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 = ΔθΔt
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 ω (Greek small letter omega)
■ So we can write Eq.4.28: ω=ΔθΔt
4. The straight line between P and P’ is the 'displacement during Δt'
♦ It is denoted as →Δr
♦ It is a vector quantity
• But the arc length PP’ is the 'distance traveled during Δt'
♦ It is denoted as Δs
♦ 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 |→v|
■ So we get: |→v|=ΔsΔt
6. Now, using properties of arc, we have the relation: angle=arcradius
♦ 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 |→v|=Δθ×RΔt=RΔθΔt
• But from (3), we have: ΔθΔt=ω
■ So we get Eq.4.29: speed |→v|=Rω
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 n=1T
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 |→v|, T and n
A. Relation between speed |→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 = |→v| × T
Thus we get Eq.4.31: |→v|=2πRT
B. Relation between speed |→v| and n
(i) From Eq.4.30, we have: n=1T
• From this, we get: T=1n
(ii) So we can use 1T instead of T in Eq.4.31
• We can write: |→v|=2πR1n
• So we get Eq.4.32: |→v|=2πRn
12. Finally, we will see the relation between centripetal acceleration |→ac| and ω
(i) In the previous section, we saw Eq.4.27: |→ac|=|→v|2R
(ii) In this section we derived Eq.4.29: |→v|=Rω
(iii) So we can use 'Rω' instead of |→v| in Eq.4.27
• We get: |→ac|=R2ω2R
• Thus we get Eq.4.33: |→ac|=ω2R
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 ω
■ We can write: Angular speed of Q = Angular speed of P = ω
5. Now we will see linear speed
• Let the linear speed of P be |→vP|
• Let the linear speed of Q be |→vQ|
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, |→vP| must be greater than |→vQ|
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: |→v|=Rω
(ii) The angular speed ω is constant for all points on the disc
So we can write:
• |→vP|=Rω
• |→vQ|=R1ω
(iii) R1 is lesser than R
■ So |→vQ| will be lesser than |→vP|
■ 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 |→v|
• We can use Eq.4.32: |→v|=2πRn
• Substituting the values, we get: |→v|=2π×12×0.07 = 5.28 cm s-1.
2. Once we find the linear velocity, we can use Eq.4.29 to find angular speed ω
• |→v|=Rω
• Substituting the values, we get: 5.28=12×ω
• So ω = 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: |→ac|=ω2R
• Substituting the values, we get |→ac|=0.442×12 = 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 −5ˆj 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 −5ˆj ms-1.
• The unit vector ˆj 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, |→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 →v at P(0,-4) is −5ˆi 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 →ac
• We can use Eq.4.27: |→ac|=|→v|2R
• Substituting the values, we get: |→ac| = 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 = √(−2.83)2+2.832 = 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 →v′
• This →v′ will be perpendicular to OQ
• So →v′ makes an angle 45o with the x axis
(iv) Thus we get:
• Magnitude of the x component of →v′ = |→v′| × cos 45 = 5 × cos 45 = 3.536 ms-1.
• Magnitude of the y component of →v′ = |→v′| × sin 45 = 5 × sin 45 = 3.536 ms-1.
• So →v′=3.536ˆi+3.536ˆj
(v) Direction of →v′ is obtained as follows:
• tanθ=|→v′y||→v′x|=3.5363.536=1
• So θ = tan-11 = 45°
• We can write: →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 ωm and ωh respectively, find the ratio ωmωh
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 2π3600 radians in 1 s
(iv) We can write: ωm=2π3600
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 2π12×3600 radians in 1 s
(iv) We can write: ωh=2π12×3600
3. So we get: ωmωh=2π3600÷2π12×3600=2π3600×12×36002π = 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 = ΔθΔt
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 ω (Greek small letter omega)
■ So we can write Eq.4.28: ω=ΔθΔt
4. The straight line between P and P’ is the 'displacement during Δt'
♦ It is denoted as →Δr
♦ It is a vector quantity
• But the arc length PP’ is the 'distance traveled during Δt'
♦ It is denoted as Δs
♦ 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 |→v|
■ So we get: |→v|=ΔsΔt
6. Now, using properties of arc, we have the relation: angle=arcradius
♦ 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 |→v|=Δθ×RΔt=RΔθΔt
• But from (3), we have: ΔθΔt=ω
■ So we get Eq.4.29: speed |→v|=Rω
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 n=1T
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 |→v|, T and n
A. Relation between speed |→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 = |→v| × T
Thus we get Eq.4.31: |→v|=2πRT
B. Relation between speed |→v| and n
(i) From Eq.4.30, we have: n=1T
• From this, we get: T=1n
(ii) So we can use 1T instead of T in Eq.4.31
• We can write: |→v|=2πR1n
• So we get Eq.4.32: |→v|=2πRn
12. Finally, we will see the relation between centripetal acceleration |→ac| and ω
(i) In the previous section, we saw Eq.4.27: |→ac|=|→v|2R
(ii) In this section we derived Eq.4.29: |→v|=Rω
(iii) So we can use 'Rω' instead of |→v| in Eq.4.27
• We get: |→ac|=R2ω2R
• Thus we get Eq.4.33: |→ac|=ω2R
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 ω
■ We can write: Angular speed of Q = Angular speed of P = ω
5. Now we will see linear speed
• Let the linear speed of P be |→vP|
• Let the linear speed of Q be |→vQ|
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, |→vP| must be greater than |→vQ|
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: |→v|=Rω
(ii) The angular speed ω is constant for all points on the disc
So we can write:
• |→vP|=Rω
• |→vQ|=R1ω
(iii) R1 is lesser than R
■ So |→vQ| will be lesser than |→vP|
■ 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 |→v|
• We can use Eq.4.32: |→v|=2πRn
• Substituting the values, we get: |→v|=2π×12×0.07 = 5.28 cm s-1.
2. Once we find the linear velocity, we can use Eq.4.29 to find angular speed ω
• |→v|=Rω
• Substituting the values, we get: 5.28=12×ω
• So ω = 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: |→ac|=ω2R
• Substituting the values, we get |→ac|=0.442×12 = 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 −5ˆj 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 −5ˆj ms-1.
• The unit vector ˆj 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 →v at P(0,-4) is −5ˆi 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 →ac
• We can use Eq.4.27: |→ac|=|→v|2R
• Substituting the values, we get: |→ac| = 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 = √(−2.83)2+2.832 = 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 →v′
• This →v′ will be perpendicular to OQ
• So →v′ makes an angle 45o with the x axis
(iv) Thus we get:
• Magnitude of the x component of →v′ = |→v′| × cos 45 = 5 × cos 45 = 3.536 ms-1.
• Magnitude of the y component of →v′ = |→v′| × sin 45 = 5 × sin 45 = 3.536 ms-1.
• So →v′=3.536ˆi+3.536ˆj
(v) Direction of →v′ is obtained as follows:
• tanθ=|→v′y||→v′x|=3.5363.536=1
• So θ = tan-11 = 45°
• We can write: →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 ωm and ωh respectively, find the ratio ωmωh
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 2π3600 radians in 1 s
(iv) We can write: ωm=2π3600
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 2π12×3600 radians in 1 s
(iv) We can write: ωh=2π12×3600
3. So we get: ωmωh=2π3600÷2π12×3600=2π3600×12×36002π = 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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