In the previous section we saw the basic details of uniform circular motion and centripetal acceleration. In this section, we will see Angular speed.
We will write the steps:
1. Consider the circular motion of the object that we saw in the previous section. For convenience, it is shown again in fig.4.38 below:
• In a time interval of Δt, the object travels from P to P’.
• During this time interval, the line OP turns through an angle of Δθ.
♦ This Δθ is called angular distance.
2. From this 'angular distance', we can find 'the angle which the line OP turns in an interval of 1 s'.
• We simply divide the 'angular distance' by 'time required to cover that angular distance'. We get:
• The angle which the line OP turns in 1 s = $\mathbf\small{\frac{\Delta \theta }{\Delta t}}$
3. The 'angle which the line (line which joins the object to the center) turns in 1 s' is called angular speed. It is denoted by $\mathbf\small{\omega}$ (Greek small letter omega)
■ So we can write Eq.4.28: $\mathbf\small{\omega =\frac{\Delta \theta }{\Delta t}}$
4. The straight line between P and P’ is the 'displacement during Δt'
♦ It is denoted as $\mathbf\small{\overrightarrow{\Delta r}}$
♦ It is a vector quantity
• But the arc length PP’ is the 'distance traveled during Δt'
♦ It is denoted as $\mathbf\small{{\Delta s}}$
♦ It is not a vector quantity
5. We know that 'distance traveled' divided by time gives speed
• Speed is the magnitude of the velocity vector
♦ For circular motion, we denoted it as $\mathbf\small{|\vec{v}|}$
■ So we get: $\mathbf\small{|\vec{v}|=\frac{\Delta s}{\Delta t}}$
6. Now, using properties of arc, we have the relation: $\mathbf\small{angle=\frac{arc}{radius}}$
♦ Where angle is measured in radians
• So we get: arc length PP’ = Δs = Δθ×R
♦ Where R is the radius of the path
7. So we can put (Δθ×R) instead of Δs in (5). we get:
• Speed $\mathbf\small{|\vec{v}|=\frac{\Delta \theta \times R }{\Delta t}=R\frac{\Delta \theta}{\Delta t}}$
• But from (3), we have: $\mathbf\small{\frac{\Delta \theta }{\Delta t}=\omega}$
■ So we get Eq.4.29: speed $\mathbf\small{|\vec{v}|=R\, \omega}$
8. The time required for making one complete revolution around the circular path is called time period. It is denoted by T
An example:
(i) Let an object require 0.25 s to complete one revolution. Then it’s time period T = 0.25 s
(ii) This object requires only a ‘fraction of a second’ (¼ to be exact) to make one complete revolution
(iii) Obviously, It will make 4 complete revolutions in 1 s.
• We obtain this result by dividing ‘1’ by ‘T’.
• That is., we take the reciprocal of T
9. The number of revolutions in one second is called frequency. It is denoted by ‘n’
■ So we can write Eq.4.30: Frequency $\mathbf\small{n=\frac{1}{T}}$
10. While calculating 'n', the 'T' goes to the denominator.
(i) In the above example, T (which is 0.25 s) is less than 1.
• So when we calculate n, since T is in the denominator, we get a value ‘4’ which is greater than 1
(ii) What if T is greater than 1?
• That is., the object may take more than 1 s to complete one revolution
• In that case, since T is in the denominator, the ‘n’ will be less than ‘1’
■ That means, the object will not complete even one revolution in 1 s
11. Relations between $\mathbf\small{|\vec{v}|}$, T and n
A. Relation between speed $\mathbf\small{|\vec{v}|}$ and T
(i) During an interval of ‘T’ s the object will travel a distance of 2πR along the circumference
(ii) We have the familiar relation for uniform speed: Distance = speed × time
(iii) So we get: 2πR = $\mathbf\small{|\vec{v}|}$ × T
Thus we get Eq.4.31: $\mathbf\small{|\vec{v}|=\frac{2\pi R}{T}}$
B. Relation between speed $\mathbf\small{|\vec{v}|}$ and n
(i) From Eq.4.30, we have: $\mathbf\small{n=\frac{1}{T}}$
• From this, we get: $\mathbf\small{T=\frac{1}{n}}$
(ii) So we can use $\mathbf\small{\frac{1}{T}}$ instead of T in Eq.4.31
• We can write: $\mathbf\small{|\vec{v}|=\frac{2\pi R}{\frac{1}{n}}}$
• So we get Eq.4.32: $\mathbf\small{|\vec{v}|=2\pi Rn}$
12. Finally, we will see the relation between centripetal acceleration $\mathbf\small{|\vec{a_c}|}$ and $\mathbf\small{\omega}$
(i) In the previous section, we saw Eq.4.27: $\mathbf\small{|\vec{a_c}|=\frac{|\vec{v}|^2}{R}}$
(ii) In this section we derived Eq.4.29: $\mathbf\small{|\vec{v}|=R\, \omega}$
(iii) So we can use 'Rω' instead of $\mathbf\small{|\vec{v}|}$ in Eq.4.27
• We get: $\mathbf\small{|\vec{a_c}|=\frac{R^2 \omega ^2}{R}}$
• Thus we get Eq.4.33: $\mathbf\small{|\vec{a_c}|= \omega ^2 R}$
Now we will see an interesting case. We will write it in steps:
1. Consider a rotating circular disk of radius R cm shown in fig.4.39(a) below:
• It is rotating at a constant angular speed
2. Mark a point P on it's circumference
♦ Join P to the center O by the line OP. Then OP = R cm
• Mark another point Q on OP
♦ Let OQ = R1 cm
3. Let during an interval of Δt s, the line OP turn through an angle of Δθ.
• Then, in that interval Δt, OQ will also turn through the same angle Δθ
• This is shown in fig.b
4. Since both P and Q turn through the same angle Δθ, we can write:
• Angular speed of Q = Angular speed of P
• Let this angular speed be $\mathbf\small{\omega}$
■ We can write: Angular speed of Q = Angular speed of P = $\mathbf\small{\omega}$
5. Now we will see linear speed
• Let the linear speed of P be $\mathbf\small{|\vec{v_P}|}$
• Let the linear speed of Q be $\mathbf\small{|\vec{v_Q}|}$
6. During a time interval of Δt, P travels a distance of arc length PP'.
• During the same time interval of Δt, Q travels a distance of arc length QQ'.
7. But (arc length QQ') is less than (arc length PP')
• That is., even though the time interval is the same, the distances are not the same
■ Q travels a lesser distance in Δt
■ P travels a greater distance in Δt
■ Obviously, $\mathbf\small{|\vec{v_P}|}$ must be greater than $\mathbf\small{|\vec{v_Q}|}$
8. This can be proved in another way also:
(i) We have Eq.4.29 which gives the relation between linear speed and angular speed: $\mathbf\small{|\vec{v}|=R\, \omega}$
(ii) The angular speed $\mathbf\small{\omega}$ is constant for all points on the disc
So we can write:
• $\mathbf\small{|\vec{v_P}|=R\, \omega}$
• $\mathbf\small{|\vec{v_Q}|=R_1 \, \omega}$
(iii) R1 is lesser than R
■ So $\mathbf\small{|\vec{v_Q}|}$ will be lesser than $\mathbf\small{|\vec{v_P}|}$
■ When distance from the center decreases, the linear speed decreases
Now we will see some solved examples
Solved example 4.13
An insect trapped in a circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100 s. (a) What is the angular speed, and the linear speed of the motion? (b) Is the acceleration vector a constant vector ? What is its magnitude ?
Solution:
• Number of revolutions in 100 s = 7
• So no. of revolutions in 1 s = n = 7/100 = 0.07
Part (a):
1. First we find linear speed $\mathbf\small{|\vec{v}|}$
• We can use Eq.4.32: $\mathbf\small{|\vec{v}|=2\pi Rn}$
• Substituting the values, we get: $\mathbf\small{|\vec{v}|=2\pi \times 12 \times 0.07}$ = 5.28 cm s-1.
2. Once we find the linear velocity, we can use Eq.4.29 to find angular speed $\mathbf\small{\omega}$
• $\mathbf\small{|\vec{v}|=R\, \omega}$
• Substituting the values, we get: $\mathbf\small{5.28=12 \times \omega}$
• So $\mathbf\small{\omega}$ = 5.28/12 = 0.44 rad s-1.
Part (b):
1. The acceleration vector is always directed towards the center
• For each position of the insect, the acceleration is directed from that position to the center
■ That means, acceleration vector changes continuously. It is not a constant vector
2. But the magnitude of the acceleration vector is a constant
• To find that magnitude, we can use Eq.4.33: $\mathbf\small{|\vec{a_c}|= \omega ^2 R}$
• Substituting the values, we get $\mathbf\small{|\vec{a_c}|= 0.44 ^2 \times 12}$ = 2.32 ms-2.
Solved example 4.14
A particle is in uniform circular motion along a circle whose center is at the origin in the xy plane. When the particle is at (4,0), it's velocity is $\mathbf\small{-5\hat{j}}$ ms-1. Determine the velocity and acceleration when the particle is at P(0,-4) and Q(-2.83, 2.83)
Solution:
1. The velocity at (4,0) is given as $\mathbf\small{-5\hat{j}}$ ms-1.
• The unit vector $\mathbf\small{\hat{j}}$ indicates that it is along the y axis
• The negative sign indicates that, it is towards the 'negative side of y axis'
This situation is shown in fig.4.40(a) below:
2. The particle is said to be in uniform motion. So at any point along the path, $\mathbf\small{|\vec{v}|}$ will be 5 ms-1.
• We just need to find the directions.
3. Also given that, the center is at the origin O and (4,0) is a point on the path
• So obviously, the radius R = 4 m
4. Part (a):
(i) P(0,-4) is on the y axis.
(ii) The particle is moving in the anti-clockwise direction
• So $\mathbf\small{\vec{v}}$ at P(0,-4) is $\mathbf\small{-5\hat{i}}$ ms-1.
(iii) We are also asked to find the acceleration
• In uniform circular motion, there is no linear acceleration. There is only centripetal acceleration $\mathbf\small{\vec{a_c}}$
• We can use Eq.4.27: $\mathbf\small{|\vec{a_c}|=\frac{|\vec{v}|^2}{R}}$
• Substituting the values, we get: $\mathbf\small{|\vec{a_c}|}$ = 6.25 ms-2.
• The direction of this acceleration is towards the center O
5. Part (b):
(i) Q(-2.83, 2.83) is neither on the x axis nor on the y axis
• First let us confirm that it is on the given circle.
(ii) For that, we find it's distance from the origin.
• We have: Distance from origin = $\mathbf\small{\sqrt{(-2.83)^2+2.83^2}}$ = 4 m
• So the point is indeed on the circle
(iii) Consider the following points:
x cordinate and y ccordinate of Q are equal in magnitudes
x coordinate is negative and y cordinate is positive
So obviously, Q is on a line inclined at 45o to the x axis. And it is in the second quadrant. This is shown in fig.b
• Let the velocity vector at Q be $\mathbf\small{\vec{v'}}$
• This $\mathbf\small{\vec{v'}}$ will be perpendicular to OQ
• So $\mathbf\small{\vec{v'}}$ makes an angle 45o with the x axis
(iv) Thus we get:
• Magnitude of the x component of $\mathbf\small{\vec{v'}}$ = $\mathbf\small{|\vec{v'}|}$ × cos 45 = 5 × cos 45 = 3.536 ms-1.
• Magnitude of the y component of $\mathbf\small{\vec{v'}}$ = $\mathbf\small{|\vec{v'}|}$ × sin 45 = 5 × sin 45 = 3.536 ms-1.
• So $\mathbf\small{\vec{v'}=3.536\, \hat{i}+3.536\, \hat{j}}$
(v) Direction of $\mathbf\small{\vec{v'}}$ is obtained as follows:
• $\mathbf\small{\tan \theta =\frac{|\vec{v'_y}|}{|\vec{v'_x}|}=\frac{3.536}{3.536}=1}$
• So θ = tan-11 = 45°
• We can write: $\mathbf\small{\vec{v'}}$ makes an angle of 45° with the x axis
Solved example 4.15
If the angular speeds of the minute hand and hour hand of a watch are $\mathbf\small{\omega_m}$ and $\mathbf\small{\omega_h}$ respectively, find the ratio $\mathbf\small{\frac{\omega _m}{\omega _h}}$
Solution:
■ Angular speed is the 'angle turned in 1 s'
1. The angular speed of a minute hand:
(i) The minute hand makes one complete revolution in 60 minutes.
♦ 60 minutes is (60 × 60) = 3600 seconds
(ii) In other words, the minute hand turns through 2π radians in 3600 s
(iii) So it turns $\mathbf\small{\frac{2\pi }{3600}}$ radians in 1 s
(iv) We can write: $\mathbf\small{\omega _m=\frac{2\pi }{3600}}$
2. The angular speed of a hour hand:
(i) The hour hand makes one complete revolution in 12 hours.
♦ 12 hours is (12 × 60 × 60) = (12 ×3600) seconds
(ii) In other words, the hour hand turns through 2π radians in (12 × 3600) s
(iii) So it turns $\mathbf\small{\frac{2\pi }{12 \times 3600}}$ radians in 1 s
(iv) We can write: $\mathbf\small{\omega _h=\frac{2\pi }{12 \times 3600}}$
3. So we get: $\mathbf\small{\frac{\omega _m}{\omega _h}=\frac{2\pi }{3600}\div \frac{2\pi }{12 \times 3600}=\frac{2\pi }{3600}\times \frac{12 \times 3600 }{2 \pi} }$ = 12
■ Thus we get: angular speed of the minute hand is 12 times that of the hour hand
We will write the steps:
1. Consider the circular motion of the object that we saw in the previous section. For convenience, it is shown again in fig.4.38 below:
 |
| Fig.4.38 |
• During this time interval, the line OP turns through an angle of Δθ.
♦ This Δθ is called angular distance.
2. From this 'angular distance', we can find 'the angle which the line OP turns in an interval of 1 s'.
• We simply divide the 'angular distance' by 'time required to cover that angular distance'. We get:
• The angle which the line OP turns in 1 s = $\mathbf\small{\frac{\Delta \theta }{\Delta t}}$
3. The 'angle which the line (line which joins the object to the center) turns in 1 s' is called angular speed. It is denoted by $\mathbf\small{\omega}$ (Greek small letter omega)
■ So we can write Eq.4.28: $\mathbf\small{\omega =\frac{\Delta \theta }{\Delta t}}$
4. The straight line between P and P’ is the 'displacement during Δt'
♦ It is denoted as $\mathbf\small{\overrightarrow{\Delta r}}$
♦ It is a vector quantity
• But the arc length PP’ is the 'distance traveled during Δt'
♦ It is denoted as $\mathbf\small{{\Delta s}}$
♦ It is not a vector quantity
5. We know that 'distance traveled' divided by time gives speed
• Speed is the magnitude of the velocity vector
♦ For circular motion, we denoted it as $\mathbf\small{|\vec{v}|}$
■ So we get: $\mathbf\small{|\vec{v}|=\frac{\Delta s}{\Delta t}}$
6. Now, using properties of arc, we have the relation: $\mathbf\small{angle=\frac{arc}{radius}}$
♦ Where angle is measured in radians
• So we get: arc length PP’ = Δs = Δθ×R
♦ Where R is the radius of the path
7. So we can put (Δθ×R) instead of Δs in (5). we get:
• Speed $\mathbf\small{|\vec{v}|=\frac{\Delta \theta \times R }{\Delta t}=R\frac{\Delta \theta}{\Delta t}}$
• But from (3), we have: $\mathbf\small{\frac{\Delta \theta }{\Delta t}=\omega}$
■ So we get Eq.4.29: speed $\mathbf\small{|\vec{v}|=R\, \omega}$
8. The time required for making one complete revolution around the circular path is called time period. It is denoted by T
An example:
(i) Let an object require 0.25 s to complete one revolution. Then it’s time period T = 0.25 s
(ii) This object requires only a ‘fraction of a second’ (¼ to be exact) to make one complete revolution
(iii) Obviously, It will make 4 complete revolutions in 1 s.
• We obtain this result by dividing ‘1’ by ‘T’.
• That is., we take the reciprocal of T
9. The number of revolutions in one second is called frequency. It is denoted by ‘n’
■ So we can write Eq.4.30: Frequency $\mathbf\small{n=\frac{1}{T}}$
10. While calculating 'n', the 'T' goes to the denominator.
(i) In the above example, T (which is 0.25 s) is less than 1.
• So when we calculate n, since T is in the denominator, we get a value ‘4’ which is greater than 1
(ii) What if T is greater than 1?
• That is., the object may take more than 1 s to complete one revolution
• In that case, since T is in the denominator, the ‘n’ will be less than ‘1’
■ That means, the object will not complete even one revolution in 1 s
11. Relations between $\mathbf\small{|\vec{v}|}$, T and n
A. Relation between speed $\mathbf\small{|\vec{v}|}$ and T
(i) During an interval of ‘T’ s the object will travel a distance of 2πR along the circumference
(ii) We have the familiar relation for uniform speed: Distance = speed × time
(iii) So we get: 2πR = $\mathbf\small{|\vec{v}|}$ × T
Thus we get Eq.4.31: $\mathbf\small{|\vec{v}|=\frac{2\pi R}{T}}$
B. Relation between speed $\mathbf\small{|\vec{v}|}$ and n
(i) From Eq.4.30, we have: $\mathbf\small{n=\frac{1}{T}}$
• From this, we get: $\mathbf\small{T=\frac{1}{n}}$
(ii) So we can use $\mathbf\small{\frac{1}{T}}$ instead of T in Eq.4.31
• We can write: $\mathbf\small{|\vec{v}|=\frac{2\pi R}{\frac{1}{n}}}$
• So we get Eq.4.32: $\mathbf\small{|\vec{v}|=2\pi Rn}$
12. Finally, we will see the relation between centripetal acceleration $\mathbf\small{|\vec{a_c}|}$ and $\mathbf\small{\omega}$
(i) In the previous section, we saw Eq.4.27: $\mathbf\small{|\vec{a_c}|=\frac{|\vec{v}|^2}{R}}$
(ii) In this section we derived Eq.4.29: $\mathbf\small{|\vec{v}|=R\, \omega}$
(iii) So we can use 'Rω' instead of $\mathbf\small{|\vec{v}|}$ in Eq.4.27
• We get: $\mathbf\small{|\vec{a_c}|=\frac{R^2 \omega ^2}{R}}$
• Thus we get Eq.4.33: $\mathbf\small{|\vec{a_c}|= \omega ^2 R}$
Now we will see an interesting case. We will write it in steps:
1. Consider a rotating circular disk of radius R cm shown in fig.4.39(a) below:
 |
| Fig.4.39 |
2. Mark a point P on it's circumference
♦ Join P to the center O by the line OP. Then OP = R cm
• Mark another point Q on OP
♦ Let OQ = R1 cm
3. Let during an interval of Δt s, the line OP turn through an angle of Δθ.
• Then, in that interval Δt, OQ will also turn through the same angle Δθ
• This is shown in fig.b
4. Since both P and Q turn through the same angle Δθ, we can write:
• Angular speed of Q = Angular speed of P
• Let this angular speed be $\mathbf\small{\omega}$
■ We can write: Angular speed of Q = Angular speed of P = $\mathbf\small{\omega}$
5. Now we will see linear speed
• Let the linear speed of P be $\mathbf\small{|\vec{v_P}|}$
• Let the linear speed of Q be $\mathbf\small{|\vec{v_Q}|}$
6. During a time interval of Δt, P travels a distance of arc length PP'.
• During the same time interval of Δt, Q travels a distance of arc length QQ'.
7. But (arc length QQ') is less than (arc length PP')
• That is., even though the time interval is the same, the distances are not the same
■ Q travels a lesser distance in Δt
■ P travels a greater distance in Δt
■ Obviously, $\mathbf\small{|\vec{v_P}|}$ must be greater than $\mathbf\small{|\vec{v_Q}|}$
8. This can be proved in another way also:
(i) We have Eq.4.29 which gives the relation between linear speed and angular speed: $\mathbf\small{|\vec{v}|=R\, \omega}$
(ii) The angular speed $\mathbf\small{\omega}$ is constant for all points on the disc
So we can write:
• $\mathbf\small{|\vec{v_P}|=R\, \omega}$
• $\mathbf\small{|\vec{v_Q}|=R_1 \, \omega}$
(iii) R1 is lesser than R
■ So $\mathbf\small{|\vec{v_Q}|}$ will be lesser than $\mathbf\small{|\vec{v_P}|}$
■ When distance from the center decreases, the linear speed decreases
Now we will see some solved examples
Solved example 4.13
An insect trapped in a circular groove of radius 12 cm moves along the groove steadily and completes 7 revolutions in 100 s. (a) What is the angular speed, and the linear speed of the motion? (b) Is the acceleration vector a constant vector ? What is its magnitude ?
Solution:
• Number of revolutions in 100 s = 7
• So no. of revolutions in 1 s = n = 7/100 = 0.07
Part (a):
1. First we find linear speed $\mathbf\small{|\vec{v}|}$
• We can use Eq.4.32: $\mathbf\small{|\vec{v}|=2\pi Rn}$
• Substituting the values, we get: $\mathbf\small{|\vec{v}|=2\pi \times 12 \times 0.07}$ = 5.28 cm s-1.
2. Once we find the linear velocity, we can use Eq.4.29 to find angular speed $\mathbf\small{\omega}$
• $\mathbf\small{|\vec{v}|=R\, \omega}$
• Substituting the values, we get: $\mathbf\small{5.28=12 \times \omega}$
• So $\mathbf\small{\omega}$ = 5.28/12 = 0.44 rad s-1.
Part (b):
1. The acceleration vector is always directed towards the center
• For each position of the insect, the acceleration is directed from that position to the center
■ That means, acceleration vector changes continuously. It is not a constant vector
2. But the magnitude of the acceleration vector is a constant
• To find that magnitude, we can use Eq.4.33: $\mathbf\small{|\vec{a_c}|= \omega ^2 R}$
• Substituting the values, we get $\mathbf\small{|\vec{a_c}|= 0.44 ^2 \times 12}$ = 2.32 ms-2.
Solved example 4.14
A particle is in uniform circular motion along a circle whose center is at the origin in the xy plane. When the particle is at (4,0), it's velocity is $\mathbf\small{-5\hat{j}}$ ms-1. Determine the velocity and acceleration when the particle is at P(0,-4) and Q(-2.83, 2.83)
Solution:
1. The velocity at (4,0) is given as $\mathbf\small{-5\hat{j}}$ ms-1.
• The unit vector $\mathbf\small{\hat{j}}$ indicates that it is along the y axis
• The negative sign indicates that, it is towards the 'negative side of y axis'
This situation is shown in fig.4.40(a) below:
 |
| Fig.4.40 |
• We just need to find the directions.
3. Also given that, the center is at the origin O and (4,0) is a point on the path
• So obviously, the radius R = 4 m
4. Part (a):
(i) P(0,-4) is on the y axis.
(ii) The particle is moving in the anti-clockwise direction
• So $\mathbf\small{\vec{v}}$ at P(0,-4) is $\mathbf\small{-5\hat{i}}$ ms-1.
(iii) We are also asked to find the acceleration
• In uniform circular motion, there is no linear acceleration. There is only centripetal acceleration $\mathbf\small{\vec{a_c}}$
• We can use Eq.4.27: $\mathbf\small{|\vec{a_c}|=\frac{|\vec{v}|^2}{R}}$
• Substituting the values, we get: $\mathbf\small{|\vec{a_c}|}$ = 6.25 ms-2.
• The direction of this acceleration is towards the center O
5. Part (b):
(i) Q(-2.83, 2.83) is neither on the x axis nor on the y axis
• First let us confirm that it is on the given circle.
(ii) For that, we find it's distance from the origin.
• We have: Distance from origin = $\mathbf\small{\sqrt{(-2.83)^2+2.83^2}}$ = 4 m
• So the point is indeed on the circle
(iii) Consider the following points:
x cordinate and y ccordinate of Q are equal in magnitudes
x coordinate is negative and y cordinate is positive
So obviously, Q is on a line inclined at 45o to the x axis. And it is in the second quadrant. This is shown in fig.b
• Let the velocity vector at Q be $\mathbf\small{\vec{v'}}$
• This $\mathbf\small{\vec{v'}}$ will be perpendicular to OQ
• So $\mathbf\small{\vec{v'}}$ makes an angle 45o with the x axis
(iv) Thus we get:
• Magnitude of the x component of $\mathbf\small{\vec{v'}}$ = $\mathbf\small{|\vec{v'}|}$ × cos 45 = 5 × cos 45 = 3.536 ms-1.
• Magnitude of the y component of $\mathbf\small{\vec{v'}}$ = $\mathbf\small{|\vec{v'}|}$ × sin 45 = 5 × sin 45 = 3.536 ms-1.
• So $\mathbf\small{\vec{v'}=3.536\, \hat{i}+3.536\, \hat{j}}$
(v) Direction of $\mathbf\small{\vec{v'}}$ is obtained as follows:
• $\mathbf\small{\tan \theta =\frac{|\vec{v'_y}|}{|\vec{v'_x}|}=\frac{3.536}{3.536}=1}$
• So θ = tan-11 = 45°
• We can write: $\mathbf\small{\vec{v'}}$ makes an angle of 45° with the x axis
Solved example 4.15
If the angular speeds of the minute hand and hour hand of a watch are $\mathbf\small{\omega_m}$ and $\mathbf\small{\omega_h}$ respectively, find the ratio $\mathbf\small{\frac{\omega _m}{\omega _h}}$
Solution:
■ Angular speed is the 'angle turned in 1 s'
1. The angular speed of a minute hand:
(i) The minute hand makes one complete revolution in 60 minutes.
♦ 60 minutes is (60 × 60) = 3600 seconds
(ii) In other words, the minute hand turns through 2π radians in 3600 s
(iii) So it turns $\mathbf\small{\frac{2\pi }{3600}}$ radians in 1 s
(iv) We can write: $\mathbf\small{\omega _m=\frac{2\pi }{3600}}$
2. The angular speed of a hour hand:
(i) The hour hand makes one complete revolution in 12 hours.
♦ 12 hours is (12 × 60 × 60) = (12 ×3600) seconds
(ii) In other words, the hour hand turns through 2π radians in (12 × 3600) s
(iii) So it turns $\mathbf\small{\frac{2\pi }{12 \times 3600}}$ radians in 1 s
(iv) We can write: $\mathbf\small{\omega _h=\frac{2\pi }{12 \times 3600}}$
3. So we get: $\mathbf\small{\frac{\omega _m}{\omega _h}=\frac{2\pi }{3600}\div \frac{2\pi }{12 \times 3600}=\frac{2\pi }{3600}\times \frac{12 \times 3600 }{2 \pi} }$ = 12
■ Thus we get: angular speed of the minute hand is 12 times that of the hour hand
In the next section, we will see Relative velocity in two dimensional motion.
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