PHYS 218 Chapter 10

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Torque

Vectors at play in torque

(See MATH 152 Chapter 11.3→)

Effectiveness of a force to cause a rotational movement around an axis. Denoted by Greek letter Tau (τ). Measured in "Newton Meters" [Nm] NOT Joules

${\begin{aligned}\tau &=\mathbf {r} \times \mathbf {F} \\|\tau |&=|\mathbf {r} ||\mathbf {F} |\sin {\theta }\end{aligned}}$

Origin will always be at axis

Torque and Angular Acceleration

$\sum \tau =I\alpha$ (equivalent to Newton's second law: $\sum F=ma$

Motion of a Rigid Body

Combination of:

• Translation ${\tfrac {1}{2}}mv^{2}$ (velocity of center of mass)
• Rotation ${\tfrac {1}{2}}I\omega ^{2}$ (about an axis)

Now when solving problems:

• Free body diagrams must have all forces exactly where they are applied (not necessarily at center of mass of object
• Add the sum of all torques in ${\hat {z}}$ to the sum of all forces in ${\hat {x}}$ and ${\hat {y}}$
• Relate the accelerations to each other (assume rope/disk does not slip)

Example

Is the moon rotating? YES! (around earth and on its own axis since same side always faces us)

Rolling without Slipping

Rotation and translation are related:

• $v_{cm}=\pm R\omega$
• $a=\pm R\alpha$

(Plus/minus depends on problem; use right-hand rule)

Example

Total Kinetic energy (translational and rotational) of a rolling disk/wheel.

From the center of the disk:

${\frac {1}{2}}mv^{2}+{\frac {1}{2}}I\omega ^{2}$

From the point of contact with the ground:

${\frac {1}{2}}I_{p}\omega ^{2}={\frac {1}{2}}\left(I_{cm}+mR^{2}\right)\omega ^{2}$

These two formulas are the same when simplified.

Problem

Disk rolling down an inclined plane

$v={\sqrt {\frac {2gh}{1+{\frac {I}{mR^{2}}}}}}={\sqrt {\frac {2gh}{1+c}}}$

Work done by Torque

Work done by a force applied to a rotating object (W is still measured in Joules):

$W=\int _{\theta _{1}}^{\theta _{2}}\tau \mathrm {d} \theta$

Corollary

Work of net force:

$W_{N}={\frac {1}{2}}I\omega _{f}^{2}-{\frac {1}{2}}I\omega _{i}^{2}$

Power of Torque

(Use second equation if torque is constant.)

${\begin{aligned}P&={\frac {\Delta W}{\Delta t}}={\frac {\mathrm {d} }{\mathrm {d} t}}\int _{\theta _{1}}^{\theta _{2}}\tau \mathrm {d} \theta \\&=\tau \omega \end{aligned}}$

Angular Momentum

Angular momentum is represented by $\mathbf {L}$. Let $\mathbf {r} _{q}$ represent the position with respect to the coordinate origin point $q$. $\mathbf {L}$ is the cross product of the position vector and $q$'s linear momentum:

${\begin{aligned}\mathbf {L} _{q}&=\mathbf {r} _{q}\times \mathbf {P} =\mathbf {r} _{q}\times m\mathbf {v} \\L&=rmv\end{aligned}}$

Note. Position vector can be broken into components parallel and perpendicular to velocity vector. The magnitude of $\mathbf {L}$ only depends on the perpendicular component

Note

Angular Momentum if spinning around its center of mass for any rigid body is completely independent of $q$!

Object has intrinsic angular momentum

$\mathbf {L} =I_{cm}\mathbf {\omega }$

Conservation of Angular Momentum

If there is no external torque applied to a rotating body, then the angular momentum does not change and is therefore conserved.

This also applies to each component: no ext. torque in ${\hat {z}}$ means that $z$ component of angular momentum remains the same.

$L_{i,q}=L_{f,q}\,\!$

• Gravity/weight performs no work and thus angular momentum is conserved.

Example

A small disk attached to a big disk $d$ meters away from the big disk's axis, both of which are spinning with a positive angular velocity:

$L=I_{s}\omega _{s}+(I_{s}+M_{s}d^{2})\omega _{b}$