`●` 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.
`●` 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.