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SS1: PHYSICS – 1ST TERM

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  1. Introduction to Physics | Week 1
    4Topics
    |
    1 Quiz
  2. Measurement | Week 2
    3Topics
  3. Measurement of Mass | Week 3
    6Topics
    |
    1 Quiz
  4. Motion | Week 4
    5Topics
    |
    1 Quiz
  5. Velocity-Time Graph | Week 5
    4Topics
    |
    1 Quiz
  6. Causes of Motion | Week 6
    5Topics
    |
    1 Quiz
  7. Work, Energy & Power | Week 7
    3Topics
  8. Energy Transformation / Power | Week 8
    3Topics
    |
    1 Quiz
  9. Heat Energy | Week 9
    5Topics
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    1 Quiz
  10. Linear Expansion | Week 10
    6Topics
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    1 Quiz
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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.

angular speed and velocity

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.

radian and arc of a circle

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 = \( \frac{15}{0.15} \)

 Angular velocity, ω = 100 radians per second (rad/s)

(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 = 2π rad

360 degrees = 2 x 3.142

360 degrees = 6.284 radians

Now we will calculate how many degrees make 1 radian.

6.284 rad = 360°

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

=2π x 10 radians

= 20π radians

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

\( \scriptsize s = vt \\ \scriptsize = 0.942 \: \times \: 5 \\ \scriptsize = 4.71m \)

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