Rotational Dynamics Concepts Page - 4

Definition

Couple

The direction of torque is always perpendicular to plane of rotation of body as a cross product is in perpendicular plane to r$r$ and F$F$ vectors and the torques produced by two forces of couple are in same direction to each other.

Formula to calculate couple is:
Couple=F×l$Couple=F\times l$
Where l$l$ is perpendicular distance between forces.

Definition

Principle of Moments

If algebraic sum of moments of all the forces acting on the body about the axis of rotation is zero, the body is in equilibrium. 
According to the principle of moments, in equilibrium:
Sum of anticlockwise Moments = Sum of clockwise moments

Definition

Factors Affecting Turning of a Body

The turning of a body by a force depends on the following two factors:
1. The magnitude of the force applied
2. The distance between the line of action of force and the axis of rotation (or pivoted point)

Diagram

Examples of turning effect of force

(1) To open and shut a door, we apply a force (push or pull) normal to the door at its handle P which is at the maximum distance from the hinge.
(2) The upper circular stone of a hand flour grinder is provided with a handle near its rim (i.e. maximum distance from the centre) so that it can easily be rotated about the iron pivot at its centre by applying a small force at the handle.
(3) For turning a steering wheel, a force is applied tangentially on the rim of the wheel.
(4) In a bicycle, to turn the wheel anticlockwise, a small force is applied on the pedal of a toothed wheel.

Example

Principle of Moments

When slightly different weights are placed on the two pans of a beam balance, the beam comes to rest at an angle with the horizontal. The beam is supported at a fixed single point by a pivot. The net torque about this fixed single point due to the two weights is nonzero at the equilibrium position. The whole system does not continue to rotate about the fixed point because the moment is balanced.

Example

Condition for rotational equilibrium

Example: A right triangular plate ABC of mass m$m$ is free to rotate in the
vertical plane about a fixed horizontal axis through A. It is supported by a string such that the side AB is horizontal. What is the reaction at the support A in equilibrium?

Solution: The distance of Centre Of Mass of given right angled triangle is
2L3${\displaystyle \frac{2L}{3}}$ along BA and L3${\displaystyle \frac{L}{3}}$ along AC from the point B.
Force of magnitude mg$mg$ is acting downwards at its COM. Moment balance around B gives:
mg(2L3)−FA(L)=0$mg({\displaystyle \frac{2L}{3}})-F_A(L)=0$  
(Moment= r⃗ ×F⃗ =rFsin(θ)=F(rsin(θ))=Fr⊥$\overrightarrow{r}\times \overrightarrow{F}=rFsin(\theta )=F(rsin(\theta ))=F{r}_{\mathrm{\perp }}$)
∴FA=23mg$\therefore F_A={\displaystyle \frac{2}{3}}mg$

Rotation Equilibrium: An object in rotational equilibrium has no net external torque.

Definition

Discuss the condition for rotational and translational equilibrium

Translational Equilibrium: ∑F=0$\sum F=0$ (that is, Fnet=0$F_{net}=0$).
An object may be rotating, even rotating at a changing rate, but may be in translational equilibrium if the acceleration of the center of mass of the object is still zero.
Rotational Equilibrium: ∑τ=0$\sum \tau =0$; (τnet=0${\tau }_{net}=0$). 
An object may be accelerating in a linear fashion (along a straight line and or even turning at a constant rate; an object in rotational equilibrium will NOT be accelerating in a rotational sense (ie. the angular momentum of an object in rotational equilibrium will be constant).

Example

Problem on total equilibrium in which all the forces are not coplanar

Example: A uniform thin cylindrical disk of mass M$\mathrm{M}$ and radius
R$\mathrm{R}$ is attached to two identical massless springs of spring constant k$\mathrm{k}$ which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance d$\mathrm{d}$ from its centre. The axle is massless and both the springs and the axle are in horizontal plane. The unstretched length of each spring is L$\mathrm{L}$. The disk is initially at its equilibrium position with its centre of mass (CM$\mathrm{C}\mathrm{M}$) at a distance L$\mathrm{L}$ from the wall.
The disk rolls without slipping with velocity V⃗ 0=V0i^${\overrightarrow{\mathrm{V}}}_0=V_0\hat{i}$. The coefficient of friction is μ$\mu$.

Solution:
let the cylinder is displaced by x distance
2kx−f=ma$2kx-f=ma$
also fR=Iα$fR=I\alpha$
a=αR$a=\alpha R$

ma=4kx/3$ma=4kx/3$
F1R=Iα$F_{1}R=I\alpha$

4kRθ×R/3=mR2α$4kR\theta \times R/3=m{R}^2\alpha$α=4k3mθ$\alpha ={\displaystyle \frac{4k}{3m}}\theta$
ω=4k3m−−−−√

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