Rotational Mechanics Notes

Notes for Rotational Mechanics

Centre of mass:

The centre of mass of an object is also its centre of weight – the position of the fulcrum for which there is no rotation when suspended in a uniform gravitational field.

Centre of mass in one dimension:

If C is the centre of mass, rotation about C in a uniform field of strength g is zero. Thus the moments about C sum to zero such that:

m₁gl₁m₂gl₂=0

m₁l₁=m₂l₂

Since l₁=x₀x₁, and l₂=x₂x₀:

m₁(x₀x₁)=m₂(x₂x₀)

$$\therefore x_{0} = \frac{m_{1}x_{1} + m_{2}x_{2}}{m_{1} + m_{2}} = \frac{m_{1}x_{1} + m_{2}x_{2}}{M}$$

Where M is total mass.

This definition can be extended to a one-dimensional body of N discrete masses δm adding to a total mass of M, and to a continuous body of mass M in the limit that N:

Centre of mass in three dimensions:

The position of the centre of mass for discrete bodies in three dimensions can be found by fulfilling three separate one-dimensional conditions:

Which combine to produce the vector forms:

Where r_i is the position vector of the ith mass, r_i=(x_i,y_i,z_i), R is the position vector of the centre of mass, R=(x₀,y₀,z₀), and ′body′ refers to a function which specifies the shape of the body.

The integral is calculated by expressing dm in terms of variables in which r can be written. In one dimension, dm=ρ_ldl, in two dimensions, dm=ρ_ada and in three dimensions, dm=ρ_VdV, where dl,da and dV are length, area and volume elements, respectively.

Centre of mass of bodies of non-uniform shape:

Since R is a linear function, the principle of superposition can be used to find the centre of mass of a non-uniform body in terms of the centres of mass of several different uniform parts. The centre of mass of each part is found separately, then summed linearly to obtain the overall centre of mass.

Application to the lever balance:

For a one-dimensional lever balance, maintaining a state of equilibrium depends on both a linear mechanical and a rotational mechanical condition:

• Linear mechanical condition: Net force on the system is zero:

$$\mathbf{F}_{ext} = \sum_{i}^{\ }\mathbf{F}_{i} = \mathbf{0}$$

• Rotational mechanical condition: The fulcrum is at the centre of mass of the lever, such that all turning moments sum to zero:

$$\mathbf{G}_{ext} = \sum_{i}^{\ }\mathbf{G}_{i} = \sum_{i}^{\ }{\mathbf{r}_{i} \times \mathbf{F}_{i}} = \mathbf{0}$$

Circular motion:

The general motion of a body consists of both an angular and a radial part. Circular motion is concerned only with the angular part of the motion:

The particle moves from P₁ to P₂ relative to O. Its initial position is described by the vector r, its final position by the vector r+δr, hence its movement is described by δr.

If the angle of rotation θ is small (θ1), then it can be described by a vector θ which points along the axis of rotation and has a magnitude of θ.

The equivalence of θ and θ for small angles can be seen by considering a segment of the circle AB on a sphere of radius a:

For small angles, AB is approximately straight. Thus it can be represented by the vector s, which has length aθ.

If three rotations are performed, the vectors s₁,s₂ and s₃ can be summed to give a closed triangle.

Since aθ₁,aθ₂ and aθ₃ sum to give a closed triangle, and their lengths are proportional to θ₁,θ₂ and θ₃ with identical constants of proportionality, it follows that θ₁,θ₂ and θ₃ can also be summed to give a closed triangle.

Thus, angles θ can be represented by vectors θ.

Sign conventions for angles and vectors:

The axis of a rotation is always defined in the right-handed sense, such that:

• A negative rotation is one for which θ is antiparallel to the axis of rotation, hence the rotation is left-handed

• A positive rotation is one for which θ is parallel to the axis of rotation, hence the rotation is right-handed

Angular quantities:

Angular velocity and acceleration:

Angular speed, velocity (ω) and acceleration ($\dot{\mathbf{\omega}}$) are the rotational analogues of linear speed, velocity and acceleration in linear mechanics.

If a body at point P rotates a small angle δθ in time increment δt, then the angular speed is given by:

$$\frac{\delta\theta}{\delta t}$$

The angular velocity is the vector form of this expression:

$$\mathbf{\omega} = \frac{\delta\mathbf{\theta}}{\delta t}$$

And in the limiting case as δθdθ:

$$\boxed{\mathbf{\omega} = \frac{d\mathbf{\theta}}{dt} = \dot{\mathbf{\theta}}}$$

Similarly, the angular acceleration is given by:

$$ \boxed{\dot{\mathbf{\omega}} = \frac{d^{2}\mathbf{\theta}}{dt^{2}} = \ddot{\mathbf{\theta}}}$$

Rotational moments of linear quantities:

If a given linear quantity λ_L produces rotational motion, thus has a rotational equivalent λ_R, then λ_R is the rotational moment of λ_L, given by the scalar product of the vector distance r between the point of application and the axis of rotation, and the value of the linear quantity λ_L, such that:

$$\boxed{\mathbf{\lambda}_{R} = \mathbf{r} \times \mathbf{\lambda}_{L}}$$

This definition is justified as follows:

• The component of λ_L which is tangential to the rotational motion, hence which contributes to it, has magnitude λ_Lsinθ

• When multiplied by the magnitude of r, this gives the rotational moment a magnitude of λ_Lrsinθ and a direction parallel to θ, hence perpendicular to r and λ_L

• Thus λ_R=rλ_L

Linear quantities and their angular moments:

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