### Rotational Dynamics - with Problems -Angular Position, Displacement, Velocity, Momentum, Acceleration; Moment of Inertia

Basic Mechanics Tutorials- At a glance Vectors; Rectilinear and Projectile MotionVectors and Projectile MotionNewton's Laws of MotionWork, Force and EnergySimple Harmonic MotionRotational DynamicsFluid Mechanics

## A Quick Summary of what we'll study in this chapter and the kind of problems we'll solve

### Introduction

When we hear the word rotation, we immediately think of the ceiling fan and the wheels of a moving car, though the motion of the wheels is combined rotation and translation, also called rolling. Now, when we talk of rotational motion, we cannot assume the object to be a particle of the same mass located at the center of mass of the object because as we see in a fan, the center of mass does not move at all. Thus in rotation, we have to consider the dimensions or the shape and size of the object. Considering the rotating fan again, we see that the wings of the fan rotate about a vertical axis passing though the center of mass of the fan but it is not always necessary that an object rotates about an axis passing through its center of mass or anywhere through its body. The axis of rotation may as well lie outside the object like whirling a stone tied to a string in a circle. The stone is rotating about the axis passing through the end of the string which is held stationary in hand. In this case, the center of mass is not stationary, it is also in rotational motion like every other particle of the stone.

There is a very important assumption while we try to follow the rotational motion of a body. We assume that all the particles of the body in motion are stationary w.r.t each other. Such objects are called rigid bodies and no bodies in nature are rigid. But this is a very logical assumption to make as we won’t be able to follow the motion of the body if its parts were moving w.r.t. each other as the mathematics would become too complicated for us to solve and this assumption is mostly harmless as the motion of the particles w.r.t. each other is very negligible mostly. But this kind of assumption should not be made when very accurate results for the position of a body are needed.

Just like for following the translational motion of a body, we have defined certain variables and we have laws or principles governing the changes in those variables w.r.t each other; we have similar variables defined for following the rotational motion of a body and similar kinematic equations governing the changes in them.

### Angular Position

To describe the rotation of a rigid body about a fixed axis, called the rotation axis, we assume a reference line is fixed in the body, perpendicular to that axis and rotating with the body. We measure the angular position θ of this line relative to a fixed direction. When θ is measured in radians, l = θ/r where l  is the arc length of a circular path of radius r and angle θ.

### Angular Displacement

A body that rotates about a rotation axis, changing its angular position from θ1 to θ2, undergoes an angular displacement

∆θ = θ2 - θ1

where ∆θ is positive for counterclockwise rotation and negative for clockwise rotation.

### Angular Velocity and Speed

If a body rotates through an angular displacement ∆θ in a time interval ∆t, its average angular velocity ωavg is

ωavg = ∆θ/∆t

The instantaneous angular velocity ω of the body is

ω = dθ/dt

Both ωavg and ω are vectors, with directions given by the right-hand rule. They are positive for counterclockwise rotation and negative for clockwise rotation. The magnitude of the body’s angular velocity is the angular speed.

### Angular Acceleration

If the angular velocity of a body changes from ω1 to ω2 in a time interval ∆t = t2 - t1, the average angular acceleration αavg of the body is

αavg  = (ω2 – ω1)/(t2 – t1) = ∆ω/∆t

The instantaneous angular acceleration a of the body is

α = dω/dt

Both αavg and α are vectors.

### The Kinematic equations for constant angular acceleration

Constant angular acceleration (α = constant) is an important special case of rotational motion. The appropriate kinematic equations are

ω = ωo + αt

θ – θo = ωt + ½ αt2

ω2 = ωo2 + 2α(θ – θo)

### Relation between linear and angular variables

A point in a rigid rotating body, at a perpendicular distance r from the rotation axis, moves in a circle with radius r. If the body rotates through an angle θ, the point moves along an arc with length l given by

l = θr where θ is in radians. The linear velocity of the point is tangent to the circle; the point’s linear speed v is given by v = ωr where ω is the angular speed (in radians per second) of the body. This is evident from the fact that all the particles complete one circle in the same time irrespective of the radius of the circle in which they are moving. The linear acceleration of the point has both tangential and radial components. The tangential component is

at = αr where α is the magnitude of the angular acceleration (in radians per second-squared) of the body. The radial component of is ar = v2/r = ω2r If the point moves in uniform circular motion, the period T of the motion for the point and the body is T = 2пr/v = 2п/ω

### Rotational Dynamics

There is a very important concept to understand while studying rotational motion of a body, i.e. a body can have rotational motion even if the total external forces acting on the body add up to zero but if the applied force is zero, there can be no rotation. It is slightly confusing but the concept here is that angular acceleration is caused by torque (just like linear acceleration is caused by force) and torque of a force about a rotation axis is defined as

τ = r X F = rFsinθ

It is the twisting action of a force on a body about the rotation axis. Like when we try to shut an almirah, we need to apply less force if we hold it from the handle than if we try to push the part closer to the hinged end of the almirah door. Thus, the torque of a force depends on the perpendicular distance of the point of application of the force from the rotation axis i.e. r. The SI unit of torque is the newton-meter (N.m). A torque, τ is positive if it tends to rotate a body at rest counterclockwise and negative if it tends to rotate the body clockwise.

The torque will be there only if an external force is applied on the body, the sum of the external forces can be zero or non-zero. Thus, a body can rotate even if the sum of external forces on it is zero. In this case, the center of mass of the body will be in equilibrium and thus, we can conclude that the axis of rotation will pass through the center of mass of the body like in the case of a rotating fan. If the sum of external forces on the body is non-zero, its center of mass will be in motion like in the case of the wheel of a moving car.

### Newton’s second law in angular form

The rotational analog of Newton’s second law is

τnet = Iα where τnet is the net torque acting on a particle or rigid body, I is the rotational inertia of the particle or body about the rotation axis, and a is the resulting angular acceleration about that axis.

### Angular momentum

Angular momentum of a particle about a point O is defined as l = r X p
where p is the linear momentum and r is the position vector of the particle from the given point O.
The angular momentum of a system of particles is the vector sum of the angular momenta of the particles of the system, i.e.

L = ∑li = ∑(ri X pi)

### Conservation of angular momentum

Angular momentum is defined as L = ∑(ri X pi)
Differentiating w.r.t. time;

dL/dt = d[∑(ri X pi)]/dt = ∑[ (dri/dt) X pi + (dpi/dt) X ri] = ∑[vi X mvi + ri X Fi]
Now, cross product of a vector with itself is zero
Therefore, dL/dt = ∑(ri X Fi) = τtotal
where Fi is the total force acting on the ith particle. This includes external as well as internal forces on the particle by all the other particles. When summation is taken over all the particles, the internal torques add to zero. Thus,
dL/dt =  τext
where τext is the total torque on the system due to all the external forces acting on the system.
From this equation, we can conclude that “If the total external torque on a system is zero, its angular momentum remains constant” which is known as the principle of conservation of angular momentum.

### Rotational kinetic energy

Suppose a body is rotating about an axis AB with angular speed ω. The ith particle is rotating in a circle of radius ‘ri’ with linear speed equal to ωri. The kinetic energy of this particle is equal to ½ miv2 = ½ mi(ω2ri2). The kinetic energy of the whole body would then be

∑½ mi(ω2ri2) = ½ ∑(miri22 = ½ Iω2

Thus rotational kinetic energy of the body is basically the sum of ½ mv2 of all the particles of the body.

### Moment of Inertia

The moment of inertia is the equivalent of mass in rotational motion. It is defined as

I = ∑miri2

where mi is the mass of the ith particle and ri is its perpendicular distance from the given axis of rotation.

### Theorem of parallel axes

This theorem states that if we know the moment of inertia of a body of mass ‘m’ about a given axis, then the moment of inertia of the body about another axis parallel to the given axis can be calculated as I = I0 + md2 Proof: In the given figure, let O be the center of mass of the body and let O be the origin of the coordinate system. Let I and Io be the moments of inertia of the body about AB and OQ respectively.
Let P be any arbitrary particle of the body with coordinates (xi, yi, zi). Also let AB be an axis parallel to OQ where A is at a distance d from O. PR and PQ be the perpendicular distance of particle P to the axes AB and OQ respectively.
Now, coordinates of point A are (d, 0, 0) and we have CQ = AR = zi. thus coordinates of point Q are (0, 0, zi) and that of point R are (d, 0, zi).
I = ∑mi(PR)2 = ∑mi[(xi - d)2 + (yi - 0)2 + (zi – zi)2] = ∑mi(xi2 + yi2 + d2 – 2xid)
= ∑mi(xi2 + yi2) + ∑mid2 -2d∑mixi
We have ∑mixi = MXCOM = 0
Therefore, I = ∑mi(xi2 + yi2) + ∑mid2   …(1)
The moment of inertia about OQ is
Io = = ∑mi(PQ)2 = ∑mi[(xi - 0)2 + (yi - 0)2 + (zi – zi)2] = ∑mi(xi2 + yi2)   …(2)
From (1) and (2);
I = Io + ∑mid2 = Io + Md2

### Theorem of perpendicular axes This theorem is applicable only to planar (or two-dimensional) bodies. Let x and Y axes lie in the plane of the body and z-axis be perpendicular to it. Then this theorem states that Iz = Ix + Iy
Consider an arbitrary particle P of the body. Let PQ and PR be the perpendiculars from P on the x and y axes respectively. PO is the perpendicular from P on the z-axis. Thus the moment of inertia of the body about the z-axis is
I = = ∑mi(PO)2 = ∑mi[(PQ)2 + (OQ)2] = ∑mi[(PQ)2 + (PR)2] =∑mi(PQ)2 + ∑mi(PR)2 = Ix + Iy

### Rolling

Rolling is combined rotation and translation like the wheel of a moving car or bicycle. By the time, the wheel completes a full rotation, its center of mass moves through a distance ‘2пr’. This is called pure rolling. In this case, the velocity of the point of contact is zero. The velocity of the center of mass is vCOM = Rω and that of the topmost point is vtop = 2Rω = 2vCOM. If the wheel moves a distance greater or less than 2пr in one full rotation, slipping occurs at the point of contact and thus friction force acts on the body, until pure rolling starts and no slipping occurs.

## Here are some of the problems solved in this tutorial :

Q: A disc of radius 5 cm is rotating about its axis at an angular speed of 15 rad/s. Find the linear speed of (a) a point on the rim (b) the middle point of a radius.

Q: A wheel of radius 8 cm can rotate freely about its center. A string is wrapped over its rim and is pulled by a force of 10 N. It is found that the torque produces an angular acceleration of 2.5 rad/s2 in the wheel. Calculate the moment of inertia of the wheel.

Q: Find the moment of inertia of a pair of spheres, each having a mass m and radius R, kept in contact about the tangent passing through the point of contact.

Q: The uniform solid block has mass, m = 1 kg and edge lengths a = 4 cm, b = 8.4 cm, and c = 1.5 cm. Calculate its rotational inertia about an axis through one corner and perpendicular to the large faces.

Q: The surface density (mass/area) of a circular disc of radius R depends on the distance from the center as ρ(r) = A + Br. Find its moment of inertia about the line perpendicular to the plane of the disc through its center.

Q: A uniform spherical shell of mass M = 5 kg and radius R = 10 cm can rotate about a  vertical axis on frictionless bearings. A massless cord passes around the equator of the shell, over a pulley of rotational inertia I = 3 X 10-3 kgm2 and radius r = 5 cm, and is attached to a small object of mass m = 0.5 kg. There is no friction on the pulley’s axle; the cord does not slip on the pulley. What is the speed of the object when it has fallen 80 cm after being released from rest? Take g = 10 ms-2.

Q: A small disk of radius 2 cm is glued to the edge of a larger disk of radius 5 cm so that the disks lie in the same plane. The disks can be rotated around a perpendicular axis through point O at the center of the larger disk. The disks both have a uniform density (mass per unit volume) of 1.4 X 103 kg/m3 and a uniform thickness of 5 mm. What is the rotational inertia of the two-disk assembly about the rotation axis through O?

Q: A thin uniform rod has length 2 m and can pivot about a horizontal, frictionless pin through one end. It is released from rest at angle θ = 45° above the horizontal. Determine the angular speed of the rod as it passes through the horizontal position.

Q: A wheel of radius 0.25 m is mounted on a frictionless horizontal axle. A massless cord is wrapped around the wheel and attached to a 1 kg box that slides on a frictionless surface inclined at angle θ = 30° with the horizontal. The box accelerates down the surface at 2.5 m/s2. What is the rotational inertia of the wheel about the axle? Take g = 10 ms-2.

Q: Two thin rods (each of mass 1 kg) are joined together to form a rigid body as shown in figure. One of the rods has length L1 = 1 m, and the other has length L2 = 0.8 m. What is the rotational inertia of this rigid body about (a) an axis that is perpendicular to the plane of the paper and passes through the center of the shorter rod and (b) an axis that is perpendicular to the plane of the paper and passes through the center of the longer rod?

Q: A particle of mass ‘m’ is projected with a speed ‘u’ at an angle ‘θ’ to the horizontal. Find the torque of the weight of the particle about the point of projection when the particle is at the highest point.

Q: A string is wrapped on a wheel of moment of inertia 0.25 kgm2 and radius 25 cm and goes through a light pulley to support a block of mass 2 kg as shown in figure. Find the acceleration of the block. Take g = 10 ms-2.

Q: A kid of mass 25 kg stands at the edge of a disc shaped horizontal platform of radius 2 m which can be freely rotated about its axis.  The moment of inertia of the platform is 5 kgm2. The system is at rest when a friend throws a ball of mass 0.25 kg and the kid catches it. If the velocity of the ball is 20 m/s horizontally along the tangent to the edge of the platform when it was caught by the kid, find the angular speed of the platform after the event.

Q: A string is wrapped over the edge of a uniform disc and the free end is fixed with the ceiling. The disc moves down, unwinding the string. Find the downward acceleration of the disc.

Q: A solid sphere of mass ‘m’ is released from rest from the rim of a hemispherical cup so that it rolls along the surface. If the rim of the hemisphere is kept horizontal, find the normal force exerted by the cup on the ball when the ball reaches the bottom of the cup.

Q: A small m = 50 g block slides down a frictionless surface through height h = 25 cm and then sticks to a uniform rod of mass 100 g and length, d = 50 cm. The rod pivots about point O through angle θ before momentarily stopping. Find θ.

Q: A constant horizontal force of magnitude 12 N is applied to a uniform solid cylinder by fishing line wrapped around the cylinder. The mass of the cylinder is 15 kg, its radius is 0.2 m, and the cylinder rolls smoothly on the horizontal surface. (a) What is the magnitude of the acceleration of the center of mass of the cylinder? (b) What is the magnitude of the angular acceleration of the cylinder about the center of mass? (c) What is the frictional force acting on the cylinder?

Q: A uniform ladder of length ‘L’ and mass 15 kg leans against a smooth vertical wall making an angle of 53o with it. The other end rests on a rough horizontal surface. Find the normal force and the frictional force that the floor exerts on the ladder.

Q: A uniform rod of mass ‘M’ and length ‘L’ lies on a smooth horizontal plane. A particle of mass ‘m’ moving at a speed ‘v0’ perpendicular to the length of the rod strikes it at a distance L/4 from the center and stops after the collision. Find (a) the velocity of the center of the rod and (b) the angular velocity of the rod about its center just after the collision.
Q: A force F = 25 N acts tangentially at the highest point of a sphere of mass m = 3 kg kept on a rough horizontal plane. If the sphere rolls without slipping, find the acceleration of the center of the sphere.

Q: A solid sphere of mass ‘m’ is set into motion on a rough horizontal surface with a linear speed ‘v’ in the forward direction and an angular speed v/r in the anticlockwise direction. Find the linear speed of the sphere (a) when it stops rotating (b) when slipping finally ceases and pure rolling starts.

Q: A solid sphere rolling on a rough horizontal surface with a linear speed ‘v’ collides elastically with a fixed, smooth, vertical wall. Find the speed of the sphere after it has started pure rolling in the backward direction.

## MCQ Quiz on Rotational Dynamics

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