(ii) What we are looking for isθ. We know the value forω and t, so we can use the formula,

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

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

Makeθ 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 55.29°.

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

The angle θ which the object swept through = 687480°.

We can change this 687480° to numbers of revolutions by dividing it by 360° because 360° 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 by360° or 2π 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