Kinematics of rotation of rigid body

Torque of a force about the axis of rotation

Angular momentum

Conservation of angular momentum

Work done by a torque

Power delivered by a torque

Moment of inertia

Moment of inertia theorems

Theorem of parallel axes

Theorem of perpendicular axes

Combined rotation and translation

Rolling

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**Rotation of a rigid body**

If each particle of a rigid body moves in a circle, with centres of all the circles on a straight line and with planes of the circles perpendicular to this line we say that the body is rotating about this line. The straight line is called the axis of rotation. (Particle makes circular motion. Rigid body makes rotation.)

**Kinematics of rotation of rigid body**

For a rigid body let the axis of rotation be Z-axis.

At time t =0, let particle P be at P

_{0}.

Perpendicular to axis of rotation from P

_{0}be PQ. (Q is on the axis)

If at time t Particle P moves P

_{1}and angle P

_{0}Q P

_{1}= θ.

Hence the particle has rotated through θ.

All particles have rotated through θ.

We can say the whole rigid body has rotated through angle θ.

The angular position of the body at time t is θ.

If P has made a complete revolution its circular path, every particle will do so and hence rigid body has done so. We can say rigid body made a complete revolution and it has rotated through an angle of 2 π radians.

Hence, rotation of a rigid body is measured by the rotation of a line PQ (P is a particle on the rigid body and Q is a point on the axis of rotation and PQ is perpendicular from P to the axis of rotation).

As the rotation of rigid body is defined in terms of the circular motion of a particle on the rigid body, kinematics of circular motion of particle becomes applicable to rotation.

**Angular variables**

θ = angular position of the particle

ω = angular velocity = d θ/dt = lim∆t→0 ∆θ/∆t

α = angular acceleration = d ω/dt = d²θ/dt²

If the angular acceleration is constant, formulas similar in form to linear formulas can be used to find the angular variables:

θ = ω

_{0}t + ½ αt²

ω = ω

_{0}+ αt

ω² = ω

_{0}² + 2 α θ

Where

ω

_{0}is velocity at time t = 0.

Given the axis of rotation, the body can rotate in two directions – clockwise or anticlockwise. One of the directions has to be defined as positive direction according to the convenience of the problem.

The SI unit for angular velocity is radian/sec (rad/s).

One revolution/sec = 2 π radian/sec

Similar to the circular motion of the particle, in rotation for a particle P

s = Linear distance traveled by the particle in circular motion

∆s = Linear distance traveled by the particle in circular motion in time ∆t

∆s = r∆θ

Where

r = radius of the circle over which the particle is moving

∆θ = angular displacement in time ∆t

∆s/∆t = r∆θ/∆t

v = r ω

where

v = linear speed of the particle

a

_{t}= rate of change of speed of the particle in circular motion

a

_{t}= dv/dt = rdω/dt = r α

**Rotational dynamics**

In rotation of a body the resultant force due to external forces is zero, but the resultant of

Torque produced by the external forces is nonzero and this torque produces rotation motion.

Torque of a force about the axis of rotation

First we define torque a force about a point.

For a force

**F**acting on a particle P, to find torque about a point O, define the position vector of P with respect to O. Let this position vector be

**r**.

Torque of force F about O =

**Γ**=

**F**×

**r**.

This is vector product of two vectors hence a vector quantity, as per the rules of vector product, the direction of

**Γ**will be perpendicular to to

**F**and

**r**.

When the torque about an axis of rotation is to be determined, select a point on the axis of the rotation and find the torque of the force acting on a particle about this point. Find the angle between the axis and the line joining the point on axis (about which the torque is calculated) and the particle (one which the force is acting). Let the angle be θ.

Torque about the axis due to a force is the component along the axis, of the torque of the force about a point on the axis.

Magnitude of the torque = |

**F**×

**r**| cos θ

The torque about the axis is same, even if different points are chosen along the axis for determining the torque of the force about those points.

Some special cases of relation between force and the axis of rotation.

1. Force is parallel to the axis of rotation.

Torque along the axis is zero.

2.

**F**and

**r**are collinear. The torque about O is zero and the torque about axis is zero.

3. Force and axis are perpendicular but they do not intersect. In three dimensions, two lines may be perpendicular without intersecting. Example: A vertical line on

a wall and a horizontal line on the opposite wall.

In this case torque about the axis is equal to Force multiplied by the perpendicular to axis from the force direction (line along which the force is acting).

Torque produced by forces on a particle

A particle having circular motion will have two forces acting on it.

One force produces tangential acceleration dv/dt in it. Hence the force named tangential force is 'ma' = mdv/dt = mrα

This force creates a torque of mr²α

the other force creates radial acceleration or centripetal acceleration ω²r. Hence the force named radial force is mω²r

As intersects the axis of rotation, the torque produced by it is zero.

Hence total torque produced by a rigid body consisting of n particles is

Г(total) = Σm

_{i}r

_{i}²α

= αΣm

_{i}r

_{i}² as α is same for all particles

Let I = Σm

_{i}r

_{i}²

Г(total) = Iα

Quantity I is called moment of inertia of the body about the axis of rotation.

I = Σm

_{i}r

_{i}²

where

m

_{i}= mass of the ith particle

r

_{i}= perpendicular distance of ith particle from the axis of rotation.

Angular momentum

Conservation of angular momentum

Work done by a torque

Power delivered by a torque

Moment of inertia

Moment of Inertia

Torque created by external forces in rotating motion

When a particle is rotating it has tangential acceleration and radial acceleration.

Radial acceleration = ω²r

Hence radial force acting on it = m ω²r

Tangential acceleration = dv/dt = rdω/dt = r α

ω = angular velocity

α = angular acceleration

Tangential force = mrα

Torque created by radial force is zero as the force intersects the axis of rotation.

Tangential force and axis are skew (they do not intersect) and they are perpendicular. Thus the resultant torque is Force*radius = mr²α

In the case of a body having ‘n’ particles and rotating, the total torque is equal to torque acting on each of the particles

Total torque on the body = Γ(total) = Σ m

_{i}r

_{i}²α

Σ m

_{i}r

_{i}² is called as moment of inertia.

Moment of inertia can be calculated using the above formula for collection of discrete particles.

If the body is a continuous, the technique of integration needs to be used. We consider a small element of the body with mass dm and having a perpendicular distance from the axis or line about which moment of inertial is to be calculated.

We find ∫ r²dm under proper limits to get the moment of inertia of the body.

r²dm is the moment of inertia of the small element.

Determination of moment of inertia of representative bodies.

1. Uniform rod about a perpendicular

M = total mass of the body

l = length of the body

I = Ml²/12

I is obtained by taking a small element dx at a distance x from the centre of the rod.

Mass dm of the element = (M/l)*dx (M/l give mass per unit length)

dI = (M/l)*dx*x²

I = ∫(M/l)*dx*x² (limits are from –l/2 to l/2: these limits cover the entire rod).

= M/l[x³/3] from –l/2 to l/2

= M/3l[l³/8 + l³/8] = M/3l(l³/40) = Ml²/12

2. Moment of inertia of uniform rectangular plate about a line parallel to an edge and passing through the centre.

M = total mass of the body

Plate measurements l,b

Axis or line is parallel to b

I = Ml²/12

If the axis of line is parallel to l

I = Mb²/12

3. Circular ring

M = total mass of the body

R = radius

I = MR²

4. Uniform circular plate

M = total mass of the body

R = radius

I = MR²/2

5. Hollow cylinder about its axis

M = total mass of the body

R = radius

I = MR²

6. Uniform solid cylinder about its axis.

M = total mass of the body

R = radius

I = MR²/2

7. Hollow sphere about a diameter

M = total mass of the body

R = radius

I =(2/3) MR²

8. Uniform solid sphere about a diameter

M = total mass of the body

R = radius

I = (2/5) MR²

Moment of inertia theorems

Theorem of parallel axes

Thoerem of perpendicular axes

Combined rotation and translation

Rolling

## Formula Sheet – Rotational Mechanics

Rotational kinematics

**Angular variables**

θ = angular position of the particle

ω = angular velocity = dθ/dt = lim∆t→0 ∆θ/∆t

α = angular acceleration = dω/dt = d²θ/dt²

If the angular acceleration is constant, formulas similar in form to linear formulas can be used to find the angular variables:

θ = ω

_{0}t + ½ αt²

ω = ω

_{0}+ αt

ω² = ω

_{0}² + 2 α θ

where

ω

_{0}= angular velocity at the beginning

Relation between the linear motion of a particle of a rigid body and rotation of the rigid body

v = r ω

where

v = linear speed of the particle

a

_{t}= rate of change of speed of the particle in circular motion

a

_{t}= dv/dt = rdω/dt = r α (These relations are from the chapter of circular motion)

Updated 10 Jan 2016, 7 May 2008

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