Angular Kinematics and Torque

Angular Kinematics Discussion

Relating linear and angular stuff to each other
Pure rotation: points on an object rotate with the same angular displacement, but different linear displacements based on their distance from the axis of rotation.

Arc Length

The relationship between arc length $(s)$ , radius $(r)$ , and angular displacement $\theta$ : $s = r \theta$ Where:
- $s$ is the length of the arc on the circular path.
- $r$ is the radius of the circular path.
- $\theta$ is the angular displacement in radians.
This equation is a rearrangement of the definition of a radian: $\text{radian} = \frac{s}{r}$
Angular kinematic variables ($\theta $,$ \omega $,$ \alpha $) must be in radians for the equations to work correctly.</li><li>Radians are technically unitless, while degrees are a unit. This is why radians are required to maintain correct unit consistency.</li></ul><h4 id="668b93c4-071a-40b0-ae6f-afa9d25543a2" data-toc-id="668b93c4-071a-40b0-ae6f-afa9d25543a2" collapsed="false" seolevelmigrated="true">Relationship Between Linear and Angular Quantities</h4><ul><li>$ s = r \theta where:
- s = arc length
- r = radius
- $\theta$ = angular displacement (in radians)
The angular kinematic variable must be in radians.
Radians is a unitless quantity. If using degrees, degrees are a unit so you'd end up with a meter degrees, which does not match what you want out of it.

Velocity on a Circular Path

Instantaneous velocity is tangent to the circular path.
v_t = r \omega $where:<ul><li>$ v_t = magnitude of the linear velocity (tangential)
$\omega$ = magnitude of angular velocity
r = radius

Use positive values for angular velocity and r $. Only looking for the length of that arc length.</li></ul><h4 id="324dbaa4-cd3a-4a33-bbfc-10d1bf40ce43" data-toc-id="324dbaa4-cd3a-4a33-bbfc-10d1bf40ce43" collapsed="false" seolevelmigrated="true">Linear Accelerations</h4><ul><li>Two perpendicular components:<ul><li>Tangent to the path ($ a_t $)</li><li>Perpendicular/normal to the path ($ a_n $)</li></ul></li></ul><h5 id="dafc08bb-7d92-402f-b72d-d206dcd9949d" data-toc-id="dafc08bb-7d92-402f-b72d-d206dcd9949d" collapsed="false" seolevelmigrated="true">Tangential Acceleration</h5><ul><li>$ a_t = r \alpha where:

$\alpha$ = magnitude of angular acceleration

If angular velocity is constant (\alpha = 0 $), then$ a_t = 0 $.</li></ul><h5 id="ae42491f-3ab4-46e5-9465-8ff50cd756aa" data-toc-id="ae42491f-3ab4-46e5-9465-8ff50cd756aa" collapsed="false" seolevelmigrated="true">Normal/Centripetal/Radial Acceleration</h5><ul><li>Always directed towards the center of the circle.</li><li>$ an = ac = a_r = r \omega^2 $</li><li>Alternative form:$ a_n = \frac{v^2}{r} $, where v is the tangential velocity.</li><li>Finding the magnitude of linear acceleration is:$ \sqrt{at^2 + an^2} $</li><li>As the radius increases, linear attributes also increase.</li><li>motion = normal/centripetal acceleration</li></ul><h4 id="6d95356b-c649-44cf-86af-952434631c16" data-toc-id="6d95356b-c649-44cf-86af-952434631c16" collapsed="false" seolevelmigrated="true">Centripetal vs. Centrifugal Force</h4><ul><li>Centripetal Force:<ul><li>A force inward is a thing that exists.</li><li>Causes centripetal acceleration.</li></ul></li><li>Centrifugal Force:<ul><li>Doesn't exist.</li><li>The phenomenon that we observe because of the lack of a centripetal force to keep it on that circular path.</li></ul></li><li>If restraints are removed, the object will want to get further away from the center because the velocity is tangent to the path.</li><li>Centrifuge: works off of the lack of the centripetal force on that object that's suspended in there.</li></ul><h4 id="b02298bc-7b1f-4f10-a864-1cf93393f97e" data-toc-id="b02298bc-7b1f-4f10-a864-1cf93393f97e" collapsed="false" seolevelmigrated="true">Biceps Curl Example</h4><ul><li>Analysis of$ an $and$ at $during a biceps curl.</li><li>$ an $is proportional to$ \omega^2 $and$ at $is proportional to$ \alpha $.</li><li>The forearm segment angle increases as we flex the joint.</li><li>Magnitude of angular velocity squared will be hill hill.</li><li>Magnitude of angular acceleration becomes a hill hill and a hill hill profile.</li><li>Angular profiles of the forearm segment are:<ul><li>Angle: Increases during flexion, decreases during extension.</li><li>Angular Velocity$ \omega $: Hill-valley profile.</li><li>Angular Acceleration$ \alpha $: Hill-valley-valley-hill profile.</li></ul></li><li>The equations for$ an $and$ at $:<ul><li>$ a_n = r \omega^2 $</li><li>$ a_t = r \alpha $</li></ul></li></ul><h5 id="9cf4fd5a-bcff-427b-9e32-83a908fcfd95" data-toc-id="9cf4fd5a-bcff-427b-9e32-83a908fcfd95" collapsed="false" seolevelmigrated="true">Analysis of$ \mathbf{a_n} $</h5><ul><li>Since we only care about the magnitude:<ul><li>Angular velocity, which had a hill-valley profile, becomes hill-hill after being squared. Larger velocity values become more accentuated with squaring.</li><li>$ a_n $has a hill-hill shape. As the angular velocity increases, more centripetal force is needed.</li></ul></li></ul><h5 id="3fbb7fc9-277a-4713-907c-80bd7218ff29" data-toc-id="3fbb7fc9-277a-4713-907c-80bd7218ff29" collapsed="false" seolevelmigrated="true">Analysis of$ \mathbf{a_t}

The angular acceleration ($\alpha) profile, which is hill-valley-valley-hill, becomes hill-hill-hill-hill after taking the magnitude.

Halfway Point of Up Phase

Vertical acceleration is zero.
Tangential acceleration is zero.
Horizontal acceleration is large and negative.
Normal/centripetal/radial acceleration is large.

Effect of Repetition Speed

Changing repetition speed alters $\omega$ by changing $\Delta t$ .
Acceleration has a squared effect.
Forearm and hand length have linear effects, not squared effects like changes in time.

Rolling Ball Example

Relationship between linear and angular velocity:
- $v{cm} = r \omega$ or $\omega = \frac{v{cm}}{r}$
At any instant, the point on the ball touching the ground has zero linear velocity if the ball is not slipping.

Torque

Main construct of angular kinetics.
Analogous to force in linear kinetics.
Capital T is the variable, and it is a vector quantity.

Definition

The rotational analog to force.
The effect of a force that tends to cause a change in a body's state of angular position or motion.
Units: Newton-meter (N·m) or foot-pound (ft·lb) in the English system
- 1 N·m = 0.736 ft·lb
Biomechanists prefer the term "moment" or "moment of force".

Calculation

Torque is the product of the magnitude of a force and its moment arm.
A vector about an axis perpendicular to the plane of the force and moment arm.

Convention

Counterclockwise rotation: positive torque
Clockwise rotation: negative torque
Magnitude of torque = (magnitude of force) * (length of moment arm).
The moment arm is the perpendicular distance from the line of action of the force to the axis of rotation.

Subscripts

Use a subscript on $T$ to connect it with the force (e.g., $T_F$ ).
Sometimes a subscript is associated with the axis of rotation (e.g., $T_A$ ).

Example: Free Body Diagram of Forearm and Hand

Forces:
- Bone-on-bone contact force
- Muscle force ( $F_m$ )
- Weight of forearm and hand
- Force from barbell

Torque Calculations

Torque from muscle: $Tm = |Fm| * r_m$
Torque from forearm and hand (weight): $Tw = -|W| * rw$
Torque from force in hand: $TH = -|FH| * r_H$
Torque from bone on bone is zero because the line of action goes through the axis of rotation, so arm is 0.
These torques have different signs because some promote rotation in different directions.

Lever Arm vs. Moment Arm

Avoid the term "lever arm" because it has two definitions:
- Synonymous with moment arm (perpendicular distance).
- Straight-line distance from the axis to the point of force application.
Leverage: conveys the notion of increasing the moment arm.
The moment arm is the perpendicular distance from the line of action of the force to the axis of rotation.

Torque as a Vector Cross Product

Avoid this approach.
The way Torque is a vector cross product is $T = F \times \Delta p$
$T = F \times \Delta P \sin(\theta)$
$\Delta P$: A displacement to the point of application from the axis which is just ends up being the length.
But instead the preferred way is $T = F \times R$