Energy

Energy is defined as the capacity to do work. In other words, work done is the manifestation of energy. That is why work and energy have the same dimension \( \mathrm{(ML^2T^{-2})} \).

The important aspect of energy is that for an isolated system, the sum of all forms of energy i.e., the total energy remains the same in any process irrespective of whatever internal changes may take place. This means that the energy disappearing in one form reappears in another form. This is known as the law of conservation of energy. In this chapter we shall take up only the mechanical energy for discussion.

In a broader sense, mechanical energy is classified into two types

1. Kinetic energy
2. Potential energy

The energy possessed by a body due to its motion is called kinetic energy. The energy possessed by the body by virtue of its position is called potential energy.

The SI unit of energy is the same as that of work done i.e., \( \mathrm{N\,m} \) (or) joule (J). The dimension of energy is also the same as that of work done. It is given by \( \mathrm{[ML^2T^{-2}]} \). The other units of energy and their SI equivalent values are given in Table 4.2.

Table 4.2 SI equivalent of other units of energy

4.2.1 Kinetic energy#

Kinetic energy is the energy possessed by a body by virtue of its motion. All moving objects have kinetic energy. A body that is in motion has the ability to do work. For example a hammer kept at rest on a nail does not push the nail into the wood. Whereas the same hammer when it strikes the nail, draws the nail into the wood as shown in Figure 4.7. Kinetic energy is measured by the amount of work that the body can perform before it comes to rest. The amount of work done by a moving body depends both on the mass of the body and the magnitude of its velocity. A body which is not in motion does not have kinetic energy.

Figure 4.7 Demonstration of kinetic energy

4.2.2 Work-Kinetic Energy Theorem#

Work and energy are equivalents. This is true in the case of kinetic energy also. To prove this, let us consider a body of mass m at rest on a frictionless horizontal surface.

The work (W) done by the constant force (F) for a displacement (s) in the same direction is,

The constant force is given by the equation,

The third equation of motion can be written as,

Substituting for a in the force equation,

Substituting this in the work equation,

The expression for kinetic energy:

The term \( \left(\frac{1}{2}\mathrm{mv}^{2}\right) \) in the above equation is the kinetic energy of the body of mass (m) moving with velocity (v).

Kinetic energy of the body is always positive.

Thus, \( \mathrm{W} = \Delta \mathrm{KE} \)

1. If the work done by the force on the body is positive then its kinetic energy increases.
2. If the work done by the force on the body is negative then its kinetic energy decreases.
3. If there is no work done by the force on the body then there is no change in its kinetic energy, which means that the body has moved at constant speed provided its mass remains constant.

4.2.3 Relation between Momentum and Kinetic Energy#

Consider an object of mass m moving with a velocity \( \bar{\mathbf{v}} \). Then its linear momentum is

and its kinetic energy,

Multiplying both the numerator and denominator by mass, m

where \( \left|\bar{\mathbf{p}}\right| \) is the magnitude of the momentum. The magnitude of the linear momentum can be obtained by

Note that if kinetic energy and mass are given, only the magnitude of the momentum can be calculated but not the direction of momentum. It is because the kinetic energy and mass are scalars.

EXAMPLE 4.7

Two objects of masses \( 2 \, \mathrm{kg} \) and \( 4 \, \mathrm{kg} \) are moving with the same momentum of \( 20 \, \mathrm{kg}\,\mathrm{m}\,\mathrm{s}^{-1} \)

(a) Will they have same kinetic energy?

(b) Will they have same speed?

Solution

(a) The kinetic energy of the mass is given by \( \mathrm{KE} = \frac{\mathbf{p}^2}{2\mathbf{m}} \)

For the object of mass \( 2 \, \mathrm{kg} \), kinetic energy is \( \mathrm{KE}_1 = \frac{(20)^2}{2 \times 2} = \frac{400}{4} = 100 \, \mathrm{J} \)

For the object of mass \( 4 \, \mathrm{kg} \), kinetic energy is \( \mathrm{KE}_2 = \frac{(20)^2}{2 \times 4} = \frac{400}{8} = 50 \, \mathrm{J} \)

Note that \( \mathrm{KE}_1 \neq \mathrm{KE}_2 \) i.e., even though both are having the same momentum, the kinetic energy of both masses is not the same. The kinetic energy of the heavier object has lesser kinetic energy than smaller mass. It is because the kinetic energy is inversely proportional to the mass \( (\mathrm{KE} \propto \frac{1}{m}) \) for a given momentum.

(b) As the momentum, \( p = mv \), the two objects will not have same speed.

4.2.4 Potential Energy#

The potential energy of a body is associated with its position and configuration with respect to its surroundings. This is because the various forces acting on the body also depends on position and configuration.

Potential energy of an object at a point \( P \) is defined as the amount of work done by an external force in moving the object at constant velocity from the point \( O \) (initial location) to the point \( P \) (final location). At initial point \( O \) potential energy can be taken as zero.

Mathematically, potential energy is defined as

where the limit of integration ranges from initial location point \( O \) to final location point \( P \).

We have various types of potential energies. Each type is associated with a particular force. For example,

(i) The energy possessed by the body due to gravitational force gives rise to gravitational potential energy.

(ii) The energy due to spring force and other similar forces give rise to elastic potential energy.

(iii) The energy due to electrostatic force on charges gives rise to electrostatic potential energy.

4.2.5 Potential energy near the surface of the Earth#

The gravitational potential energy (U) at some height \( h \) is equal to the amount of work required to take the object from ground to that height \( h \) with constant velocity.

Let us consider a body of mass \( m \) being moved from ground to the height \( h \) against the gravitational force as shown in Figure 4.8.

Figure 4.8 Gravitational potential energy

The gravitational force \( \bar{F}_{g} \) acting on the body is, \( \bar{F}_{g} = - mg \,\hat{j} \) (as the force is in \( y \) direction, unit vector \( \hat{j} \) is used). Here, negative sign implies that the force is acting vertically downwards. In order to move the body without acceleration (or with constant velocity), an external applied force \( \bar{F}_{a} \) equal in magnitude but opposite to that of gravitational force \( \bar{F}_{\mathrm{g}} \) has to be applied on the body i.e., \( \bar{F}_{\mathrm{a}} = -\bar{F}_{\mathrm{g}} \). This implies that \( \bar{F}_{a} = +\mathrm{mg}\,\hat{j} \). The positive sign implies that the applied force is in vertically upward direction. Hence, when the body is lifted up its velocity remains unchanged and thus its kinetic energy also remains constant.

The gravitational potential energy (U) at some height \( h \) is equal to the amount of work required to take the object from the ground to that height \( h \).

Since the displacement and the applied force are in the same upward direction, the angle between them, \( \theta = 0^{\circ} \). Hence, \( \cos 0^{\circ} = 1 \) and \( \left|\bar{F}_{\mathrm{a}}\right| = \mathrm{mg} \) and \( \left|\mathrm{d}\bar{r}\right| = \mathrm{d}r \).

Note that the potential energy stored in the object is defined through work done by the external force which is positive. Physically this implies that the agency which is applying the external force is transferring the energy to the object which is then stored as potential energy. If the object is allowed to fall from a height \( h \) then the stored potential energy is converted into kinetic energy.

Why should the object be moved at constant velocity when we define potential energy? If the object does not move at constant velocity, then it will have different velocities at the initial and final locations. According to work-kinetic energy theorem, the external force will impart some extra kinetic energy. But we associate potential energy to the forces like gravitational force, spring force and coulomb force. So the external agency should not impart any kinetic energy when the object is taken from initial to final location.

EXAMPLE 4.8

An object of mass \( 2 \, \mathrm{kg} \) is taken to a height \( 5 \, \mathrm{m} \) from the ground \( \left(g = 10 \, \mathrm{m}\mathrm{s}^{-2}\right) \).

(a) Calculate the potential energy stored in the object.

(b) Where does this potential energy come from?

(c) What external force must act to bring the mass to that height?

(d) What is the net force that acts on the object while the object is taken to the height ‘h’?

Solution

(a) \( \mathrm{U} = mgh = 2 \times 10 \times 5 = 100 \, \mathrm{J} \)

(b) The potential energy comes from the external agency which lifts the object.

(c) The external force must be equal to \( mg \) but in the opposite direction i.e., \( F_{ext} = mg = 20 \, \mathrm{N} \) upwards.

(d) Since the external force and gravitational force are equal and opposite, the net force is zero.

4.2.6 Elastic Potential Energy#

When a spring is elongated, it develops a restoring force. The potential energy possessed by a spring due to a deforming force which stretches or compresses the spring is termed as elastic potential energy. The work done by the applied force against the restoring force of the spring is stored as the elastic potential energy in the spring.

Consider a spring-mass system. Let us assume a mass, \( m \) lying on a smooth horizontal table as shown in Figure 4.9. Here, \( x = 0 \) is the equilibrium position. One end of the spring is attached to a rigid wall and the other end to the mass.

Figure 4.9 Potential energy of the spring (elastic potential energy)

As long as the spring remains in equilibrium position, its potential energy is zero. Now an external force \( \bar{F}_{\mathrm{a}} \) is applied so that it is stretched by a distance \( x \) in the direction of the force.

There is a restoring force called spring force \( \bar{F}_{\mathrm{s}} \) developed in the spring which tries to bring the mass back to its original position. This applied force and the spring force are equal in magnitude but opposite in direction i.e., \( \bar{F}_{\mathrm{a}} = -\bar{F}_{\mathrm{s}} \). According to Hooke’s law, the restoring force developed in the spring is

The negative sign in the above expression implies that the spring force is always opposite to that of displacement \( \bar{x} \) and \( k \) is the spring constant.

Work done by the spring force

The work done by the spring force for a displacement from \( x_i \) to \( x_f \) is given by

If we take the displacement from \( x_i = 0 \) to \( x_f = x \), then

Elastic potential energy

The work done by the external applied force is stored as potential energy in the spring.

EXAMPLE 4.9

Let the two springs A and B be such that \( k_A > k_B \). On which spring will more work has to be done if they are stretched by the same force?

Solution

The work done on the springs are stored as potential energy in the springs.

\( k_A > k_B \) implies that \( U_B > U_A \). Thus, more work is done on B than A.

EXAMPLE 4.10

A body of mass m is attached to the spring which is elongated to 25 cm by an applied force from its equilibrium position.

(a) Calculate the potential energy stored in the spring-mass system?

(b) What is the work done by the spring force in this elongation?

(c) Suppose the spring is compressed to the same 25 cm, calculate the potential energy stored and also the work done by the spring force during compression. (The spring constant, \( k = 0.1 \, \mathrm{N\,m^{-1}} \)).

Solution

The spring constant, \( k = 0.1 \, \mathrm{N\,m^{-1}} \)

The displacement, \( x = 25 \, \mathrm{cm} = 0.25 \, \mathrm{m} \)

(a) The potential energy stored in the spring is given by

(b) The work done \( W_s \) by the spring force is given by,

Note that the potential energy is defined through the work done by the external agency. The positive sign in the potential energy implies that the energy is transferred from the agency to the object. But the work done by the restoring force in this case is negative since restoring force is in the opposite direction to the displacement direction.

(c) During compression also the potential energy stored in the object is the same.

Work done by the restoring spring force during compression is given by

Potential energy-displacement graph for a spring

A compressed or extended spring will transfer its stored potential energy into kinetic energy of the mass attached to the spring. The potential energy-displacement graph is shown in Figure 4.11.

Figure 4.11 Potential energy–displacement graph for a spring-mass system

In a frictionless environment, the energy gets transferred from kinetic to potential and potential to kinetic repeatedly such that the total energy of the system remains constant. At the mean position,

4.2.7 Conservative and non-conservative forces#

Conservative force

A force is said to be a conservative force if the work done by or against the force in moving the body depends only on the initial and final positions of the body and not on the nature of the path followed between the initial and final positions.

Let us consider an object at point A on the Earth. It can be taken to another point B at a height \( h \) above the surface of the Earth by three paths as shown in Figure 4.12.

Whatever may be the path, the work done against the gravitational force is the same as long as the initial and final positions are the same. This is the reason why gravitational force is a conservative force. Conservative force is equal to the negative gradient of the potential energy. In one dimensional case,

Examples for conservative forces are elastic spring force, electrostatic force, magnetic force, gravitational force, etc.

Figure 4.12 Conservative force

Properties of conservative forces:

1. Work done is independent of the path
2. Work done in a round trip is zero
3. Total energy remains constant
4. Work done is completely recoverable
5. Force is the negative gradient of potential energy

Non-conservative force

A force is said to be non-conservative if the work done by or against the force in moving a body depends upon the path between the initial and final positions. This means that the value of work done is different in different paths.

1. Frictional forces are non-conservative forces as the work done against friction depends on the length of the path moved by the body.
2. The force due to air resistance, viscous force are also non-conservative forces as the work done by or against these forces depends upon the velocity of motion.

The properties of conservative and non-conservative forces are summarized in the Table 4.3.

EXAMPLE 4.11

Compute the work done by the gravitational force for the following cases

Solution

Displacement vector \( d\vec{r} = dx\hat{i} + dy\hat{j} \)

(As the displacement is in two dimension; unit vectors \( \hat{i} \) and \( \hat{j} \) are used)

(a) Since the motion is only vertical, horizontal displacement component dx is zero. Hence, work done by the force along path 1 (of distance h).

Total work done for path 2 is

But

Therefore, the total work done by the force along the path 2 is

Note that the work done by the conservative force is independent of the path.

EXAMPLE 4.12

Consider an object of mass \( 2 \, \mathrm{kg} \) moved by an external force \( 20 \, \mathrm{N} \) in a surface having coefficient of kinetic friction 0.9 to a distance \( 10 \, \mathrm{m} \). What is the work done by the external force and kinetic friction? Comment on the result. (Assume \( \mathrm{g} = 10 \, \mathrm{ms}^{-2} \))

Solution

When an object is in motion on the horizontal surface, it experiences two forces.

(a) External force, \( F_{ext} = 20 \, \mathrm{N} \)

(b) Kinetic friction, \( f_{k} = \mu_{k} mg = 0.9 \times 2 \times 10 = 18 \, \mathrm{N} \)

The work done by the external force \( W_{ext} = Fd = 20 \times 10 = 200 \, \mathrm{J} \)

The work done by the force of kinetic friction \( W_{k} = f_{k}d = (-18) \times 10 = -180 \, \mathrm{J} \). Here the negative sign implies that the force of kinetic friction is opposite to the direction of displacement.

The total work done on the object \( W_{total} = W_{ext} + W_{k} = 200 \, \mathrm{J} - 180 \, \mathrm{J} = 20 \, \mathrm{J} \).

Since the friction is a non-conservative force, out of \( 200 \, \mathrm{J} \) given by the external force, the \( 180 \, \mathrm{J} \) is lost and it cannot be recovered.

4.2.8 Law of conservation of energy#

When an object is thrown upwards its kinetic energy goes on decreasing and consequently its potential energy keeps increasing (neglecting air resistance). When it reaches the highest point its energy is completely potential. Similarly, when the object falls back from a height its kinetic energy increases whereas its potential energy decreases. When it touches the ground its energy is completely kinetic. At the intermediate points the energy is both kinetic and potential as shown in Figure 4.13. When the body reaches the ground the kinetic energy is completely dissipated into some other form of energy like sound, heat, light and deformation of the body etc.

Figure 4.13 Conservation of energy

The law of conservation of energy states that energy can neither be created nor destroyed. It may be transformed from one form to another but the total energy of an isolated system remains constant.

Figure 4.13 illustrates that, if an object starts from rest at height \( h \), the total energy is purely potential energy \( \mathrm{(U = mgh)} \) and the kinetic energy (KE) is zero at \( h \). When the object falls at some distance \( y \), the potential energy and the kinetic energy are not zero whereas, the total energy remains same as measured at height \( h \). When the object is about to touch the ground, the potential energy is zero and total energy is purely kinetic.

EXAMPLE 4.13

An object of mass \( 1 \, \mathrm{kg} \) is falling from the height \( h = 10 \, \mathrm{m} \). Calculate

(a) The total energy of an object at \( h = 10 \, \mathrm{m} \)

(b) Potential energy of the object when it is at \( h = 4 \, \mathrm{m} \)

(c) Kinetic energy of the object when it is at \( h = 4 \, \mathrm{m} \)

(d) What will be the speed of the object when it hits the ground?

Assume \( g = 10 \, \mathrm{m}\mathrm{s}^{-2} \)

Solution

(a) The gravitational force is a conservative force. So the total energy remains constant throughout the motion. At \( h = 10 \, \mathrm{m} \), the total energy \( E \) is entirely potential energy.

(b) The potential energy of the object at \( h = 4 \, \mathrm{m} \) is

(c) Since the total energy is constant throughout the motion, the kinetic energy at \( h = 4 \, \mathrm{m} \) must be \( KE = E - U = 100 - 40 = 60 \, \mathrm{J} \)

Alternatively, the kinetic energy could also be found from velocity of the object at \( 4 \, \mathrm{m} \). At the height \( 4 \, \mathrm{m} \), the object has fallen through a height of \( 6 \, \mathrm{m} \).

The velocity after falling \( 6 \, \mathrm{m} \) is calculated from the equation of motion,

The kinetic energy is \( \mathrm{KE} = \frac{1}{2} \mathrm{mv}^{2} = \frac{1}{2} \times 1 \times 120 = 60 \, \mathrm{J} \)

(d) When the object is just about to hit the ground, the total energy is completely kinetic and the potential energy, \( U = 0 \).

EXAMPLE 4.14

A body of mass \( 100 \, \mathrm{kg} \) is lifted to a height 10 m from the ground in two different ways as shown in the figure. What is the work done by the gravity in both the cases? Why is it easier to take the object through a ramp?

Path (1) straight up

Path (2) along the ramp

Solution

\( \mathrm{m} = 100 \, \mathrm{kg}, \quad \mathrm{h} = 10 \, \mathrm{m} \)

Along path (1):

The minimum force \( F_{1} \) required to move the object to the height of \( 10 \, \mathrm{m} \) should be equal to the gravitational force, \( F_{1} = mg = 100 \times 10 = 1000 \, \mathrm{N} \)

The distance moved along path (1) is, \( h = 10 \, \mathrm{m} \)

The work done on the object along path (1) is

Along path (2):

In the case of the ramp, the minimum force \( F_{2} \) that we apply on the object to take it up is not equal to \( mg \), it is rather equal to \( mg\sin \theta \). \( (\mathrm{mg}\sin \theta < mg) \)

Here, angle \( \theta = 30^{\circ} \)

Therefore, \( \mathrm{F}_{2} = \mathrm{mg}\sin \theta = 100 \times 10 \times \sin 30^{\circ} = 100 \times 10 \times 0.5 = 500 \, \mathrm{N} \)

Hence, \( (\mathrm{mg}\sin \theta < mg) \)

The path covered along the ramp is, \( l = \frac{h}{\sin 30^{\circ}} = \frac{10}{0.5} = 20 \, \mathrm{m} \)

The work done on the object along path (2) is, \( \mathrm{W} = \mathrm{F}_{2}l = 500 \times 20 = 10000 \, \mathrm{J} \)

Since the gravitational force is a conservative force, the work done by gravity on the object is independent of the path taken.

In both the paths the work done by the gravitational force is 10,000 J

Along path (1): more force needs to be applied against gravity to cover lesser distance.

Along path (2): lesser force needs to be applied against the gravity to cover more distance.

As the force needs to be applied along the ramp is less, it is easier to move the object along the ramp.

EXAMPLE 4.15

An object of mass m is projected from the ground with initial speed \( \mathbf{v}_0 \). Find the speed at height h.

Solution

Since the gravitational force is conservative; the total energy is conserved throughout the motion.

By law of conservation of energy, the initial and final total energies are the same.

Note that similar result is obtained using kinematic equation based on calculus method. However, calculation through energy conservation method is much easier than calculus method.

EXAMPLE 4.16

An object of mass 2 kg attached to a spring is moved to a distance \( x = 10 \, \mathrm{m} \) from its equilibrium position. The spring constant \( k = 1 \, \mathrm{N} \, \mathrm{m}^{-1} \) and assume that the surface is frictionless.

(a) When the mass crosses the equilibrium position, what is the speed of the mass?

(b) What is the force that acts on the object when the mass crosses the equilibrium position and extremum position \( x = \pm 10 \, \mathrm{m} \).

Solution

(a) Since the spring force is a conservative force, the total energy is constant. At \( x = 10 \, \mathrm{m} \), the total energy is purely potential.

When the mass crosses the equilibrium position \( (x = 0) \), the potential energy

The entire energy is purely kinetic energy at this position.

The speed

(b) Since the restoring spring force is \( \mathrm{F} = -\mathrm{kx} \), when the object crosses the equilibrium position, it experiences no force. Note that at equilibrium position, the object moves very fast. When the object is at \( x = +10 \, \mathrm{m} \) (elongation), the force \( \mathrm{F} = -\mathrm{kx} = -(1)(10) = -10 \, \mathrm{N} \). Here the negative sign implies that the force is towards equilibrium i.e., towards negative \( x \)-axis and when the object is at \( x = -10 \, \mathrm{m} \) (compression), it experiences a force \( \mathrm{F} = -(1)(-10) = +10 \, \mathrm{N} \). Here the positive sign implies that the force points towards positive \( x \)-axis. The object comes to momentary rest at \( x = \pm 10 \, \mathrm{m} \) even though it experiences a maximum force at both these points.

4.2.9 Motion in a vertical circle#

Imagine that a body of mass (m) attached to one end of a massless and inextensible string executes circular motion in a vertical plane with the other end of the string fixed. The length of the string becomes the radius \( (\vec{r}) \) of the circular path (Figure 4.14).

Let us discuss the motion of the body by taking the free body diagram (FBD) at a position where the position vector \( (\vec{r}) \) makes an angle \( \theta \) with the vertically downward direction and the instantaneous velocity is as shown in Figure 4.14. There are two forces acting on the mass: 1. Gravitational force which acts downward, 2. Tension along the string. Applying Newton’s second law on the mass,

In the tangential direction,

where, \( \mathrm{a}_{\mathrm{t}} = -\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}t} \) is tangential retardation

In the radial direction,

where, \( \mathrm{a}_{\mathrm{r}} = \frac{\mathrm{v}^{2}}{\mathrm{r}} \) is the centripetal acceleration.

Figure 4.14 Motion in vertical circle

The circle can be divided into four sections A, B, C, D for better understanding of the motion. The four important facts to be understood from the two equations are as follows:

(i) The mass is having tangential acceleration (g sin \( \theta \)) for all values of \( \theta \) (except \( \theta = 0^{\circ} \)), it is clear that this vertical circular motion is not a uniform circular motion.

(ii) From the equations it is understood that as the magnitude of velocity is not a constant in the course of motion, the tension in the string is also not constant.

(iii) The equation \( \mathrm{T} = \mathrm{mg}\cos \theta + \frac{\mathrm{mv}^{2}}{\mathrm{r}} \) highlights that in sections A and D of the circle \( \left(\text{for } -\frac{\pi}{2} < \theta < \frac{\pi}{2}; \cos \theta \text{ is positive}\right) \), the term mg \( \cos \theta \) is always greater than zero. Hence the tension cannot vanish even when the velocity vanishes.

(iv) The equation \( \frac{\mathrm{mv}^{2}}{\mathrm{r}} = \mathrm{T} - \mathrm{mg}\cos \theta \) further highlights that in sections B and C of the circle \( \left(\text{for } \frac{\pi}{2} < \theta < \frac{3\pi}{2}; \cos \theta \text{ is negative}\right) \), the second term is always greater than zero. Hence velocity cannot vanish, even when the tension vanishes.

To start with let us consider only two positions, say the lowest point 1 and the highest point 2 as shown in Figure 4.15 for further analysis. Let the velocity of the body at the lowest point 1 be \( \vec{\nu}_{1} \), at the highest point 2 be \( \vec{\nu}_{2} \). The direction of velocity is tangential to the circular path at all points. Let \( \vec{T}_{1} \) be the tension in the string at the lowest point and \( \vec{T}_{2} \) be the tension at the highest point. Tension at each point acts towards the centre. The tensions and velocities at these two points can be found by applying the law of conservation of energy.

Figure 4.15 Motion in vertical circle shown for lowest and highest points

For the lowest point (1)

When the body is at the lowest point 1, the gravitational force \( m\vec{g} \) acts vertically downwards, and the tension \( \vec{T}_{1} \) acts vertically upwards, i.e. towards the centre. From the radial equation, we get

For the highest point (2)

At the highest point 2, both the gravitational force \( m\vec{g} \) on the body and the tension \( \vec{T}_{2} \) act downwards, i.e. towards the centre again.

From the above equations, it is understood that \( \mathrm{T}_1 > \mathrm{T}_2 \). The difference in tension \( \mathrm{T}_1 - \mathrm{T}_2 \) is obtained by subtracting the second equation from the first.

The term \( (v_1^2 - v_2^2) \) can be found easily by applying law of conservation of energy at point 1 and also at point 2.

Total Energy at point 1 \( (E_{1}) \) is same as the total energy at point 2 \( (E_{2}) \)

Potential Energy at point 1, \( U_{1} = 0 \) (by taking reference as point 1)

Kinetic Energy at point 1, \( KE_{1} = \frac{1}{2} mv_{1}^{2} \)

Total Energy at point 1, \( E_{1} = U_{1} + KE_{1} = 0 + \frac{1}{2}\mathrm{mv}_1^2 = \frac{1}{2}\mathrm{mv}_1^2 \)

Similarly, Potential Energy at point 2, \( U_{2} = \mathrm{mg}(2r) \) (h is 2r from point 1)

Kinetic Energy at point 2, \( KE_{2} = \frac{1}{2}\mathrm{mv}_{2}^{2} \)

Total Energy at point 2, \( E_{2} = U_{2} + KE_{2} = 2\mathrm{mg}\mathrm{r} + \frac{1}{2}\mathrm{mv}_{2}^{2} \)

From the law of conservation of energy, we get

After rearranging,

Substituting this in the difference in tension equation we get,

Therefore, the difference in tension is

Minimum speed at the highest point (2)

The body must have a minimum speed at point 2 otherwise, the string will slack before reaching point 2 and the body will not loop the circle. To find this minimum speed let us take the tension \( \mathrm{T}_{2} = 0 \) in the equation for \( T_2 \).

The body must have a speed at point 2, \( v_2 \geq \sqrt{\mathrm{gr}} \) to stay in the circular path.

Minimum speed at the lowest point 1

To have this minimum speed \( (v_2 = \sqrt{\mathrm{gr}}) \) at point 2, the body must have minimum speed also at point 1.

By making use of the equation \( v_1^2 - v_2^2 = 4\mathrm{gr} \),

The body must have a speed at point 1, \( v_1 \geq \sqrt{5\mathrm{gr}} \) to stay in the circular path.

From the above equations, it is clear that the minimum speed at the lowest point 1 should be \( \sqrt{5} \) times more than the minimum speed at the highest point 2, so that the body loops without leaving the circle.

EXAMPLE 4.17

Water in a bucket tied with rope is whirled around in a vertical circle of radius \( 0.5 \, \mathrm{m} \). Calculate the minimum velocity at the lowest point so that the water does not spill from it in the course of motion. \( (\mathrm{g} = 10 \, \mathrm{ms}^{-2}) \)

Solution

Radius of circle \( \mathrm{r} = 0.5 \, \mathrm{m} \)

The required speed at the highest point \( v_2 = \sqrt{\mathrm{gr}} = \sqrt{10 \times 0.5} = \sqrt{5} \, \mathrm{ms}^{-1} \)

The speed at lowest point \( v_1 = \sqrt{5\mathrm{gr}} = \sqrt{5} \times \sqrt{\mathrm{gr}} = \sqrt{5} \times \sqrt{5} = 5 \, \mathrm{ms}^{-1} \)