Rotational Motion

Physics of Rotational Motion

The laws and equations that govern nature and natural phenomena are described by physics. One prime focus of physics is the study of motion. We have dealt in detail about translational motion (objects that move along a straight or curved line) in the previous chapters, and now we will expand our view towards other types of motions as well.

We see rotational motion in almost everything around us. Every machine, celestial bodies, most of the fun games in amusement parks and if you are a FIFA fan, and when you watch David Beckham’s familiar shot, the ball is actually executing rotational motion.

Objects turn about an axis. All the particles and the mass center do not undergo identical motions. All the particles of the body undergo identical motion. By definition, it becomes essential for us to explore how the different particles of a rigid body move when the body rotates.

Also Read: Moment of Inertia

Rotational Kinematics

In rotational kinematics, we will investigate the relation between kinematical parameters of rotation. We shall now revisit angular equivalents of the linear quantities: position, displacement, velocity and acceleration which we have already dealt in a circular motion.

Linear Kinematic Parameters Angular Kinematic Parameters
Position s Angular position θ
Displacement Δs=s1s2\Delta s={{s}_{1}}-{{s}_{2}} Angular displacement Δθ=θ1θ2\Delta \theta ={{\theta }_{1}}-{{\theta }_{2}}
Average velocity vavg= ΔsΔt{{v}_{avg}}=~\frac{\Delta s}{\Delta t} Average angular velocity ωavg= ΔθΔt{{\omega }_{avg}}=~\frac{\Delta \theta }{\Delta t}
Instantaneous velocity vinslimt0ΔsΔt=dsdt{{v}_{ins}}\,\underset{\triangle t\to 0}{\mathop{\lim }}\,\frac{\Delta s}{\Delta t}=\frac{ds}{dt} Instantaneous angular velocity ωinslimΔt0ΔθΔt=dθdt\underset{\Delta t\to 0}{\mathop{{{\omega }_{ins}}\lim }}\,\frac{\Delta \theta }{\Delta t}=\frac{d\theta }{dt}
Average acceleration aavg= ΔvΔt{{a}_{avg}}=~\frac{\Delta v}{\Delta t} Average angular acceleration αavg= ΔsΔt{{\alpha }_{avg}}=~\frac{\Delta s}{\Delta t}
Instantaneous acceleration

ainslimΔt0ΔvΔt=dvdt{{a}_{ins}}\,\underset{\Delta t\to 0}{\mathop{\lim }}\,\frac{\Delta v}{\Delta t}=\frac{dv}{dt}

 

Instantaneous angular acceleration

αinslimΔt0ΔωΔt=dωdt{{\alpha }_{ins}}\,\underset{\Delta t\to 0}{\mathop{\lim }}\,\frac{\Delta \omega }{\Delta t}=\frac{d\omega }{dt}

A case of constant angular acceleration is of great importance and a parallel set of equations holds for this case just as in constant linear acceleration.

Linear Equations of Motion Angular Equations of Motion
v=vo+atv = v_{o} + at ω=ω0+αt\omega ={{\omega }_{0}}+\alpha t
xx0= v0+12at2x-{{x}_{0}}=~{{v}_{0}}+\frac{1}{2}a{{t}^{2}} θθ0= ω0+12αt2\theta -{{\theta }_{0}}=~{{\omega }_{0}}+\frac{1}{2}\alpha {{t}^{2}}
v2=v02+2a(xx0){{v}^{2}}=v_{0}^{2}+2a\left( x-{{x}_{0}} \right) ω2=ω02+2α(θθ0){{\omega }^{2}}=\omega _{0}^{2}+2\alpha \left( \theta -{{\theta }_{0}} \right)

Axis of Rotation

A rigid body of an arbitrary shape in rotation about a fixed axis (axis that does not move) called axis of rotation or rotation axis is shown in the figure

Axis of Rotation

Types of Motion involving Rotation

  1. Rotation about a fixed axis (Pure rotation)
  2. Rotation about an axis of rotation (Combined translational and rotational motion)
  3. Rotation about an axis in the rotation (rotating axis – out of the scope of JEE)

Rotation About a Fixed Axis

Rotation of a ceiling fan, opening and closing of the door, rotation of our planet, rotation of hour and minute hands in analogue clocks are few examples of this type.

Rotation About a Fixed Axis

Rotation about an axis of rotation

Rolling is an example of this category. Arguably, the most important application of rotational physics is in the rolling of wheels and wheels like objects as our world now is filled with automobiles and other rolling vehicles.

Rolling Motion of a body is a combination of both translational and rotational motion of a round-shaped body placed on a surface. When a body is set in a rolling motion, every particle of the body has two velocities – one due to its rotational motion and the other due to its translational motion (of the centre of mass), and the resulting effect is the vector sum of both velocities at all particles Rotation about an axis of rotation

Check your understanding

There is a wheel that rotates with an angular acceleration which is given by α = 4at3 — 3bt2, where t is the time and a and b are constants. If the wheel has initial angular speed  ω0. Give the equations for (a) angular speed (b) angular displacement.

1. α=dωdtdω=αdt\alpha = \frac{d\omega}{dt} \Rightarrow d\omega = \alpha dt

ωoω=otαdt=ot(4at33bt2)dt\Rightarrow \int_{\omega_{o}}^{\omega} = \int_{o}^{t} \alpha dt = \int_{o}^{t} (4at^{3}- 3bt^{2})dt

ω=ωo+at4bt3\Rightarrow \omega = \omega _{o} + at^{4} – bt^{3}

2. Further,

ω=dΘdtdΘ=ωdt\omega = \frac{d\Theta }{dt} \Rightarrow d\Theta = \omega dt

oΘ=otωdt=ot(ωo+at4bt3)dt\Rightarrow \int_{o}^{\Theta } = \int_{o}^{t} \omega dt = \int_{o}^{t} (\omega _{o}+at^{4}- bt^{3})dt

Θ=ωo+t+at55bt44\Rightarrow \Theta = \omega _{o} + t + \frac{at^{5}}{5} – \frac{bt^{4}}{4}

Kinetic Energy of Rotation

The rapidly rotating blades of a table saw machine and the blades of a fan certainly have kinetic energy due to the rotation. If we apply the familiar equation to the saw machine as a whole, it would give us kinetic energy of its centre of mass only, which is zero.

Image result for ceiling fan rotational motion

The right approach:

We shall treat the saw machine or any rotating rigid body as a collection of particles at different speeds. We shall sum up all the kinetic energy of the particles to find the rotational kinetic energy of the whole body.

If m1, m2,…. mn are the masses of the constituent particles moving on circular paths with radii r1, r2,…. ro with velocities v1, v2,…. vo, then the kinetic energy of the body is given by;

KE=1n12mivi2=1n12miri2ω2=12ω21nmiri2=12Iω2KE = \sum_{1}^{n}\frac{1}{2}m_{i}v_{i}^{2} = \sum_{1}^{n}\frac{1}{2}m_{i}r_{i}^{2}\omega ^{2} = \frac{1}{2}\omega ^{2} \sum_{1}^{n}m_{i}r_{i}^{2} = \frac{1}{2}I\omega ^{2}

The term 1nmiri2\sum_{1}^{n}m_{i}r_{i}^{2} is called rotational inertia or moment of inertia of the system of particles. For a continuous distribution of mass I=Bodyr2dmI = \int_{Body} r^{2}dm where dm is the mass of a particle at a distance r from the axis of rotation.

What is Torque

Torque is a rotational analogue of force and expresses the tendency of a force applied to an object that causes the object to rotate about a given point.

If you want to open a door, you will apply a force on the doorknob which is located as far as possible from the hinges of the door. If you try to apply the force nearer to the hinge line than the knob, or at any other angle other than 90ᴼ to the plane of the door, you must apply greater force than the former to rotate the door.

Image result for torque in door

To determine how the applied force results in a rotation of the body about an axis, we resolve the Force (F) into two components. The tangential component (Fsinθ) is perpendicular to r and it does cause rotation whereas the radial component (Fcosθ) does not cause rotation because it acts along the line that intersects with the axis or pivot point.

Torque

The ability to rotate the body depends on the magnitude of the tangential component and also on how far from the axis the force (r – moment of an arm) is applied. Therefore, mathematically it can be represented as τ=r×F\vec{\tau }=\vec{r}\times \vec{F}

SI unit of torque is Nm.

To find the direction of τ,\vec{\tau }, we use right hand thumb rule sweeping the fingers from τ\vec{\tau } (the first vector in the product) into F\vec{F} (the second vector in the product),the outstretched thumb will give the direction of τ.\vec{\tau }.

Image result for torque and right hand thumb rule

Stability Of The Object

  • Greater the rotational inertia of the object, an object is more stable because it is difficult to move.
  • The stability of an object depends on the torques produced by its weight. Larger the torques are produced if the mass is from COM, therefore more force is required to alter the stability.

Example: If you consider disk and bike wheel of the same mass, a bike wheel is more stable.

  • When the object’s rotation is greater it is more stable.

Newton’s Second Law of Rotation

If the net torque acting on a body about any inertial axis is τ\vec{\tau } and the moment of inertia about that axis is I, then the angular acceleration of the body is given by the relation:

 τ=Iα \overrightarrow{~\tau }=I\overrightarrow{\alpha ~}

Rotational Equilibrium

The centre of mass of a body remains in equilibrium if the total external force acting on the body is zero. This follows from the equation F = Ma.

Similarly, a body remains in rotational equilibrium if the total external torque acting on the body is zero. This follows from the equation τ = Iα. Therefore a  body in rotational equilibrium must either be in rest or rotation with constant angular velocity.

Thus, if a body remains at rest in an inertial frame, the total external force acting on the body should be zero in any direction and the total external torque should be zero about any line.

Under the action of several coplanar forces, the net torque is zero for rotational equilibrium. Rotational Equilibrium

Note: If the net force on the body is zero, then the net torque may or may not be zero.

Ex. Determine the point of application of third force for which the body is in equilibrium when forces of 20N & 30N are acting on the rod as shown in the figure.

Solution: 

Let the magnitude of the third force is F, is applied in upward direction then the body is in the equilibrium when

(i) Fnet=0\vec{F}_{net}=0 (Translational Equilibrium)

⇒ 20 + F = 30 ⇒ F = 10N

So the body is in translational equilibrium when 10N force acts on it in the upward direction. 

 

(ii) Let us assume that this 10N force act.

Then keep the body in rotational equilibrium.

So, torque about C = 0

i.e τC = 0

⇒ 30 × 20 = 10x

x = 60cm

So, 10N force is applied at 70cm from point A to keep the body in equilibrium.

Ex: A stationary uniform rod of mass ‘m’, length ‘l’ leans against a smooth vertical wall making an angle q with a rough horizontal floor. Find the normal force and frictional force that is exerted by the floor on the rod.

Solution: 

As the rod is stationary, the linear acceleration and angular acceleration of the rod are 0. 

i.e. acm = 0; ɑ = 0

N2 = f

N1 = mg [∵ acm = 0]

The torque about any point of the rod should also be zero. 

∴ ɑ = 0

τA = 0 ⇒ mg cos θ (l / 2) + fl sin θ = N1 cos θ. l

N1 cos θ = sin θ f + (mg cos θ) / 2

f = [mg cos θ] / [2 sin θ] = mg cot θ / 2 

 

Ex: The ladder shown in the figure has negligible mass and rests on a frictionless floor. The crossbar connects the two legs of the ladder in the middle. The angle between the two legs is 60o. The fat person sitting on the ladder has a mass of 80 kg. Find the contact force exerted by the floor on each leg and the tension in the crossbar.

Solution: 

The forces acting on different parts are shown in the figure. Consider the vertical equilibrium of “the ladder plus the person” system. The forces acting on this system are its weight (80 kg) g and the contact force N + N = 2N due to the floor. Thus 2N = (80 kg) g or N = (40 kg) (9.8 m/s2) = 392 N

Next, consider the equilibrium of the left leg of the ladder. Taking torques of the forces acting on it about the upper end, 

N (2m) tan 30o = T (1m) or 

T = N (2/√3) = [(392) N] * (2/√3) = 450N

Ex. The pulley shown in the figure has a moment of inertia l about the axis and radius R. Find the acceleration of the two blocks. Assume that the string is light and does not slip on the pulley.

Solution: 

Suppose the tension in the left string is T1 and that in the right string is T2. Suppose the block of mass M goes down with an acceleration a and the other block moves up with the same acceleration. This is also the tangential acceleration of the rim of the wheel as the string does not slip over the rim.

The angular acceleration of the wheel ɑ = a / R.

The equations of motion for the mass M, the mass m and the pulley are as follows; 

Mg – T1 = Ma —- (i)

T2 – mg = ma —- (ii)

T1R – T2R = Iɑ = Ia / R —- (iii)

Substituting for T1 and T2 from equations (i) and (ii) in equation (iii)

[M (g – a)] – m (g + a)] R = Ia / R

Solving we get, a = [(M – m)gR2] / [I + (M + m)R2]

 

Angular momentum of a particle describing circular motion:

L=r×p\vec{L}=\vec{r}\times\vec{p}; as linear momentum p\vec{p} is along with the tangent, hence r×p=rpn\vec{r}\times\vec{p}=rpn, where n is the unit vector perpendicular to the plane of the circle. 

L=mvr=mωr2, where p=mv=mωr\vec{|L|}=mvr=m\omega r^{2},\text \ where \ \vec{|p|}=mv=m\omega r

 

Angular momentum of a rigid in a fixed axis rotation

In a fixed axis rotation, all the constituent particles describe circular motion. Hence, the angular momentum of the particles about corresponding centres are 

L1 = m1𝛚r12, L2 = m2𝛚r22

and similarly, Ln = mn𝛚rn2 where 𝛚 is the angular speed of the body.

Since the angular momentum of all the particles have same direction, therefore angular momentum of the whole body is given by

L = L1 + L2 +…… + Ln

= (m1r12 + m2r22 +….. + mn𝛚rn2) 𝛚

⇒ L = I𝛚

Angular Momentum

Angular Momentum

The concept of linear momentum and conservation of linear momentum are extremely powerful tools to predict the collision of two objects without any other details of collision. Thus, the angular counterpart, angular momentum plays a crucial role in orbital mechanics.

Angular momentum of a particle about a given point is given by,

l=r×p=m(r×v)\vec{l}=\vec{r}\times \vec{p}=m\left( \vec{r}\times \vec{v} \right)

The direction of angular momentum is also given by right hand rule. (refer torque)

Newton’s law in angular form:

The vector sum of all the torques acting on a particle is equal to the time rate of change of the angular momentum of that particle.

 τnet= dldt{{\overrightarrow{~\tau }}_{net}}=~\frac{d\vec{l}}{dt}

 

Conservation of Angular Momentum

By the definition of torque,

 τnet= dldt,if τnet=0, then dldt=0,l=constant{{\overrightarrow{~\tau }}_{net}}=~\frac{d\vec{l}}{dt},if{{\overrightarrow{~\tau }}_{net}}=0,~then~\frac{d\vec{l}}{dt}=0,\vec{l}=constant

When the resultant torque acting on a system is zero, then the total vector angular momentum of the system remains constant. This is called as principle of conservation of angular momentum.

Examples of conservation of angular momentum

Combined Translational and Rotational Motion

Rolling motion is one such example.

JEE FOCUS

Example:

Frequently Asked Questions

1. What is rotational motion? Give an example.

Answer:

Rotational motion is a type of motion in which the body follows the circular path.

An example is car wheel.

2. What is the reason for rotational motion?

Answer:

The torque or rotational analogue force is a reason for rotational motion. When torque is applied to the system of the particle about to its axis it gives a twist and this is the reason for rotational motion.

3. Is circular motion is the same as rotational motion? Explain

Answer:

Circular motion: It is a motion of the body around a fixed point. In this case, a fixed point lies outside the body. Here the centripetal force is the reason for the circular motion.
Rotational motion: In the case of rotational motion, the fixed point lies inside the body. Rotational motion is due to the torque acting on the system of the particles.

 

 

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