In the previous section, we saw that the critical velocity at the topmost point of the vertical circle is $\mathbf\small{\sqrt{rg}}$. We also saw that the direction of the reaction at the top most point can be upwards or downwards. In this section we will see some practical cases.
In fig.6.53 below, a sphere of mass 1.2 kg rolls down a track shown in yellow color.
It starts rolling from the point A. It reaches the lowest point B and then rolls upwards on the other side. We must ensure that the sphere does not fly off while passing through the highest point C on the other side. How can we ensure that? Assume that the surfaces are smooth. [g = 10 ms-2]
Let us analyze:
1. Let the sphere roll down from a height 'h'
• That is., the point 'A' is at a height of 'h' from the datum
• When it reaches the lowest point 'B', it will have a large velocity
2. But when it rolls upwards on the other side, the velocity will begin to decrease.
• The velocity will reach the smallest possible value when it reaches 'C'
• However, since 'C' is at a lower level than 'A', vC will have a significant value
3. If vC is very large, the sphere will fly off at 'C'
• We have to find the safe velocity at C so that 'flying off' is avoided
4. For that, we will draw the FBD of the sphere at C
• It is shown in fig.6.54 below:
• The normal reaction from the track is FN(C)
5. So the net force acting on the sphere is (mg-FN(C))
• This net force must be equal to the centripetal force
• So we can write: $\mathbf\small{mg-F_{N(C)}=\frac{mv_C^2}{r}}$
$\mathbf\small{\Longrightarrow F_{N(C)}=m \left(g-\frac{v_C^2}{r}\right)}$
6. If the normal reaction FN(C) is zero, then it would mean that, the sphere is not touching the track
• That is., if the normal reaction FN(C) is zero, the sphere is about to fly off
• Then the equation in (5) becomes: $\mathbf\small{0=m \left(g-\frac{v_C^2}{r}\right)}$
7. Obviously, for the left side to become zero, $\mathbf\small{\frac{v_C^2}{r}}$ must be equal to g
• So we get: $\mathbf\small{v_C=\sqrt{rg}}$
• We can write: If vC is equal to $\mathbf\small{\sqrt{rg}}$, the sphere will be at the 'point of flying off'
8. We can prevent such a situation by reducing the speed
• That is., vC must be less than $\mathbf\small{\sqrt{rg}}$
9. In our present case,
• If the elevation of point A is high, vC will be high
• If the elevation of point A is low, vC will be low
10. Let us find the limiting height:
• For that, we use the Law of conservation of energy
• Let us apply it to the points A and C
■ By the Law of conservation of energy, the total energy must be the same at those two points
• We can write EA = EC
11. Let us write the various energies:
(i) Total energy EA:
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh}$
■ So total energy EA = $\mathbf\small{mgh}$
(ii) Total energy EC
• Kinetic energy = $\mathbf\small{\frac{m\,v_C^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mg \times 3}$
■ So total energy EC = $\mathbf\small{\frac{m\,v_C^2}{2}+3mg}$
12. Equating the two energies we get:
$\mathbf\small{mgh=\frac{m\,v_C^2}{2}+3mg}$
• Multiplying both sides by $\mathbf\small{\frac{2}{m}}$, we get:
$\mathbf\small{2gh=v_C^2+6g}$
$\mathbf\small{\Longrightarrow 20h=v_C^2+60}$
13. But vC must not exceed $\mathbf\small{\sqrt{rg}}$
• That is., vC must not exceed $\mathbf\small{\sqrt{1.5 \times 10} = \sqrt{15}}$
• Substituting this in (12), we get:
$\mathbf\small{20h=15+60=75}$
Thus we get: h = 75⁄20 = 3.75 m
14. If the sphere is released from a greater height, the speed vC will exceed $\mathbf\small{\sqrt{15}}$ the and it will fly off at C
• So we must release it only from a height less than 3.75 m
Another example:
In fig.6.55 below, a sphere of mass 1.2 kg rolls down a track shown in yellow color.
It starts rolling from the point A. It reaches the lowest point B and then rolls upwards into a vertical circular track on the other side. We must ensure that the sphere does not fall while passing through the highest point C on the circular track. How can we ensure that? Assume that the surfaces are smooth. [g = 10 ms-2]
Let us analyze:
1. Let the sphere roll down from a height 'h'
• That is., the point 'A' is at a height of 'h' from the datum
• When it reaches the lowest point B, it will have a large velocity
2. But when it rolls upwards on vertical circular track the other side, the velocity will begin to decrease.
• The velocity will reach the smallest possible value when it reaches 'C'
• However, since 'C' is at a lower level than 'A', vC will have a significant value
3. Even then, if vC is very small, the sphere will fall down at 'C'
• We have to find the safe velocity at C
4. For that, we will draw the FBD of the sphere at C
• It is shown in fig.6.56 below:
• The normal reaction from the track is FN(C)
5. So the net force acting on the sphere is (mg+FN(C))
• This net force must be equal to the centripetal force
• So we can write: $\mathbf\small{mg+F_{N(C)}=\frac{mv_C^2}{r}}$
$\mathbf\small{\Longrightarrow F_{N(C)}=m \left(\frac{v_C^2}{r}-g\right)}$
6. If the normal reaction FN(C) is zero, then it would mean that, the sphere is not touching the track
• That is., if the normal reaction FN(C) is zero, the sphere will fall down
• Then the equation in (5) becomes: $\mathbf\small{0=m \left(\frac{v_C^2}{r}-g\right)}$
7. Obviously, for the left side to become zero, $\mathbf\small{\frac{v_C^2}{r}}$ must be equal to g
• So we get: $\mathbf\small{v_C=\sqrt{rg}}$
• We can write: If vC is equal to $\mathbf\small{\sqrt{rg}}$, the sphere will fall down
8. We can prevent such a situation by increasing the speed
• That is., vC must be greater than $\mathbf\small{\sqrt{rg}}$
9. In our present case,
• If the elevation of point A is high, vC will be high
• If the elevation of point A is low, vC will be low
10. Let us find the limiting height:
• For that, we use the Law of conservation of energy
• Let us apply it to the points A and C
■ By the Law of conservation of energy, the total energy must be the same at those two points
• We can write EA = EC
11. Let us write the various energies:
(i) Total energy EA:
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh}$
■ So total energy EA = $\mathbf\small{mgh}$
(ii) Total energy EC
• Kinetic energy = $\mathbf\small{\frac{m\,v_C^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mg \times 3}$
■ So total energy EC = $\mathbf\small{\frac{m\,v_C^2}{2}+3mg}$
12. Equating the two energies we get:
$\mathbf\small{mgh=\frac{m\,v_C^2}{2}+3mg}$
• Multiplying both sides by $\mathbf\small{\frac{2}{m}}$, we get:
$\mathbf\small{2gh=v_C^2+6g}$
$\mathbf\small{\Longrightarrow 20h=v_C^2+60}$
13. But vC must exceed $\mathbf\small{\sqrt{rg}}$
• That is., vC must exceed $\mathbf\small{\sqrt{1.5 \times 10} = \sqrt{15}}$
• Substituting this in (12), we get:
$\mathbf\small{20h=15+60=75}$
Thus we get: h = 75⁄20 = 3.75 m
14. If the sphere is released from a height lesser than 3.75 m, the speed vC will be lesser than $\mathbf\small{\sqrt{15}}$ and so the sphere will fall at point C
• So we must release it only from a height greater than 3.75 m
We can write a summary based on the above two examples
■ In example 1, the normal reaction from the track is given by: $\mathbf\small{F_{N(C)}=m \left(g-\frac{v_C^2}{r}\right)}$
♦ If FN(C) is not to become zero, vC must be less than $\mathbf\small{\sqrt{rg}}$
♦ This is because, $\mathbf\small{\frac{v_C^2}{r}}$ is being subtracted from g
■ In example 2, the normal reaction from the track is given by: $\mathbf\small{F_{N(C)}=m \left(\frac{v_C^2}{r}-g\right)}$
♦ If FN(C) is not to become zero, vC must be greater than $\mathbf\small{\sqrt{rg}}$
♦ This is because, g is being subtracted from $\mathbf\small{\frac{v_C^2}{r}}$
In fig.6.53 below, a sphere of mass 1.2 kg rolls down a track shown in yellow color.
 |
| Fig.6.53 |
Let us analyze:
1. Let the sphere roll down from a height 'h'
• That is., the point 'A' is at a height of 'h' from the datum
• When it reaches the lowest point 'B', it will have a large velocity
2. But when it rolls upwards on the other side, the velocity will begin to decrease.
• The velocity will reach the smallest possible value when it reaches 'C'
• However, since 'C' is at a lower level than 'A', vC will have a significant value
3. If vC is very large, the sphere will fly off at 'C'
• We have to find the safe velocity at C so that 'flying off' is avoided
4. For that, we will draw the FBD of the sphere at C
• It is shown in fig.6.54 below:
 |
| Fig.6.54 |
5. So the net force acting on the sphere is (mg-FN(C))
• This net force must be equal to the centripetal force
• So we can write: $\mathbf\small{mg-F_{N(C)}=\frac{mv_C^2}{r}}$
$\mathbf\small{\Longrightarrow F_{N(C)}=m \left(g-\frac{v_C^2}{r}\right)}$
6. If the normal reaction FN(C) is zero, then it would mean that, the sphere is not touching the track
• That is., if the normal reaction FN(C) is zero, the sphere is about to fly off
• Then the equation in (5) becomes: $\mathbf\small{0=m \left(g-\frac{v_C^2}{r}\right)}$
7. Obviously, for the left side to become zero, $\mathbf\small{\frac{v_C^2}{r}}$ must be equal to g
• So we get: $\mathbf\small{v_C=\sqrt{rg}}$
• We can write: If vC is equal to $\mathbf\small{\sqrt{rg}}$, the sphere will be at the 'point of flying off'
8. We can prevent such a situation by reducing the speed
• That is., vC must be less than $\mathbf\small{\sqrt{rg}}$
9. In our present case,
• If the elevation of point A is high, vC will be high
• If the elevation of point A is low, vC will be low
10. Let us find the limiting height:
• For that, we use the Law of conservation of energy
• Let us apply it to the points A and C
■ By the Law of conservation of energy, the total energy must be the same at those two points
• We can write EA = EC
11. Let us write the various energies:
(i) Total energy EA:
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh}$
■ So total energy EA = $\mathbf\small{mgh}$
(ii) Total energy EC
• Kinetic energy = $\mathbf\small{\frac{m\,v_C^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mg \times 3}$
■ So total energy EC = $\mathbf\small{\frac{m\,v_C^2}{2}+3mg}$
12. Equating the two energies we get:
$\mathbf\small{mgh=\frac{m\,v_C^2}{2}+3mg}$
• Multiplying both sides by $\mathbf\small{\frac{2}{m}}$, we get:
$\mathbf\small{2gh=v_C^2+6g}$
$\mathbf\small{\Longrightarrow 20h=v_C^2+60}$
13. But vC must not exceed $\mathbf\small{\sqrt{rg}}$
• That is., vC must not exceed $\mathbf\small{\sqrt{1.5 \times 10} = \sqrt{15}}$
• Substituting this in (12), we get:
$\mathbf\small{20h=15+60=75}$
Thus we get: h = 75⁄20 = 3.75 m
14. If the sphere is released from a greater height, the speed vC will exceed $\mathbf\small{\sqrt{15}}$ the and it will fly off at C
• So we must release it only from a height less than 3.75 m
Another example:
In fig.6.55 below, a sphere of mass 1.2 kg rolls down a track shown in yellow color.
 |
| Fig.6.55 |
Let us analyze:
1. Let the sphere roll down from a height 'h'
• That is., the point 'A' is at a height of 'h' from the datum
• When it reaches the lowest point B, it will have a large velocity
2. But when it rolls upwards on vertical circular track the other side, the velocity will begin to decrease.
• The velocity will reach the smallest possible value when it reaches 'C'
• However, since 'C' is at a lower level than 'A', vC will have a significant value
3. Even then, if vC is very small, the sphere will fall down at 'C'
• We have to find the safe velocity at C
4. For that, we will draw the FBD of the sphere at C
• It is shown in fig.6.56 below:
 |
| Fig.6.56 |
5. So the net force acting on the sphere is (mg+FN(C))
• This net force must be equal to the centripetal force
• So we can write: $\mathbf\small{mg+F_{N(C)}=\frac{mv_C^2}{r}}$
$\mathbf\small{\Longrightarrow F_{N(C)}=m \left(\frac{v_C^2}{r}-g\right)}$
6. If the normal reaction FN(C) is zero, then it would mean that, the sphere is not touching the track
• That is., if the normal reaction FN(C) is zero, the sphere will fall down
• Then the equation in (5) becomes: $\mathbf\small{0=m \left(\frac{v_C^2}{r}-g\right)}$
7. Obviously, for the left side to become zero, $\mathbf\small{\frac{v_C^2}{r}}$ must be equal to g
• So we get: $\mathbf\small{v_C=\sqrt{rg}}$
• We can write: If vC is equal to $\mathbf\small{\sqrt{rg}}$, the sphere will fall down
8. We can prevent such a situation by increasing the speed
• That is., vC must be greater than $\mathbf\small{\sqrt{rg}}$
9. In our present case,
• If the elevation of point A is high, vC will be high
• If the elevation of point A is low, vC will be low
10. Let us find the limiting height:
• For that, we use the Law of conservation of energy
• Let us apply it to the points A and C
■ By the Law of conservation of energy, the total energy must be the same at those two points
• We can write EA = EC
11. Let us write the various energies:
(i) Total energy EA:
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh}$
■ So total energy EA = $\mathbf\small{mgh}$
(ii) Total energy EC
• Kinetic energy = $\mathbf\small{\frac{m\,v_C^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mg \times 3}$
■ So total energy EC = $\mathbf\small{\frac{m\,v_C^2}{2}+3mg}$
12. Equating the two energies we get:
$\mathbf\small{mgh=\frac{m\,v_C^2}{2}+3mg}$
• Multiplying both sides by $\mathbf\small{\frac{2}{m}}$, we get:
$\mathbf\small{2gh=v_C^2+6g}$
$\mathbf\small{\Longrightarrow 20h=v_C^2+60}$
13. But vC must exceed $\mathbf\small{\sqrt{rg}}$
• That is., vC must exceed $\mathbf\small{\sqrt{1.5 \times 10} = \sqrt{15}}$
• Substituting this in (12), we get:
$\mathbf\small{20h=15+60=75}$
Thus we get: h = 75⁄20 = 3.75 m
14. If the sphere is released from a height lesser than 3.75 m, the speed vC will be lesser than $\mathbf\small{\sqrt{15}}$ and so the sphere will fall at point C
• So we must release it only from a height greater than 3.75 m
We can write a summary based on the above two examples
■ In example 1, the normal reaction from the track is given by: $\mathbf\small{F_{N(C)}=m \left(g-\frac{v_C^2}{r}\right)}$
♦ If FN(C) is not to become zero, vC must be less than $\mathbf\small{\sqrt{rg}}$
♦ This is because, $\mathbf\small{\frac{v_C^2}{r}}$ is being subtracted from g
■ In example 2, the normal reaction from the track is given by: $\mathbf\small{F_{N(C)}=m \left(\frac{v_C^2}{r}-g\right)}$
♦ If FN(C) is not to become zero, vC must be greater than $\mathbf\small{\sqrt{rg}}$
♦ This is because, g is being subtracted from $\mathbf\small{\frac{v_C^2}{r}}$
So we have completed a discussion on conservation of mechanical energy. Next we will see the law of conservation in other forms of energy such as heat energy, sound energy, nuclear energy etc.,
Before that, we will see a few more solved examples in the next section.
Before that, we will see a few more solved examples in the next section.
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