Chapter 7.22 - Solved examples involving Torque

In the previous section, we saw center of gravity. In this section we will see some solved examples

Solved example 7.21
The metal bar in fig.7.103(a) is 0.70 m long and has a mass of 4 kg. It is supported on two knife edges placed 0.10 m from each end. A 6 kg mass is suspended from a point P, which is 0.20 m from the left support

Fig.7.103

Find the reactions at the supports. Assume the bar is of uniform cross section and homogeneous. Take g = 9.8 ms^-2
Solution:
1. Let us name the supports as A and B
■ The weight of the rod acts at it’s CG
• Given that, the metal bar is of uniform cross section and homogeneous.
■ So the CG of the rod will be at it’s geometric center
2. The detailed measurements are shown in fig.b
The CG is marked as G
3. Since the bar is in translational equilibrium, we have:
R_A + R_B - 6g - 4g = 0
⇒ R_A + R_B = 10g = 98.0 N.
4. Since the bar is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about the support A
(i) Torque created by R_A about A = zero
(ii) Torque created by 6g about A = 6g × 0.2 = 1.2g Nm (clockwise)
(iii) Torque created by 4g about A = 4g × 0.25 = 1.0g Nm (clockwise)
(iv) Torque created by R_B about A = R_B × 0.5 Nm (anti clockwise)
5. So applying the condition, we get:
1.2g + g - 0.5R_B = 0
⇒ R_B = 43.12 N
• Substituting this value of R_B in (3), we get: R_A = (98-43.12) = 54.88 N

Solved example 7.22
A 3 m long ladder having a mass of 20 kg, leans on a frictionless wall. It’s feet rest on the ground 1 m from the wall as shown in fig.7.104(a) below:

Fig.7.104

Find the reaction forces on the wall and the floor. Take g = 9.8 ms^-2
Solution:
1. Let the end points of the ladder be A and B
• A is 1 m from the wall. This is shown in fig.b
• Let C be the foot of the wall
2. The reaction from the floor at A will be normal to the floor
• This reaction is denoted as R_A
3. The frictional force prevents the point A from moving away from C
• This frictional force is denoted as F. It pulls the ladder towards C. Other wise the ladder will slip
• Also recall that, the frictional force is always parallel to the surface
4. The reaction from the wall at B will be normal to the wall
• This reaction is denoted as R_B
5. Given that, the wall is frictionless
• If there was friction, a force F would have acted parallel to the wall in the upward direction
• This force would have made some contribution towards: 'preventing the movement of B towards C'   • In other words, this force would have made some contribution towards: 'preventing the ladder from slipping'
• In addition to that, this force would have made some contribution towards resisting the vertical force (20 g) of the ladder
    ♦ Where 'g' is the acceleration due to gravity
• But in this problem there is no such force
    ♦ The slipping is prevented entirely by the horizontal force F at A
    ♦ The vertical load 20 g is resisted entirely by the vertical force R_A at A
6. The weight of the ladder is 20 g
• It acts downwards at the CG of the ladder
• The CG is marked as G in fig.b
7. The above steps gives us all the 4 forces acting on the ladder
• Now we can apply the conditions of equilibrium
• Since the ladder is in translational equilibrium, we have:
(i) In the x direction: F - R_B = 0
(ii) In the y direction: R_A - 20 g = 0
⇒ R_A = 20 g = 196 N
8. Since the bar is in rotational equilibrium, we have: net torque = 0
Let us take the torques about A
(i) Torque created by R_A about A = zero
(ii) Torque created by F about A = zero
(iii) Torque created by R_B about A = R_B × BC
So we want the length of BC
• Applying Pythagoras theorem to the right triangle ABC, we get:
$\mathbf\small{BC=\sqrt{AB^2-AC^2}=\sqrt{3^2-1^2}=2\sqrt{2}\;\text{m}}$
Thus the required torque = $\mathbf\small{2\sqrt{2}R_B\;\text{N m}}$ (clockwise) 
(iv) Torque created by 20 g about A = 20 g × AD
• So we want the length of AD
• D is the foot of the perpendicular drawn from G
• Consider the similar triangles ABC and AGD
• We have: $\mathbf\small{\frac{AD}{AC}=\frac{AG}{AB}}$
$\mathbf\small{\Rightarrow \frac{AD}{1}=\frac{1.5}{3}}$
⇒ AD = 0.5 m
• Thus the torque = 20 g × 0.5 × 9.8 = 98 N m (anti clockwise)
9. Applying the condition, we get:
$\mathbf\small{2\sqrt{2}R_B-98=0}$
⇒ R_B = 34.65 N
10. Substituting this value of R_B in 7(i), we get: F = 34.65 N.
11. At the point A, two forces are acting on the ladder:
• R_A vertically and F horizontally. They are shown in fig.c
• The resultant of the two forces = $\mathbf\small{\sqrt{(R_A)^2+F^2}=\sqrt{196^2+34.65^2}=}$ 199.04 N
• Let this resultant make an angle α with the horizontal
• Then $\mathbf\small{\alpha=\tan^{-1}\frac{R_A}{F}=\tan^{-1}\frac{196}{34.65}=}$ 79.97^o

Solved example 7.23 
A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in fig.7.105(a) below:

Fig.7.105

The angles made by the strings with the vertical are 36.9^o and 53.1^o respectively. The bar is 2 m long. Calculate the distance d of the center of gravity of the bar from it’s left end
Solution:
1. The free body diagram is shown in fig.b
• Let T₁ and T₂ be the tensions in the strings
• In fig.a, we see that, the left string makes an angle of 36.9^o with the vertical wall
    ♦ In fig.b, we see that, this string makes the same angle with the vertical dotted line
    ♦ The angles are same because, they are alternate angles 
• In fig.a, we see that, the right string makes an angle of 53.1^o with the vertical wall
    ♦ In fig.b, we see that, this string makes the same angle with the vertical dotted line
    ♦ The angles are same because, they are alternate angles
• Now we can resolve the tensions into horizontal and vertical components. This is shown in fig.c
2. Since the bar is in translational equilibrium, we have:
(i) In the x direction: - T₁ sin θ₁ + T₂ sin θ₂  = 0
⇒ T₁ sin θ₁ = T₂ sin θ₂.
⇒ T₁ sin 36.9 = T₂ sin 53.1 = T₂ cos (90 - 36.9) = T₂ cos 36.9
⇒ $\mathbf\small{\frac{T_2}{T_1}=\tan36.9 =0.75}$
⇒ T₂ = 0.75 T₁
(ii) In the y direction: T₁ cos θ₁ + T₂ cos θ₂ - W = 0
⇒ T₁ cos 36.9 + T₂ cos 53.1 = W
⇒0.8 T₁ + 0.6 T₂ = W
3. Since the bar is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about G
(i) Torque created by T₁ sin θ₁ about G = zero
(ii) Torque created by T₁ cos θ₁ about G = T₁ cos θ₁ × d (clockwise)
(iii) Torque created by W about G = zero
(iv) Torque created by T₂ cos θ₂ about G = T₂ cos θ₂ × (2-d) (anti clockwise)
(i) Torque created by T₂ sin θ₂ about G = zero
4. Applying the condition, we get:
[T₁ cos θ₁ × d] - [T₂ cos θ₂ × (2-d)] = 0
⇒ [T₁ × cos 36.9 × d] - [0.75 T₁ × cos 53.1 × (2-d)] = 0
(∵ from 2(i), we have: T₂ = 0.75 T₁)
⇒ [0.8 T₁ d] - [0.45 T₁ (2-d)] = 0
⇒ 0.8 T₁ d - 0.9 T₁ + 0.45 T₁ d = 0
⇒ 1.25 T₁ d = 0.9 T₁
⇒ 1.25 d = 0.9
⇒ d = 0.72 m

Solved example 7.24
A car weighs 1800 kg. The distance between it’s front and back axles is 1.8 m. It’s center of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel
Solution:
1. Fig.7.106 below shows the schematic diagram

Fig.106

• The small grey circles at the center of the wheels denote the axles
• The front and rear axles are named as A and B respectively
2. When the car is in translational equilibrium, we have:
R_A + R_B -1800g = 0
3. When the car is in rotational equilibrium, we have: net torque = 0
Let us take the torques about G
(i) Torque created by R_A about G = 1.05 R_A (clockwise) 
(ii) Torque created by 1800g about G = zero
(iii) Torque created by R_B about G = 0.75 R_B (anticlockwise)
4. Applying the condition, we get:
1.05 R_A - 0.75 R_B = 0
1.05 R_A = 0.75 R_B
R_B = 1.4 R_A
5. Substituting this in (2), we get:
R_A + 1.4R_A = 1800 × 9.8
⇒ R_A = 7350 N
• So R_B = 1.4 × 7350 = 10290 N
6. The reaction on the front axle A is 7350 N
• But the load from this axle is resisted by two front wheels
• So reaction on each of the front wheels = ⁷³⁵⁰⁄₂ = 3675 N
7. The reaction on the back axle B is 10290 N
• But the load from this axle is resisted by two back wheels
• So reaction on each of the back wheels = ¹⁰²⁹⁰⁄₂ = 5145 N

Solved example 7.25
As shown in fig.7.107(a), the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE 0.5 m long is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. Take g = 9.8 ms^-2 [Hint: Consider the equilibrium of each side of the ladder separately]

Fig.7.107

Solution:
1. The detailed measurements are shown in fig.b
• Given:
    ♦ BD = 0.8 m
    ♦ DF = FA = 0.4 m
    ♦ DE = 0.5 m
• Let ∠ABC = θ
    ♦ Since DE is parallel to the ground, we get: ∠ADE = θ   
2. Let us calculate the other required dimensions:
• A perpendicular is dropped from A onto BC
    ♦ This is shown as red dashed line 
    ♦ The foot of the perpendicular on BC is O
    ♦ The foot of the perpendicular on DE is G
• Applying Pythagoras theorem to the right triangle ADG, we get:
$\mathbf\small{AG=\sqrt{AD^2-DG^2}=\sqrt{0.8^2-0.25^2}=0.76\;\text{m}}$
3. In the similar triangles ABO and ADG, we get:
• $\mathbf\small{\frac{AD}{AB}=\frac{AG}{AO}}$
$\mathbf\small{\Rightarrow \frac{0.8}{1.6}=\frac{0.76}{AO}}$
⇒ AO = 1.52 m
4. Again in the same similar triangles ABO and ADG, we get:
$\mathbf\small{\frac{AD}{AB}=\frac{DG}{BO}}$
$\mathbf\small{\Rightarrow \frac{0.8}{1.6}=\frac{0.25}{BO}}$
⇒ BO = 0.5 m
5. A perpendicular is dropped from F onto AG
• The foot of this perpendicular is H
• In the similar triangles AFH and ADG, we get:
$\mathbf\small{\frac{AF}{AD}=\frac{FH}{DG}}$
$\mathbf\small{\Rightarrow \frac{0.4}{0.8}=\frac{FH}{0.25}}$
⇒ FH = 0.125 m
6. Let us write the above distances together:
    ♦ BD = 0.8 m
    ♦ DF = FA = 0.4 m
    ♦ DE = 0.5 m
    ♦ AG =0.76 m
    ♦ AO = 1.52 m
    ♦ BO = 0.5 m
    ♦ FH = 0.125 m
• Thus we obtained all the distances that we will soon require for torque calculations
7. The forces are shown in fig.c below:

Fig.7.107 (c) & (d)

• R_B is the normal reaction at B
• R_C is the normal reaction at C
• Note that the tensions T in the string are internal forces and so will cancel each other
8. For translational equilibrium, we have:
• R_B + R_C - (40 × 9.8) = 0
⇒ R_B + R_C - 392 = 0
⇒ R_B + R_C = 392
9. For rotational equilibrium, we need to calculate the torques first:
• Let us find the torques about O
(i) Torque created by R_B about O = R_B × BO = R_B × 0.5 = 0.5R_B Nm (clockwise)
(ii) Torque created by the 40 kg load about O = (40 × 9.8) × FH = 392 0.125 = 49 Nm (anti clockwise)
(iii) Torque created by R_C about O = R_C × CO = R_C × 0.5 = 0.5R_C Nm (anti clockwise)
10. Applying the condition, we get:
⇒ 0.5R_B - 49 - 0.5R_C = 0
⇒ 0.5R_B - 0.5R_C = 49
⇒ R_B - R_C = 98
11. Solving the equations in (8) and (10), we get:
R_B = 245 N, R_C = 147 N
12. Next we want to find T
• In the fig.d above, the FBD of side AB is shown separately
• The forces acting are: R_B, T and the load of 40 kg
• Let us calculate the torques about A:
(i) Torque created by R_B about A = R_B × BO = R_B × 0.5 = 0.5R_B Nm (clockwise)
(ii) Torque created by T about A = T × AG = T × 0.76 = 0.76T Nm (anti clockwise)
(ii) Torque created by the 40 kg load about A = (40 × 9.8) × FH = 392 × 0.125 = 49 Nm (anti clockwise)
13. Applying the condition, we get:
0.5R_B - 0.76T - 49 = 0
0.5 × 245 - 0.76T - 49 = 0
T = 96.7 N

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Two more solved examples can be seen here

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In the next section, we will see moment of inertia

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Chapter 6.14 - Tension in String

In the previous section, we saw that the velocity of an ‘object in vertical circular motion’ will be different at different points along the path. We saw the method to find the velocity at any point. 
• When an object is tied to a string and whirled in a vertical circle, not only the velocity, but also the ‘tension in the string’ will vary. 
• In this section we will see the reason for such a variation. We will also see the method to find the tension at any point

1. In fig.6.45(a) below, an object of mass ‘m’ is tied to a string and whirled in  vertical circle. 

Fig.6.45

• ‘O’ is the center of the circle and ‘r’ is the radius 
2. The FBD of the object when it is at A is shown in fig.b
• We know that, the force in a string will always be tensile in nature (Details here). This tension will be pulling at both the ends of the strings 
• So the object will experience a pulling through the string. This pulling force by the string is shown in fig.b as ‘T_A’
3. Also the object will experience a pull towards the center of the earth 
• This force by the earth is shown in fig.b as ‘mg’
4. When the object is at A, 
    ♦ the force ‘T_A’ will be vertically upwards
    ♦ the force ‘mg’ will be vertically downwards
• So both the forces act along the same line but opposite in direction
• The resultant force is (T_A-mg)
■ The sign convention is: • Forces directed towards the center of the circle are positive • Forces directed away from the center of the circle are negative
5. For circular motion (both horizontal and vertical), centripetal force is required. 
• The person who whirls the object has to provide this force
• For the person, the only contact with the object is through the string. So the centripetal force should be provided through the string
• We see that, the net force in the direction of the string is (T_A-mg) 
6. By Newton's second law, the net force must be equal to (mass × acceleration)
• The acceleration here is the 'centripetal acceleration' given by $\mathbf\small{\frac{v_A^2}{r}}$  
• So (mass × acceleration) = $\mathbf\small{\frac{mv_A^2}{r}}$
• This is the centripetal force at point A. 
• So we can write: $\mathbf\small{T_A-mg=\frac{mv_A^2}{r}}$
$\mathbf\small{\Longrightarrow T_A=m \left(\frac{v_A^2}{r}+g \right)}$
• Centripetal force is always directed towards the center. So it is positive
■ Thus, if we know the velocity at A, we can easily calculate the tension at A

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7. Next we will calculate T_B, the tension at B
• The FBD of the object when it is at B is shown in fig.c
• We see that the net force in the direction of the string is T_B itself
• This is because, 'mg' is perpendicular to the string and hence have no force component along the string
8. By Newton's second law, the net force must be equal to (mass × acceleration)
• The acceleration here is the 'centripetal acceleration' given by $\mathbf\small{\frac{v_B^2}{r}}$  
• So (mass × acceleration) = $\mathbf\small{\frac{mv_B^2}{r}}$
• This is the centripetal force at point B. 
• So we can write: $\mathbf\small{T_B=\frac{mv_B^2}{r}}$
■ Thus, if we know the velocity (v_B) at B, we can easily calculate the tension at B
9. How do we calculate v_B?
• If we know the velocity v_A at the lowest point A, we can calculate v_B
• For that, we use the equation obtained in (10) in the previous section
    ♦ Here we will write it again: $\mathbf\small{v_B=\sqrt{v_A^2-2gr}}$
■ Once we calculate v_B, we can easily calculate the tension at B using the equation in (8) above 

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10. Next we will calculate T_C, the tension at C
• The FBD of the object when it is at C is shown in fig.d
• We see that the net force in the direction of the string is (T_C+mg)
11. By Newton's second law, the net force must be equal to (mass × acceleration)
• The acceleration here is the 'centripetal acceleration' given by $\mathbf\small{\frac{v_C^2}{r}}$  
• So (mass × acceleration) = $\mathbf\small{\frac{mv_C^2}{r}}$
• This is the centripetal force at point C 
• So we can write: $\mathbf\small{T_C+mg=\frac{mv_C^2}{r}}$
$\mathbf\small{\Longrightarrow T_C=m \left(\frac{v_C^2}{r}-g \right)}$
■ Thus, if we know the velocity at C, we can easily calculate the tension at C
12. If we know the velocity v_A at the lowest point A, we can calculate the velocity v_C at C
• How do we calculate v_C?
• If we know the velocity v_A at the lowest point A, we can calculate v_C
• For that, we use the equation obtained in (13) in the previous section
    ♦ Here we will write it again: $\mathbf\small{v_C=\sqrt{v_A^2-4gr}}$
■ Once we calculate v_C, we can easily calculate the tension at C using the equation in (11) above

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13. Next we will calculate T_D, the tension at D
• The FBD of the object when it is at D is shown in fig.e
• We see that the net force in the direction of the string is T_D itself
• This is because, 'mg' is perpendicular to the string and hence have no force component along the string
14. By Newton's second law, the net force must be equal to (mass × acceleration)
• The acceleration here is the 'centripetal acceleration' given by $\mathbf\small{\frac{v_D^2}{r}}$  
• So (mass × acceleration) = $\mathbf\small{\frac{mv_D^2}{r}}$
• This is the centripetal force at point D 
• So we can write: $\mathbf\small{T_D=\frac{mv_D^2}{r}}$
■ Thus, if we know the velocity at D, we can easily calculate the tension at D
15. If we know the velocity v_A at the lowest point A, we can calculate the velocity v_D at D
• How do we calculate v_D?
• If we know the velocity v_A at the lowest point A, we can calculate v_D
• For that, we use the equation obtained in (16) in the previous section
    ♦ Here we will write it again: $\mathbf\small{v_D=\sqrt{v_A^2-2gr}}$
■ Once we calculate v_D, we can easily calculate the tension at D using the equation in (14) above  
■ It is interesting to note that, at B and D, just like the 'magnitudes of velocities', the 'magnitudes of tensions' are also the same

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• So we successfully calculated the tensions at all the four quadrant points. But what about the intermediate points?
• For that, we will have to apply 'resolution of forces' (Details here). Let us see how it is done:
16. In fig.6.46(a) below, the object is at 'P'

Fig.6.46

• At that instant, the string makes an angle θ with the vertical
17. In fig.b, the portion of the object alone is shown
• In this fig.b, a vertical is drawn through the object. 
• The following two verticals will be parallel:
    ♦ Vertical through O
    ♦ Vertical through P
• So, if we extend the string downwards along the same line, that extension will make the same angle θ with the vertical
18. This 'same angle' is our clue 
• We know that, the 'component which is adjacent to the angle' will get the cosine. And the other component will get the sine
• So the component of the weight 'mg' which acts along the line of the string is mg cosθ
• This is shown in the FBD in fig.c
• We see that the net force in the direction of the string is (T_P -mg cosθ)
19. By Newton's second law, the net force must be equal to (mass × acceleration)
• The acceleration here is the 'centripetal acceleration' given by $\mathbf\small{\frac{v_P^2}{r}}$  
• So (mass × acceleration) = $\mathbf\small{\frac{mv_P^2}{r}}$
• This is the centripetal force at point P. 
• So we can write: $\mathbf\small{T_P-mg \cos \theta =\frac{mv_P^2}{r}}$
• $\mathbf\small{\Longrightarrow T_P=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
20. If we know the velocity v_A at the lowest point A, we can calculate the velocity v_P at P
• For that, we use the equation obtained in (23) in the previous section
    ♦ Here we will write it again: $\mathbf\small{v_P=\sqrt{v_A^2-2gr(1-\cos \theta)}}$
■ Once we calculate v_P, we can easily calculate the tension at P using the equation in (19) above
■ So, if we know the velocity v_P at any point P, we can easily obtain the tension T_P at that point
For that, we use the equation in (19) above

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• If we can use the equation derived in (19) above for 'any point', it must be applicable to points A, B, C and D also. Let us check:
21. First we will check the equation at point 'A' 
• We have: $\mathbf\small{T_P=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at A, θ = 0^o
    ♦ So cos θ = cos 0 = 1
• Substituting the known values, we get:
$\mathbf\small{T_A=m \left(\frac{v_A^2}{r}+g \right)}$
■ This is the same equation that we obtained in (7) above
22. Next we will check the equation at point 'B' 
• We have: $\mathbf\small{T_P=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at B, θ = 90^o
    ♦ So cos θ = cos 90 = 0
• Substituting the known values, we get: $\mathbf\small{T_B=m \left(\frac{v_B^2}{r}+g \times 0 \right)}$
$\mathbf\small{\Longrightarrow T_B=\frac{mv_B^2}{r}}$
■ This is the same equation that we obtained in (9) above
23. Next we will check the equation at point 'C' 
• We have: $\mathbf\small{T_P=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at C, θ = 180^o
    ♦ So cos θ = cos 180 = -1
• Substituting the known values, we get:
$\mathbf\small{T_C=m \left(\frac{v_C^2}{r}-g \right)}$
■ This is the same equation that we obtained in (11) above
24. Finally, we will check the equation at point 'D' 
• We have: $\mathbf\small{T_P=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$
• When the object is at D, θ = 270^o
    ♦ So cos θ = cos 270 = 0
• Substituting the known values, we get: $\mathbf\small{T_D=m \left(\frac{v_D^2}{r}+g \times 0 \right)}$
$\mathbf\small{\Longrightarrow T_D=\frac{mv_D^2}{r}}$
■ This is the same equation that we obtained in (14) above

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■ So it is confirmed:

• We can use the equation $\mathbf\small{T_P=m \left(\frac{v_P^2}{r}+g \cos \theta \right)}$ to find the tension at any point along the vertical circle
• But of course, we need to find the velocity at that point first
    ♦ For that, we use the equation $\mathbf\small{v_P=\sqrt{v_A^2-2gr(1-\cos \theta)}}$ which we derived in (23) of the previous section
    ♦ Where v_A is the velocity at the lowest point A

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Minimum velocity required at the lowest point A

1. In the fig.6.47(a) below, the object is at the highest point C

Fig.6.47

• We know that the tension in the string at the instant when the object is at C is $\mathbf\small{T_C=m \left(\frac{v_C^2}{r}-g \right)}$
2. If $\mathbf\small{\frac{v_C^2}{r}}$ becomes equal to g, the tension will become zero
• We know that, when the object is being whirled, the string will be in a ‘pulled tight’ condition
    ♦ If the string is in a ‘pulled tight’ condition, it means that it is experiencing tension
• Conversely, if there is tension in the string, then it would be in a 'pulled tight' condition
• So if there is no tension, the string will not be in a ‘pulled tight’ condition
    ♦ In other words, if there is no tension, the string will slacken. This is shown in fig.6.47(b)
3. So we can write: $\mathbf\small{\frac{v_C^2}{r}}$ must be greater than g
• But r and g are constants. $\mathbf\small{v_C}$ is the only variable
■ So we can write: $\mathbf\small{v_C^2}$ must be greater than $\mathbf\small{rg}$
That is., $\mathbf\small{v_C}$ must be greater than $\mathbf\small{\sqrt{rg}}$
4. But $\mathbf\small{v_C}$ depends on the velocity $\mathbf\small{v_A}$ at the lowest point A
• The relation is: $\mathbf\small{v_C=\sqrt{v_A^2-4gr}}$   
$\mathbf\small{\Longrightarrow v_C^2={v_A^2-4gr}}$
• Substituting this result in (3), we get:
$\mathbf\small{v_A^2-4gr}$ must be greater than $\mathbf\small{rg}$  
• That is., $\mathbf\small{v_A^2}$ must be greater than $\mathbf\small{rg+4rg}$  
• That is., $\mathbf\small{v_A^2}$ must be greater than $\mathbf\small{5rg}$
5. Note that we cannot blame v_C if the string slackens at C
• This is because, v_C depends on v_A 
• We have to give the sufficient velocity at the lowest point A. Only then will the object have the required velocity at the highest point C
6. We saw that if the string is not to slacken at C, the velocity v_C at C must be greater than $\mathbf\small{\sqrt{rg}}$
■ What if v_C is exactly equal to $\mathbf\small{\sqrt{rg}}$?
Ans: Then at that instant the tension T_C will become zero
• But v_C is not equal to zero. It is equal to $\mathbf\small{\sqrt{rg}}$
Since v_C is not equal to zero, the object will pass the point C
• Once it passes the point C, it's velocity goes on increasing. So it will complete the circular path and will return to the point A

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In the next section, we will see the motion of a bead through a vertical circular wire loop.

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