Physics

### Topic covered

star WORK DONE BY A TORQUE
star ANGULAR MOMENTUM IN CASE OF ROTATION ABOUT A FIXED AXIS
star CONSERVATION OF ANGULAR MOMENTUM

### Work done by a torque

● Figure 7.34 shows a cross-section of a rigid body rotating about a fixed axis, which is taken as the z-axis (perpendicular to the plane of the page; see Fig. 7.33).
● Let F_1 be one such typical force acting as shown on a particle of the body at point P_1 with its line of action in a plane perpendicular to the axis.
● For convenience we call this to be the x′ –y′ plane (coincident with the plane of the page). The particle at P_1 describes a circular path of radius r_1 with centre C on the axis; CP_1 = r_1.

● In time Δt, the point moves to the position P_1′ . The displacement of the particle ds_1, therefore, has magnitude ds_1 = r_1dθ and direction tangential at P_1 to the circular path as shown. Here dθ is the angular displacement of the particle, dθ = ∠P_1 CP_ ′

\color{red} ✍️ The work done by the force on the particle is
dW_1 = F_1* ds_1= F_1ds_1 cosφ_1= F_1(r_1 dθ)sinα_1

=> where φ_1 is the angle between F_1 and the tangent at P_1, and α_1 is the angle between F_1 and the radius vector OP_1; φ_1 + α_1 = 90° .

● The torque due to F_1 about the origin is OP_1 xx F_1. Now OP_1 = OC + OP_1. [Refer to Fig. 7.17(b).] Since OC is along the axis, the torque resulting from it is excluded from our consideration. The effective torque due to F_1 is τ_1= CP × F_1; it is directed along the axis of rotation and has a magnitude τ_1= r_1F_1 sinα , Therefore,

 dW_1 = τ _1dθ If there are more than one forces acting on the body, the work done by all of them can be added to give the total work done on the body. Denoting the magnitudes of the torques due to the different forces as τ_ 1, τ_ 2, … etc,

color{blue}{dW = (τ_1 +τ_2 + ...)dθ}

bbul"Remember", the forces giving rise to the torques act on different particles, but the angular displacement dθ is the same for all particles. Since all the torques considered are parallel to the fixed axis, the magnitude τ of the total torque is just the algebraic sum of the magnitudes of the torques, i.e., τ = τ_1 + τ_2 + ..... We, therefore, have

color{blue} {dW=tau d theta.................(7.41)}

● This expression gives the work done by the total (external) torque τ which acts on the body rotating about a fixed axis. Its similarity with the corresponding expression

dW= F ds for linear (translational) motion is obvious.

Dividing both sides of Eq. (7.41) by dt gives

P=(dW)/(dt)= tau (d theta)/(dt)= tau omega

or P= tau omega.................(7.42)}

● This is the instantaneous power. Compare this expression for power in the case of rotational motion about a fixed axis with the expression for power in the case of linear motion,

 P = Fv

● In a perfectly rigid body there is no internal motion. The work done by external torques is therefore, not dissipated and goes on to increase the kinetic energy of the body. The rate at which work is done on the body is given by Eq. (7.42). This is to be equated to the rate at which kinetic energy increases. The rate of increase of kinetic energy is

d/(dt) ((I omega^2)/2) =I ((2 omega))/2 (d omega)/(dt)

● We assume that the moment of inertia does not change with time. This means that the mass of the body does not change, the body remains rigid and also the axis does not change its position with respect to the body.

Since α = dω //dt, we get

d/(dt) ((I omega^2)/2)=I omega alpha

● Equating rates of work done and of increase in kinetic energy,

tau omega= I omega alpha

color{blue}{tau=I alpha.......................(7.43)}

● Eq. (7.43) is similar to Newton’s second law for linear motion expressed symbolically as

color{blue}{F = ma}

● Just as force produces acceleration, torque produces angular acceleration in a body. The angular acceleration is directly proportional to the applied torque and is inversely proportional to the moment of inertia of the body. Eq.(7.43) can be called Newton’s second law for rotation about a fixed axis.
Q 3240656513

A cord of negligible mass is wound round the rim of a fly wheel of mass 20 kg and radius 20 cm. A steady pull of 25 N is applied on the cord as shown in Fig. 7.35. The flywheel is mounted on a horizontal axle with frictionless bearings.

(a) Compute the angular acceleration of the wheel.
(b) Find the work done by the pull, when 2m of the cord is unwound.
(c) Find also the kinetic energy of the wheel at this point. Assume that the wheel starts from rest.
(d) Compare answers to parts (b) and (c).

Solution:

(a) We use color{blue}{I alpha=tau}

the torque color{blue}{tau =FR}

=25 xx 0.20  (as R = 0.20m)

= 5.0 Nm

color{blue}{I= M * I} of flywheel about its axis color{blue}{= (MR^2)/2}

=(20.0 xx (0.2)^2)/2 =0.4 kgm^2

α = angular acceleration

= 5.0 N m//0.4 kg m^2 = 12.35 s^(–2)

(b) Work done by the pull unwinding 2m of the cord

= 25 N xx 2m = 50 J

(c) Let ω be the final angular velocity. The kinetic energy gained color{blue}{= 1/ 2 Iω^2} , since the wheel starts from rest. Now,
color{blue}{ω^ =ω_0^2 + 2αθ , ω_0 = 0}

The angular displacement θ = length of unwound string / radius of wheel

= 2m/0.2 m = 10 rad

ω^ 2 = 2 ×12.5 ×10.0 = 250(rad//s)^2

color{blue}{:. "K.E. gained" =1/2 xx 0.4 xx 250 =50J}

(d) The answers are the same, i.e. the kinetic energy gained by the wheel = work done by the force. There is no loss of energy due
to friction.

### ANGULAR MOMENTUM IN CASE OF ROTATION ABOUT A FIXED AXIS

● As we know that the time rate of total angular momentum of a system of particles about a point is equal to the total external torque on the system taken about the same point.
When the total external torque is zero, the total angular momentum of the system is conserved.

colo{blue}☞ We now wish to study the angular momentum in the special case of rotation about a fixed axis. The general expression for the total angular momentum of the system is

color{blue}{L= sum_(i=1)^N r_i xx p_i.................(7.25)}

● We first consider the angular momentum of a typical particle of the rotating rigid body. We then sum up the contributions of individual particles to get L of the whole body.

● For a typical particle I = r xx p. As seen in the last section r = OP = OC + CP [Fig. 7.17(b)]. With p = m v

I=(OC xx mv)+(CP xx mv)

● The magnitude of the linear velocity v of the particle at P is given by v = wr_(bot) where r_(bot) is the length of CP or the perpendicular distance of P  from the axis of rotation. Further, v is tangential at P to the circle which the particle describes.
● Using the right-hand rule one can check that CP xx v is parallel to the fixed axis. The unit vector along the fixed axis (chosen as the z-axis) is hat k . Hence

CP xx mv= r_(bot)(mv) hat k

= m_(bot)^2 omega hat k (since v omega r_(bot) = )

● Similarly, we can check that OC xx v is perpendicular to the fixed axis. Let us denote the part of l along the fixed axis (i.e. the z-axis) by l_z, then

I_z= CP xx mv=mr_(bot)^2 omega hat k

and I=I_z+ OC xx mv

● We note that l_z is parallel to the fixed axis, but I is not. In general, for a particle, the angular momentum l is not along the axis of rotation, i.e. for a particle, l and w are not necessarily parallel. Compare this with the corresponding fact in translation. For a particle, p and v are always parallel to each other.

● For computing the total angular momentum of the whole rigid body, we add up the contribution of each particle of the body.

Thus L= sum I_i =sum l_(iz) + sum OC_i xx m_i v_i

● We denote by L_(⊥) and  L_(z) the components of L respectively perpendicular to the z-axis and along the z-axis;

color{blue}{L_(bot)= sum OC_i xx m_i v_i.................(7.44 a)}

=> where m_i and v_i are respectively the mass and the velocity of the i^(th) particle and C_i is the centre of the circle described by the particle;

and L_z = sum I_(iz) =( sum m_i r_i^2) omega hat k

or color{blue}{L_z= I omega hat k............(7.44 b)}

● The last step follows since the perpendicular distance of the i^(th) particle from the axis is r_i; and by definition the moment of inertia of the body about the axis of rotation is I= sum m_i r_i^2

Note color{blue}{L=L_z+L_(bot)....................(7.44 c)}

● The rigid bodies which we have mainly considered in this chapter are symmetric about the axis of rotation, i.e. the axis of rotation is one of their symmetry axes.
● For such bodies, for a given OC_i, for every particle which has a velocity v_i , there is another particle of velocity –v_i located diametrically opposite on the circle with centre C_i described by the particle. Together such pairs will contribute zero to  L_(bot) and as a result for symmetric bodies L_(bot) is zero, and hence

color{blue}{L=L_z=I omega hat k...............(7.44 d)}

● For bodies, which are not symmetric about the axis of rotation, L is not equal to L_z and hence L does not lie along the axis of rotation. Referring to table 7.1, can you tell in which cases L = L_z will not apply?

● Let us differentiate Eq. (7.44b). Since hat(k) is a fixed (constant) vector, we get

d/(dt)(L_z) =d/(dt) (I omega) hat k

Now, Eq. (7.28b) states

(dL)/(dt) = tau

● As we have seen in the last section, only those components of the external torques which are along the axis of rotation, need to be taken into account, when we discuss rotation about a fixed axis.
● This means we can take tau = tau hat k . Since  L = L_z + L_(bot) and the direction of L_z (vector hat k ) is fixed, it follows that for rotation about a fixed axis,

color{blue}{(dL_z)/(dt)= tau hat k..............(7.45 a)}

and color{blue}{(dL_(bot))/(dt)=0................(7.45 b)}

● Thus, for rotation about a fixed axis, the component of angular momentum perpendicular to the fixed axis is constant. As  L_z = Iomega hat k , we get from Eq. (7.45a),

color{blue}{d/(dt) (I omega)=tau....................(7.45 c)}

If the moment of inertia I does not change with time,

d/(dt)(I omega) =I (d omega)/(dt) = I alpha

and we get from Eq. (7.45c),

color{blue}{tau = I alpha.....................(7.43)}

We have already derived this equation using the work - kinetic energy route.

### Conservation of angular momentum

● We are now in a position to revisit the principle of conservation of angular momentum in the context of rotation about a fixed axis. From Eq. (7.45c), if the external torque is zero,

color{blue}{L_z = I omega = "constant" ....................... (7.46)}

=> For symmetric bodies, from Eq. (7.44d), L_z may be replaced by L .(L and L_z are respectively the magnitudes of L and L_z.)

● This then is the required form, for fixed axis rotation, of Eq. (7.29a), which expresses the general law of conservation of angular momentum of a system of particles.
● Eq. (7.46) applies to many situations that we come across in daily life. You may do this experiment with your friend. Sit on a swivel chair with your arms folded and feet not resting on, i.e., away from, the ground. Ask your friend to rotate the chair rapidly. While the chair is rotating with considerable angular speed stretch your arms horizontally.

● What happens? Your angular speed is reduced. If you bring back your arms closer to your body, the angular speed increases again. This is a situation where the principle of conservation of angular momentum is applicable.

● If friction in the rotational mechanism is neglected, there is no external torque about the axis of rotation of the chair and hence Iω is constant. Stretching the arms increases I about the axis of rotation, resulting in decreasing the angular speed ω . Bringing the arms closer to the body has the opposite effect.
● A circus acrobat and a diver take advantage of this principle. Also, skaters and classical, Indian or western, dancers performing a pirouette on the toes of one foot display ‘mastery’ over this principle. Can you explain?

● A circus acrobat and a diver take advantage of this principle. Also, skaters and classical, Indian or western, dancers performing a pirouette on the toes of one foot display ‘mastery’ over this principle.