Lecture 10b - Rotation

Introduction

In linear motion, acceleration $a$ is caused by force $F$ . In rotation, angular acceleration $\alpha$ is caused by torque $\tau$ . The two concepts are best compared side by side:

$\begin{eqnarray} F &=& ma \\ \tau &=& I \alpha \end{eqnarray}$

Torque

The torque $\tau$ is defined by:

$\tau = rF\sin\theta$

where

$r$ is distance from the axis of rotation to the point of application of the force, and

$\theta$ is the angle between

$r$ and

$F$ .

The SI unit for torque is $Nm$ .

Sign convention:
+: counterclockwise torque
-: clockwise torque

Simulation - Torque (click to show)

Drag the seesaw to change the angle.
The force can be adjusted by dragging its tip, and it can be moved by dragging its tail.

$\tau = r F \sin \theta = 20.0\times100.0\times \sin60.0^\circ = 1732.1Nm$

$\tau = r_\perp F = 17.3\times100.0=1732.1Nm$

Example - torque vs force conceptual problem

Explain why the seesaw is not balanced in the first picture, but is perfectly balanced in the second one. In other words, the value of the forces (

$F_1=m_1g < F_2=m_2g$ ) did not change, why is the seesaw balanced in one but not the other?.

Solution

Rotation of the seesaw is controled by the torque, not the force. While $F_1$ and $F_2$ did not change in both cases, their points of application are not the same.

In the first case, $m_2>m_1$ implies $|\tau_2|>|\tau_1|$ . The imblance of the torque gives rise to a net torque that causes rotation in the seesaw.

In the second case, the larger mass is applied closer to the pivot (asix of rotation), therefore the distance $r_2$ is reduced. This reduces the torque $\tau_2$ for the larger mass. If the magnitude of the torques are the same ( $|\tau_1|=|\tau_2|$ ) then the seesaw could be balanced.

Example - torque on seesaw 1

Find the net torque on the seesaw about the pivot.

Solution

Since the forces are all vertical, and the distances are horizontal, the angle

$\theta=90^\circ$ . Since

$\sin 90^\circ = 1$ , the torque simply reduces to

$\tau = rF$ .

$\begin{eqnarray} \tau_1 &=& +(2m)(10N)= +20Nm \\ \tau_2 &=& -(1.5m)(20N)= -30Nm \\ \tau_{\text{total}} &=& \tau_1 + \tau_2 \\ &=& -10Nm \end{eqnarray}$ Note that

$\tau_2$ is given a negative sign because

$F_2$ is pushing to cause a clockwise rotation. The net negative torque above means the seesaw will start rotating clockwise.

Example - torque on seesaw 2

Find the net torque on the seesaw after it is tilted

$30^\circ$ clockwise about the pivot.

Solution

The angles are no longer

$90^\circ$ after the seesaw tilted

$30^\circ$ . From the diagram, one could see that

$\theta_1$ grows by

$30^\circ$ while

$\theta_2$ shrinks by

$30^\circ$ :

$\begin{eqnarray} \theta_1 &=& 90^\circ + 30^\circ = 120^\circ \\ \theta_2 &=& 90^\circ - 30^\circ = 60^\circ \end{eqnarray}$

$\begin{eqnarray} \tau_1 &=& +(2m)(10N)\sin 120^\circ = +17.32Nm \\ \tau_2 &=& -(1.5m)(20N)\sin 60^\circ = -25.98Nm \\ \tau_{\text{total}} &=& \tau_1 + \tau_2 \\ &=& -8.66Nm \end{eqnarray}$

Note that $\tau_2$ is given a negative sign because $F_2$ is pushing to cause a clockwise rotation. The net negative torque above means the seesaw will angularly acceleration in the clockwise direction.

Try It Yourself (click to show)

Exercise - torque on a bar 1

Find the distances

$r$ and the angles

$\theta$ . If the value is not defined (the angle is not defined when

$r=0$ ), type undefined in the textbox.

$r_1=$ $m$
$r_2=$ $m$
$r_3=$ $m$
$\theta_1=$ $^\circ$
$\theta_2=$ $^\circ$
$\theta_3=$ $^\circ$

Solution

The distances between each force and the axis of rotation are shown in the figure. $r_2=0m$ because the force is acting directly on the axis of rotation. $\theta_2$ cannot be defined because the vector $\vec{r_2}$ vanishes. Also note that technically $\theta_1 = 150^\circ$ (not $30^\circ$ ). One way to see this is to ask how much you have to rotate $\vec{r_1}$ before it points in the same direction as $\vec{F_1}$ .

In this simple example, if you had mistakenly used $\theta_1 = 30^\circ$ , you will actually still get the right answer for the torque (in the next exercise) because $\sin 150^\circ$ and $\sin 30^\circ$ happen to be the same (due to the trig identity $\sin\theta = \sin (180^\circ - \theta)$ ). However, it is better to be careful because you may not always be so lucky.

Feedback will appear here.

Exercise - torque on a bar 2

Continue from the previous problem. Find the torque due to the three forces about the pivot. Use the sign convention where counterclockwise = positive; clockwise = negative.

$\tau_1=$
$\tau_2=$
$\tau_3=$
$\tau_{\text{total}}=$

Select unit:

$m$

$Nm$

$N$

$^\circ$

First figure out the distances between the axis of rotation and the forces.

Solution

$\begin{eqnarray} \tau_1 &=& -(3m)(10N) \sin 150^\circ = -15Nm \\ \tau_2 &=& 0Nm \\ \tau_3 &=& +(2m)(20N)\sin 90^\circ = +40Nm \\ \tau_{\text{total}} &=& -15+0+40 = +25Nm \end{eqnarray}$

Feedback will appear here.

Exercise - torque on a bar 3

Find the distances

$r$ and the angles

$\theta$ . If the value is not defined (the angle is not defined when

$r=0$ ), type undefined in the textbox.

$r_1=$ $m$
$r_2=$ $m$
$r_3=$ $m$
$\theta_1=$ $^\circ$
$\theta_2=$ $^\circ$
$\theta_3=$ $^\circ$

Solution

The distances between each force and the axis of rotation are shown in the figure. Pay attention to the direction of the vectors, point from the axis of rotation to where the forces are acting. $r_3=0m$ because the force is acting directly on the axis of rotation.

Note also that $r_1=5m$ because you must measure the distance starting from the axis of rotation.

Feedback will appear here.

Exercise - torque on a bar 4

Continue from the previous problem. Find the torque due to the three forces about the pivot. Use the sign convention where counterclockwise = positive; clockwise = negative.

$\tau_1=$
$\tau_2=$
$\tau_3=$
$\tau_{\text{total}}=$

Select unit:

$m$

$Nm$

$N$

$^\circ$

First figure out the distances between the axis of rotation and the forces.

Solution

$\begin{eqnarray} \tau_1 &=& -(5m)(20N) \sin 90^\circ = -100Nm \\ \tau_2 &=& -(2m)(10N)\sin 30^\circ = -10Nm \\ \tau_3 &=& 0Nm \\ \tau_{\text{total}} &=& -100-10+0 = -110Nm \end{eqnarray}$

Feedback will appear here.

Example - balancing a seesaw

Find

$r_2$ that would keep the seesaw balanced.

Solution

The total torque must be zero for the seesaw to remain balanced. This means the counterclockwise torque (

$\tau_1$ ) and clockwise torque (

$\tau_2$ ) must have the same magnitude:

$\begin{eqnarray} |\tau_1| &=& |\tau_2| \\ r_1 F_1 \sin 90^\circ &=& r_2 F_2 \sin 90^\circ \\ r_2 &=& \frac{F_1}{F_2} r_1 \\ &=& \frac{10}{20} (2) = 1m \end{eqnarray}$

Alternative Forms of Torque

Torque can be calculated in many ways besides the original definition given above:

$\begin{eqnarray} \tau &=& rF \sin\theta \\ \tau &=& r_\perp F\\ \tau &=& r F_\perp \end{eqnarray}$

The variables $F_\perp$ and $r_\perp$ are defined in the diagram below. $F_\perp$ is the component of the force perpendicular to $\vec{r}$ . $r_\perp$ is the length of the perpendicular line from the axis of rotation to the line defined by $\vec{F}$ .

Example - torque on a tilted beam

Find the torque on a titled beam about the axis of rotation.

Solution

We will use two different methods to solve this problem.

Method 1 ( $\tau = r_\perp F$ )

$\begin{eqnarray} r_\perp &=& 2m \\ \Rightarrow \tau &=& -r_\perp F \\ &=& -(2m)(10N) = -20Nm \end{eqnarray}$ The negative sign is inserted to indicate the clockwise direction of the torque.

Method 2 ( $\tau = r F \sin (angle)$ )

We can first find

$r$ :

$r = \sqrt{1^2 + 2^2} = \sqrt{5} = 2.24m$ The angle

$\theta$ is found as follows:

$\begin{eqnarray} \tan \theta &=& \frac{1}{2} \\ \theta &=& \tan^{-1} 0.5 = 26.57^\circ \end{eqnarray}$ At this point, however, it is very important to recognize that

$\theta$ is not the angle between

$r$ and

$F$ . The angle we need is the angle

$\alpha$ :

$\alpha = 90^\circ + \theta = 116.57^\circ$ Only now we have everything we need:

$\begin{eqnarray} \tau &=& - r F \sin \alpha \\ &=& -(2.24m)(10N)\sin(116.57^\circ) = -20Nm \end{eqnarray}$

Both methods give the same answer, but clearly in this case method 1 is much faster. One has to be flexible in choosing which method to use.

Newton's Second Law

The rotational version of Newton's Second Law is given by:

$\begin{eqnarray} \tau &=& I \alpha \end{eqnarray}$

Moment of inertial $I$ and angular acceleration $\alpha$ were studied in the last lecture.

Try It Yourself (click to show)

Exercise - angular acceleration of a seesaw

A seesaw with negligible mass has two objects (

$m_1 = 10kg, m_2=20kg$ ) placed on it. Find the moment of inertia, the torque, and the angular acceleration under the weight of the two objects about the pivot.

$I=$ $kgm^2$
$\tau=$ $Nm$
$\alpha=$ $rad/s^2$

Solution

Note that the forces are simply

$F_1 = m_1 g$ and

$F_2 = m_2 g$ .

$\begin{eqnarray} I &=& m_1 r_1^2 + m_2 r_2^2 \\ &=& (10)(2^2) + (20)(1.5^2) = 85kgm^2\\ \tau &=& r_1 F_1 \sin 90^\circ - r_2 F_2 \sin 90^\circ \\ &=& (2)(10)(9.8) - (1.5) (20) (9.8) = -98Nm \\ \alpha &=& \frac{\tau}{I} \\ &=& \frac{-98}{85} = -1.15rad/s^2 \end{eqnarray}$ The negative sign indicates the seesaw will start to rotate clockwise.

Feedback will appear here.

Exercise - angular acceleration of a pulley

A pulley (

$m = 10kg$ ) of radius

$0.2m$ is pulled by a constant force

$F=8N$ and rotates about the central axis. The moment of inertia of a disk is

$I_{\text{disk}} = \frac{1}{2} m r^2$ . Find the moment of inertia, the torque, and the angular acceleration about the center.

$I=$ $kgm^2$
$\tau=$ $Nm$
$\alpha=$ $rad/s^2$

Solution

$\begin{eqnarray} I &=& \frac{1}{2} m r_2^2 \\ &=& \frac{1}{2}(10)(0.2^2) = 0.2kgm^2\\ \tau &=& -r F \sin 90^\circ = -(0.2)(8) = -1.6Nm \\ \alpha &=& \frac{\tau}{I} \\ &=& \frac{-1.6}{0.2} = -8rad/s^2 \end{eqnarray}$ The negative sign indicates the pulley will start to rotate clockwise.

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Angular Momentum

The definition of angular momentum $L$ is:

$L = I \omega$

The definition is analogous to that of linear momentum $p=mv$ . The SI unit of angular momenum is $kgm^2/s$ (which is identical to $Js$ ).

Simulation - Conservation of Angular Momentum (click to hide)

Both objects have mass

$m=1kg$ and are separated from the center by distance

$r$ . Adjust

$r$ to see how the angular velocity changes, while the total angular momentu stays the same (conservation of angular momentum), obeying

$L = I \omega$ .
Note

$I = 2 m r^2$ , where the factor of 2 comes from the fact that there are 2 masses on the bar, with each mass contributing

$I_{one\ mass} = m r^2$ .

Calculations:
Moment of inertial =

$I = 2mr^2 = 2 ( 1) (1.0)^2 = 2.0kgm^2$
Angular velocity =

$\omega = \frac{L}{I} = \frac{6.0}{2.0} = 3.0rad/s$

Try It Yourself (click to show)

Exercise - compare momentum of inertia

Which objects in the picture has the larger moment of inertia?

Use

$I = m_1 r_1^2 + m_2 r_2^2$ to see how

$I$ depends on

$r$ . If you increase

$r$ how would

$I$ changes?

Solution

$I = m_1 r_1^2 + m_2 r_2^2$ means larger $r$ gives larger moment of inertia. In general the further away an object is to the axis of rotation, the larger contribution it will make to $I$ .

An application below: when a skater has her arms sketched out, her momentum of inertial $I$ is greater than the value $I'$ when her arms are folded.

$I$

$I'$

Feedback will appear here.

Exercise - a spinning skater

Embed from Getty Images

When a skater has her arms sketched out, her momentum of inertial

$I$ is greater than the value

$I'$ when her arms are folded. If

$I = 50kgm^2$ and

$I'=40kgm^2$ , and that the skater was spinning at

$\omega = 2rad/s$ in the beginning, what is the final angular velocity when she has her arms folded?

Select unit:

$Nm$

$kgm^2$

$rad/s$

$kgm^2/s$

Use

$I\omega = I' \omega'$ .

Solution

$\begin{eqnarray} L &=& L' \\ \Rightarrow I \omega &=& I' \omega' \\ \Rightarrow \omega' &=& \frac{I}{I'} \omega \\ &=& \frac{50}{40} (2) = 2.5rad/s \end{eqnarray}$

Feedback will appear here.

Video - Skater and Angular Momentum (click to show)

Example - rotational collision 1

Two rotating disks combined together and rotate at the same final angular speed

$\omega_f$ . Given

$I_{\text{disk}} = \frac{1}{2} m r^2$ , find

$\omega_f$ and the change in kinetic energy

$\Delta KE$ .

Symbol	Value
$m_A$	$2kg$
$m_B$	$4kg$

Symbol	Value
$r_A$	$0.2kg$
$r_B$	$0.1kg$

Symbol	Value
$\omega_A$	$50rad/s$
$\omega_B$	$200rad/s$

Solution

Using

$I_{\text{disk}} = \frac{1}{2} m r^2$ , we calculate:

$I_A = 0.04kgm^2, I_B = 0.02kgm^2$ Similar to conservation of linear momentum

$m_A u_A + m_B u_B = m_A v_A + m_B v_B$ , we have:

$\begin{eqnarray} I_A \omega_A + I_B \omega_B &=& I_A \omega_f + I_B \omega_f \\ \Rightarrow \omega_f &=& \frac{I_A \omega_A + I_B \omega_B}{I_A + I_B} \\ &=& \frac{(0.04)(50) + (0.02)(200)}{0.04+0.02} \\ &=& 100rad/s \end{eqnarray}$ The kinetic energy can be found using

$KE = \frac{1}{2} I \omega^2$ :

$\begin{eqnarray} KE_i &=& \frac{1}{2} I_A \omega_A^2 + \frac{1}{2} I_B \omega_B^2 \\ &=& 450J \\ KE_f &=& \frac{1}{2} I_A \omega_f^2 + \frac{1}{2} I_B \omega_f^2 \\ &=& \frac{1}{2} (I_A + I_B) \omega_f^2 \\ &=& 300J \\ \Delta KE &=& KE_f - KE_i \\ &=& -150J \end{eqnarray}$

Conservation of angular momentum leads to some interesting and sometimes counter-intuitive effects. If you are interested you could watch the videos below to find out more.

Video - Bike Wheel Gyroscope (click to show)

Video - Angular Momentum Demos (click to show)

Notations

Name	Symbol	Unit	Meaning
Torque	$\tau$	$Nm$	the cause of angular acceleration
Angular momentum	$L$	$kgm^2/s$	a conserved quantity in rotation

PHYS 170

Introduction

Torque

Simulation - Torque (click to show)

Example - torque vs force conceptual problem

Solution

Example - torque on seesaw 1

Solution

Example - torque on seesaw 2

Solution

Try It Yourself (click to show)

Exercise - torque on a bar 1

Solution

Exercise - torque on a bar 2

Solution

Exercise - torque on a bar 3

Solution

Exercise - torque on a bar 4

Solution

Example - balancing a seesaw

Solution

Alternative Forms of Torque

Example - torque on a tilted beam

Solution

Method 1 (τ=r⊥F\tau = r_\perp F)

Method 2 (τ=rFsin(angle)\tau = r F \sin (angle))

Newton's Second Law

Try It Yourself (click to show)

Exercise - angular acceleration of a seesaw

Solution

Exercise - angular acceleration of a pulley

Solution

Angular Momentum

Simulation - Conservation of Angular Momentum (click to hide)

Try It Yourself (click to show)

Exercise - compare momentum of inertia

Solution

Exercise - a spinning skater

Solution

Video - Skater and Angular Momentum (click to show)

Example - rotational collision 1

Solution

Video - Bike Wheel Gyroscope (click to show)

Video - Angular Momentum Demos (click to show)

Notations

Method 1 ( $\tau = r_\perp F$ )

Method 2 ( $\tau = r F \sin (angle)$ )