## Rotational Dynamics

Rotational kinematics relates angular position, velocity, acceleration and time. Rotational dynamics deals with rotational energy, angular momentum, and torque.

## Angular displacement, velocity, and acceleration

The relation between the arc length l , the radius R and the angle θ is

Other

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$$The formula shown is:

\[ l = \theta^r R \]

where \( l \), \( \theta \), \( r \), and \( R \) are variables or constants. If you need an explanation or context for this formula, please provide more details.$$

where is expressed in the units radians.

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Unless otherwise stated, θ is usually measured in the counterclockwise direction from the positive x -axis.  Angles can also be measured in degrees or revolutions.

$$The formula shown is the conversion between revolutions and radians/degrees:

1 revolution (rev) = 2π radians = 360 degrees (°).

This means one complete revolution around a circle corresponds to an angle of 2π radians or 360 degrees.$$

An object moves from point A to point B and cover the angular displacement Δθ =θ B -θ A in the time interval Δt.

The average angular velocity is

$$The formula shown in the image is the average angular velocity, denoted as \(\bar{\omega}\). It is given by:

\[
\bar{\omega} = \frac{\theta_B - \theta_A}{t_2 - t_1} = \frac{\Delta \theta}{\Delta t}
\]

where:
- \(\theta_B\) and \(\theta_A\) are the angular positions at times \(t_2\) and \(t_1\), respectively,
- \(\Delta \theta = \theta_B - \theta_A\) is the change in angular position,
- \(\Delta t = t_2 - t_1\) is the change in time.

This formula calculates the average rate of change of angular displacement over a time interval.$$

The image shows a circle with center O and radius R. Points A and B lie on the circumference. The arc length between points A and B is labeled as Δl. The central angle subtended by the arc AB is labeled Δθ. Lines OA and OB represent the radii of the circle. The diagram illustrates the relationship between the arc length, radius, and central angle in a circle.

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The instantaneous angular velocity ω is the average angular velocity in the limit of very short time interval.  Depending on units for θ and t , the units for ω can be rad/s, deg/s, rev/s, rev/min (rpm), etc.

The a verage angular acceleration is

$$The formula shown in the image is:

\[
\bar{\alpha} = \frac{\omega_2 - \omega_1}{t_2 - t_1} = \frac{\Delta \omega}{\Delta t}
\]

This represents the average angular acceleration, where:
- \(\bar{\alpha}\) is the average angular acceleration,
- \(\omega_1\) and \(\omega_2\) are the initial and final angular velocities,
- \(t_1\) and \(t_2\) are the initial and final times,
- \(\Delta \omega = \omega_2 - \omega_1\) is the change in angular velocity,
- \(\Delta t = t_2 - t_1\) is the change in time.$$

The instantaneous angular acceleration is α and is the average angular acceleration in the limit of very short time interval.  Depending on units for θ and t, the units for α can be rad/s 2 , deg/s 2 , rev/s 2 .

The relationships above are mathematically similar to those for motion in 1-D.  Thus, we can get the equations for constant angular acceleration from those for constant linear acceleration by appropriately changing the variables.

1-D motion with constant a Rot. motion with constant α

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v

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t

θ(

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) = θ

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) =

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)

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=

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ω

+

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(

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2

0

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+

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ω

0

)

v

v

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=

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It is important that the units in the above equations be compatible.  For example, if α is in rad/s 2 , then θ should be in rad/s, t in s, and you will get θ in rad.  Also, don't mix the left and right equations above.  For example, the equation θ( t )=θ o + v 0 t + 1 2 α t 2 makes no sense.  Also, don't confuse α and a , although they look similar.

## Example

A wheel increases its rotational velocity from 200 rpm to 300 rpm in 10 sec.

What is its angular acceleration in rad/s 2 ?

$$The formulas shown in the image are related to angular velocity and angular acceleration:

1. Initial angular velocity \(\omega_0\):
\[
\omega_0 = (200 \, \text{rev/min}) \left( \frac{1 \, \text{min}}{60 \, \text{s}} \right) \left( 2\pi \, \text{rad/rev} \right) = 20.9 \, \text{rad/s}
\]

2. Final angular velocity \(\omega\):
\[
\omega = (300 \, \text{rev/min}) \left( \frac{1 \, \text{min}}{60 \, \text{s}} \right) \left( 2\pi \, \text{rad/rev} \right) = 31.4 \, \text{rad/s}
\]

3. Angular acceleration \(\alpha\):
\[
\alpha = \frac{\Delta \omega}{\Delta t} = \frac{\omega - \omega_0}{\Delta t} = \frac{31.4 \, \text{rad/s} - 20.9 \, \text{rad/s}}{10 \, \text{s}} = 1.05 \, \text{rad/s}^2
\]

This shows the conversion from revolutions per minute to radians per second and the calculation of angular acceleration over a time interval of 10 seconds.$$

How many turns did the wheel make?

$$The formula and calculations shown in the image are related to rotational motion. Here's a breakdown:

1. The angular displacement \(\theta\) is given by the equation:
\[
\theta = \omega_0 t + \frac{1}{2} \alpha t^2
\]
where:
- \(\omega_0 = 20.9 \, \text{rad/s}\) (initial angular velocity),
- \(\alpha = 1.05 \, \text{rad/s}^2\) (angular acceleration),
- \(t = 10 \, \text{s}\) (time).

2. Substituting the values:
\[
\theta = (20.9 \, \text{rad/s})(10 \, \text{s}) + \frac{1}{2} (1.05 \, \text{rad/s}^2)(10 \, \text{s})^2 = 261.5 \, \text{rad}
\]

3. The number of turns is calculated by dividing the angular displacement by \(2\pi\):
\[
\text{number of turns} = \frac{\theta}{2\pi} = 41.6
\]

4. Alternatively, the angular displacement can be calculated using the average angular velocity:
\[
\theta = \bar{\omega} t = \frac{\omega_0 + \omega}{2} t
\]

This is a standard approach to solving problems involving angular displacement with constant angular acceleration. If you need further explanation or help with a specific part, please let me know!$$

a t

2

x

v

a

θ

ω

α

+

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## Relationships between angular and linear motion

Since l =θ r R it follows that

$$The formula shown in the image is:

\[
v_T = \frac{\Delta l}{\Delta t} = \frac{\Delta \theta}{\Delta t} R, \quad \text{or} \quad \mathbf{v_T = \omega R}
\]

This represents the tangential velocity \(v_T\) of a point on a rotating object, where:

- \(\Delta l\) is the change in arc length,
- \(\Delta t\) is the change in time,
- \(\Delta \theta\) is the change in angular displacement (in radians),
- \(R\) is the radius or distance from the axis of rotation,
- \(\omega = \frac{\Delta \theta}{\Delta t}\) is the angular velocity.

So, tangential velocity \(v_T\) is the product of angular velocity \(\omega\) and radius \(R\).$$

$$The formula shown is:

\[
a_T = \frac{\Delta v_T}{\Delta t} = \frac{\Delta \omega}{\Delta t} R, \quad \text{or} \quad a_T = \alpha R
\]

This represents the tangential acceleration \(a_T\) in rotational motion, where:

- \(a_T\) is the tangential acceleration,
- \(\Delta v_T / \Delta t\) is the change in tangential velocity over time,
- \(\Delta \omega / \Delta t\) is the change in angular velocity over time (angular acceleration \(\alpha\)),
- \(R\) is the radius or distance from the axis of rotation,
- \(\alpha\) is the angular acceleration.

The equation shows that tangential acceleration is the product of angular acceleration and the radius.$$

In the above v T is the tangential speed of a point going around a circle and aT is the component of acceleration tangent to the circle.

## Example

A merry-go-round rotates at a constant angular speed.  It takes 20 sec to make a complete revolution.  What is the speed of a rider who is 4 m from the center?

$$The formulas and calculations shown are:

1. Angular velocity (ω):
\[
\omega = \frac{\Delta \theta}{\Delta t} = \frac{2\pi \, \text{rad}}{20 \, \text{s}} = 0.314 \, \text{rad/s}
\]

2. Tangential velocity (v_t):
\[
v_t = r \omega = (4 \, m)(0.314 \, \text{rad/s}) = 1.26 \, m/s
\]

This means an object rotating with an angular displacement of \(2\pi\) radians in 20 seconds has an angular velocity of 0.314 radians per second, and if the radius is 4 meters, the tangential velocity is 1.26 meters per second.$$

The speed of a rider increases as he moves further from the center.

## Combined tangential and centripetal acceleration

We have previously shown that an object moving in a curved path can have both a tangential acceleration, as discussed above, and a radial (or centripetal) acceleration.

$$The image shows the mathematical expression "2 squared," which is written as \(2^2\). This means 2 raised to the power of 2, or 2 multiplied by itself:

\[
2^2 = 2 \times 2 = 4
\]

So, \(2^2 = 4\).$$

$$The formulas shown in the image are:

1. Tangential acceleration (\(a_t\)):
\[
a_t = \frac{\Delta v}{\Delta t}
\]
where \(\Delta v\) is the change in velocity and \(\Delta t\) is the change in time.

2. Radial (or centripetal) acceleration (\(a_r\)):
\[
a_r = \frac{v^2}{R}
\]
where \(v\) is the velocity and \(R\) is the radius of the circular path.$$

The total acceleration is the vector sum:

$$The formula shown is the vector sum of acceleration components:

\[
\vec{a} = \vec{a}_c + \vec{a}_t
\]

where:
- \(\vec{a}\) is the total acceleration vector,
- \(\vec{a}_c\) is the centripetal (or radial) acceleration vector,
- \(\vec{a}_t\) is the tangential acceleration vector.

This equation is commonly used in the context of circular motion, where the total acceleration is the vector sum of the centripetal acceleration (pointing towards the center of the circular path) and the tangential acceleration (tangent to the path, responsible for changing the speed).$$

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Since these two components of the acceleration are mutually perpendicular, then the magnitude of the total acceleration is

$$The formula shown is:

\[
a = \sqrt{a_t^2 + a_c^2} = \sqrt{R^2 \alpha^2 + R^2 \omega^4} = R \sqrt{\alpha^2 + \omega^4}
\]

This represents the total acceleration \(a\) as the vector sum of tangential acceleration \(a_t\) and centripetal acceleration \(a_c\), where:

- \(a_t = R \alpha\) (tangential acceleration, with \(\alpha\) being angular acceleration),
- \(a_c = R \omega^2\) (centripetal acceleration, with \(\omega\) being angular velocity),
- \(R\) is the radius.

The total acceleration magnitude is the square root of the sum of the squares of these two components.$$

## Example

The merry-go-round in the above example slows uniformly and comes to rest in a time of 6 seconds.

What the total acceleration of a rider when his tangential speed is 1 m/s?

$$The formulas and calculations shown in the image are related to rotational motion and acceleration. Here's a breakdown of each step:

1. Angular acceleration (α):
\[
\alpha = \frac{\Delta \omega}{\Delta t} = \frac{0 - 0.314 \, \text{rad}}{60 \, \text{s}} = -0.0523 \, \text{rad/s}^2
\]
This calculates the angular acceleration given the change in angular velocity (Δω) over time (Δt).

2. Tangential acceleration (a_t):
\[
a_t = r \alpha = (4 \, \text{m})(-0.0523 \, \text{rad/s}^2) = -0.209 \, \text{m/s}^2
\]
This calculates the tangential acceleration using the radius (r) and angular acceleration (α).

3. Centripetal acceleration (a_c):
\[
a_c = \frac{v^2}{R} = \frac{(1 \, \text{m/s})^2}{4 \, \text{m}} = 0.25 \, \text{m/s}^2
\]
This calculates the centripetal acceleration using the velocity (v) and radius (R).

4. Total acceleration (a):
\[
a = \sqrt{(0.209)^2 + (0.25)^2} = 0.326 \, \text{m/s}^2
\]
This calculates the total acceleration by combining tangential and centripetal accelerations using the Pythagorean theorem.

If you need further explanation or help with these calculations, feel free to ask!$$

The angle that the acceleration makes with the radial direction is

$$The formula shown is:

\[
\theta = \tan^{-1} \left( \frac{a_t}{a_c} \right) = \tan^{-1} \left( \frac{-0.209}{0.25} \right) = -40^\circ
\]

This represents the angle \(\theta\) calculated as the inverse tangent (arctan) of the ratio of \(a_t\) to \(a_c\), where \(a_t = -0.209\) and \(a_c = 0.25\), resulting in \(\theta = -40^\circ\).$$

The angle is in the 'backward' direction.

## Rotational Kinetic Energy

For a rigid object rotating about an axis, all the pieces that make up the object rotate in a circle with the same angular velocity ω .  Thus, the kinetic energy is

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is defined as the Moment of Inertia (or the rotational inertia) of the object.

## Example

Three masses, each 2 kg, are located at the corners of a triangle consisting of light rigid rods, as shown to the right.  The location (x, y) of each mass is shown.  What is the rotational inertia about the x-axis?

What matters here is the nearest distance of each mass from the axis of rotation.  So,

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If the triangle of masses rotates about the x-axis at 60 rpm, what is the rotational kinetic energy?

The angular velocity must be in rad/s, so

$$The formula you provided shows the calculation of angular velocity (ω) and rotational kinetic energy (K_rot):

1. Angular velocity:
\[
\omega = (60 \, rpm) \left(\frac{1 \, min}{60 \, s}\right) (2\pi \, rad/rev) = 6.28 \, rad/s
\]

2. Rotational kinetic energy:
\[
K_{rot} = \frac{1}{2} I \omega^2 = \frac{1}{2} (34 \, kg \cdot m^2) (6.28 \, rad/s)^2 = 670 \, J
\]

This shows the conversion from revolutions per minute to radians per second for angular velocity, and then the calculation of rotational kinetic energy using the moment of inertia \(I = 34 \, kg \cdot m^2\) and the angular velocity. The final kinetic energy is 670 Joules.$$

Suggested exercise:  Find the rotational inertia and rotational kinetic energy about the yaxis and about the z-axis (which is perpendicular to the page) for the same rotational velocity where:

## Moment of Inertia of a Continuous Mass Distribution

To calculate the moment of inertia of a continuous distribution of masses the use of calculus is required. Instead we look at the results for some simple objects.

## Moment of   Inertia of Various Objects

A round object that rolls also rotates about an axis through it center of mass.  Below are expressions for some such rotational inertia that have been determined by integrating over the volume of the object.

The image illustrates the moments of inertia (I) formulas for various solid objects about specified axes:

1. Solid cylinder or disc about symmetry axis: 
   \( I = \frac{1}{2} MR^2 \)

2. Hoop about symmetry axis: 
   \( I = MR^2 \)

3. Solid sphere about the symmetry axis:
   \( I = \frac{2}{5} MR^2 \)

4. Rod about center:
   \( I = \frac{1}{12} ML^2 \)

5. Solid cylinder about central diameter:
   \( I = \frac{1}{4} MR^2 + \frac{1}{12} ML^2 \)

6. Hoop about diameter:
   \( I = \frac{1}{2} MR^2 \)

7. Thin spherical shell about the symmetry axis:
   \( I = \frac{2}{3} MR^2 \)

8. Rod about end:
   \( I = \frac{1}{3} ML^2 \)

Each formula indicates the moment of inertia based on mass (M), radius (R), or length (L) in relation to the object's rotation axis shown graphically.

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Parallel axis theorem

For any other axis (in the picture to the right the zaxis) that is parallel with this axis through the center of mass CM , the moment of inertia is given by the parallel axis theorem ,

$$The formula shown is:

\[ I_z = I_{CM} + MD^2 \]

This is the parallel axis theorem in physics and engineering, which relates the moment of inertia \( I_z \) about an axis \( z \) to the moment of inertia \( I_{CM} \) about a parallel axis through the center of mass, where:

- \( I_z \) is the moment of inertia about the new axis,
- \( I_{CM} \) is the moment of inertia about the center of mass axis,
- \( M \) is the mass of the object,
- \( D \) is the perpendicular distance between the two axes.

If you need further explanation or application examples, feel free to ask!$$

where D is the distance between the two axes, M is the mass of the object and I CM the moment of inertia respect to the center of mass.

The image shows a three-dimensional coordinate system with axes labeled x, y, and z intersecting at point O. A flat irregular-shaped object is placed in the coordinate system. The object's center of mass (CM) is marked above the object along the y-axis. A blue vertical line labeled "Axis through CM" runs parallel to the z-axis through the center of mass (CM). The "Rotation axis" is labeled along the z-axis intersecting at point O, indicating it is different from the axis through the CM.

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## Rolling Objects

A rolling object has both translational and rotational kinetic energy:

$$The formula shown is the total kinetic energy \( KE \) of a rigid body, which is the sum of the translational kinetic energy \( KE_{trans} \) and the rotational kinetic energy \( KE_{rot} \):

\[
KE = KE_{trans} + KE_{rot} = \frac{1}{2} M v_{CM}^2 + \frac{1}{2} I_{CM} \omega^2
\]

Where:
- \( M \) is the mass of the body,
- \( v_{CM} \) is the velocity of the center of mass,
- \( I_{CM} \) is the moment of inertia about the center of mass,
- \( \omega \) is the angular velocity.$$

where v CM is the speed of the center of mass and I CM is the rotational inertia about the rotation axis through the center of mass.

The condition for an object to roll with without slipping is given by v CM = v t

## Example

A ball is released from rest at the top of an inclined plane of height 2 m.  What is its speed when it reaches the bottom, assuming it rolls without slipping?

$$The formulas in the image are:

\[
E_f = E_i
\]

\[
\frac{1}{2} M v^2 + \frac{1}{2} I \omega^2 = Mgh
\]

\[
\omega = \frac{v}{R}
\]

\[
\frac{1}{2} M v^2 + \frac{1}{2} I \frac{v^2}{R^2} = \frac{1}{2} \left( M + \frac{I}{R^2} \right) v^2 = Mgh
\]

\[
v = \sqrt{\frac{2 Mgh}{M + \frac{I}{R^2}}}
\]$$

For a solid ball of uniform density I = 2 5 MR 2 so

$$The formula shown is the calculation of velocity \( v \) derived from the energy conservation principle for a rolling object, likely a rolling sphere or cylinder, down an incline or from a height \( h \).

The formula is:

\[
v = \sqrt{\frac{2Mgh}{M + \frac{2}{5}M}} = \sqrt{\frac{10}{7}gh} = \sqrt{\frac{10}{7} (9.8)(2)} = 5.29 \, m/s
\]

Explanation:
- \( M \) is the mass of the object.
- \( g \) is the acceleration due to gravity (9.8 m/s²).
- \( h \) is the height (2 meters in this case).
- The denominator \( M + \frac{2}{5}M = \frac{7}{5}M \) accounts for the rotational inertia of a solid sphere (moment of inertia \( I = \frac{2}{5}MR^2 \)).

The final velocity \( v \) is calculated to be approximately 5.29 m/s.$$

## Torque

Torque can cause an object to rotate.  The torque depends on the force and on the point at which the force is applied relative to the axis of rotation.  Specifically, torque is the product of the force and the nearest distance between the line of action of the force and the axis or rotation ,

τ = Fd (units are N∙m)

If r is the distance from the point of application of the force to the pivot, then d = r sinφ (called the 'lever arm').  We can also write

τ = F ┴ r = F sin φ r F

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where F  = F sin φ is the component of F that is perpendicular to r .

In vector notation, the torque if the cross product (vector product) of r and F :

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The direction of τ is perpendicular to the plane formed by r and F .  It would be along the axis of rotation due to the torque.

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The torque due to the force shown in the diagram would produce a counter-clockwise rotation about the pivot and is assumed to be positive . The direction of the torque would be out of the plane.  A torque that would produce a clockwise rotation is negative and would be directed into the plane in the diagram.

## Example

A disk pivoted about an axis through its center is subjected to two forces, as shown to the right.  Given: F1 = 10 N, F2

= 5 N, d1 = 8 cm, d2 = 5 cm.  What is the net torque? τ = F 1 d1 - F 2 d2 = (10 N)(0.08 m) - (5 N)(0.05 m)

= 0.8 N∙m - 0.25 N∙m = 0.55 N∙m

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Torque and Angular Acceleration

For a single point mass going in a circle subject to a tangential force,

$$The formula shown is:

\[ F_t = ma_t = mr \alpha \]

This represents the tangential force \( F_t \) acting on a rotating object.

- \( m \) is the mass of the object.
- \( a_t \) is the tangential acceleration.
- \( r \) is the radius or distance from the axis of rotation.
- \( \alpha \) is the angular acceleration.

The equation states that the tangential force is equal to the mass times the tangential acceleration, which can also be expressed as the mass times the radius times the angular acceleration. This is a fundamental relation in rotational dynamics.$$

The point also has a centripetal component of force and acceleration (otherwise it won't rotates) but we are interested in the object having a non-zero angular acceleration.

The torque is then

$$The formula shown is:

\[ F_t r = m r^2 \alpha \quad , \quad \tau = m r^2 \alpha \]

This represents the relationship between torque (\(\tau\)), force (\(F_t\)), radius (\(r\)), mass (\(m\)), and angular acceleration (\(\alpha\)).

- \(F_t r\) is the torque produced by a tangential force \(F_t\) applied at a radius \(r\).
- \(m r^2 \alpha\) is the torque expressed in terms of moment of inertia (\(I = m r^2\)) and angular acceleration \(\alpha\).
- \(\tau\) is the torque.

So, the equation states that the torque \(\tau\) equals the tangential force times the radius, which also equals the moment of inertia times the angular acceleration.$$

For a rigid body consisting of many masses, each mass goes in a circle and has the same angular acceleration α .  Thus, the net torque is

$$The formula shown is:

\[
\tau = \sum_i \tau_i = \left( \sum_i m_i r_i^2 \right) \alpha,
\]

where:
- \(\tau\) is the total torque,
- \(\tau_i\) is the torque on the \(i\)-th particle,
- \(m_i\) is the mass of the \(i\)-th particle,
- \(r_i\) is the distance of the \(i\)-th particle from the axis of rotation,
- \(\alpha\) is the angular acceleration.

This formula represents the relationship between the net torque and angular acceleration for a system of particles, where the term \(\sum_i m_i r_i^2\) is the moment of inertia \(I\) of the system. Thus, it can be rewritten as:

\[
\tau = I \alpha.
\]$$

or,

$$The formula shown in the image is:

\[
\vec{\tau} = I \vec{\alpha}
\]

This represents the rotational form of Newton's second law, where:

- \(\vec{\tau}\) is the torque,
- \(I\) is the moment of inertia,
- \(\vec{\alpha}\) is the angular acceleration.$$

## Example

For the disk in the example above that is subjected to two forces, assume that the radius is R = 15 cm and the mass is 5 kg.  What is the angular acceleration of the disk?

$$The formulas shown are related to rotational motion:

1. Moment of inertia \( I \) of a point mass:
\[
I = \frac{1}{2} MR^2 = \frac{1}{2} (5\, \text{kg}) (0.15\, \text{m})^2 = 0.0563\, \text{kg} \cdot \text{m}^2
\]

2. Angular acceleration \( \alpha \):
\[
\alpha = \frac{\tau}{I} = \frac{0.55\, \text{N} \cdot \text{m}}{0.0563\, \text{kg} \cdot \text{m}^2} = 9.8\, \text{rad/s}^2
\]

Where:
- \( M = 5\, \text{kg} \) is the mass,
- \( R = 0.15\, \text{m} \) is the radius,
- \( \tau = 0.55\, \text{N} \cdot \text{m} \) is the torque applied,
- \( I \) is the moment of inertia,
- \( \alpha \) is the angular acceleration.

The calculations show the moment of inertia for a rotating object and the resulting angular acceleration when a torque is applied.$$

## Example

An Atwood machine consists of two masses hanging by a string over a pulley, as shown below.

The image consists of three parts illustrating a pulley system with two masses, m1 and m2, connected by a string passing over a pulley.

1. Left diagram: Shows a pulley with moment of inertia I and radius R. Mass m1 hangs on the left side, and mass m2 hangs on the right side of the pulley.

2. Middle diagram: Depicts the forces acting on each mass.
   - Mass 1 (m1) experiences downward gravitational force m1g and upward tension T1. It accelerates downward with acceleration a.
   - Mass 2 (m2) experiences downward gravitational force m2g and upward tension T2. It accelerates upward with acceleration a.

3. Right diagram: Shows the forces on the pulley through the tensions T1 and T2 acting at the rim but in opposite directions.

The image overall represents a physical setup commonly analyzed in mechanics to study rotational dynamics and linear accelerations involving a pulley and hanging masses.

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Determine an expression for the acceleration of the masses, assuming that friction can be neglected.  The pulley has inertia I and radius R .  We apply the 2 nd law to mass 1, mass 2, and the pulley.

$$The formulas given are:

For mass 1:
\[ m_1 g - T_1 = m_1 a \]

For mass 2:
\[ T_2 - m_2 g = m_2 a \]

For the pulley:
\[
\sum \tau = T_1 R - T_2 R = I \alpha = I \frac{a}{R}
\]

or equivalently:
\[
T_1 - T_2 = \frac{I a}{R^2}
\]

These equations describe the forces and torques acting on a system with two masses connected by a string over a pulley, where \(T_1\) and \(T_2\) are the tensions on either side of the pulley, \(m_1\) and \(m_2\) are the masses, \(g\) is the acceleration due to gravity, \(a\) is the linear acceleration of the masses, \(I\) is the moment of inertia of the pulley, \(R\) is the radius of the pulley, and \(\alpha\) is the angular acceleration of the pulley.$$

By adding the first two equations we get

$$The formula shown is:

\[
-T_1 + T_2 + (m_1 - m_2) g = (m_1 + m_2) a
\]

where:
- \(T_1\) and \(T_2\) are tensions,
- \(m_1\) and \(m_2\) are masses,
- \(g\) is the acceleration due to gravity,
- \(a\) is the acceleration of the system.

If you need help with this formula, such as solving for a variable or understanding its context, please let me know!$$

Using the third equation to eliminate T 1 - T 2 , we get

$$The formula shown is:

\[
- I a / R^2 + (m_1 - m_2) g = (m_1 + m_2) a
\]

where:
- \(I\) is the moment of inertia,
- \(a\) is the linear acceleration,
- \(R\) is the radius,
- \(m_1\) and \(m_2\) are masses,
- \(g\) is the acceleration due to gravity.

If you need help solving or rearranging this equation, please let me know!$$

Solving for a ,

$$The formula shown is:

\[
a = \frac{(m_1 - m_2) g}{m_1 + m_2 + \frac{I}{R^2}}
\]

This formula is commonly used in physics to calculate the acceleration \(a\) of a system involving two masses \(m_1\) and \(m_2\), gravitational acceleration \(g\), moment of inertia \(I\), and radius \(R\). It often appears in problems related to rotational dynamics, such as an Atwood machine with a pulley having moment of inertia.$$

## Work, Power, and Energy

A force F (red arrow) in applied to a disk which accelerates and starts to rotate.

Only the component of the force perpendicular to r F┴ = F sinφ is responsible of the rotation and therefore does work. φ is the angle between r and F .

The work done over Δ l is

$$The formula you provided is:

\[
W_R = F_{\perp} \Delta s = F \sin \varphi \, \Delta l
\]

This represents the work done \( W_R \) by a force component perpendicular to a displacement. Here:

- \( F_{\perp} \) is the component of the force perpendicular to the displacement,
- \( \Delta s \) is the displacement,
- \( F \) is the magnitude of the force,
- \( \varphi \) is the angle between the force and the displacement,
- \( \Delta l \) is the length of the displacement vector.

The work done is the product of the force component in the direction of displacement and the displacement itself. If you need further explanation or application of this formula, feel free to ask!$$

Since the τ = Fsinφ r and Δl = r Δθ then

$$The formula shown is:

\[ W_R = \tau \Delta \theta \]

where:
- \( W_R \) represents the work done by a rotational force,
- \( \tau \) (tau) is the torque applied,
- \( \Delta \theta \) is the angular displacement in radians.

This formula expresses the work done by a torque when an object rotates through an angle \( \Delta \theta \).$$

Engineering drawing

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A torque applied to a rigid object does work on the object if it rotates.

The work-energy theorem also applies to the work done by a torque.  Since

$$Given the equations:

\[
\tau = I \alpha
\]

and

\[
\omega^2 = \omega_0^2 + 2 \alpha \Delta \theta,
\]

where:
- \(\tau\) is the torque,
- \(I\) is the moment of inertia,
- \(\alpha\) is the angular acceleration,
- \(\omega\) is the final angular velocity,
- \(\omega_0\) is the initial angular velocity,
- \(\Delta \theta\) is the angular displacement,

these are standard rotational motion equations analogous to linear motion equations.

If you want to find the angular acceleration \(\alpha\) or relate these quantities further, please specify what you need!$$

$$The formula shown is related to rotational work and energy in physics:

\[
W_R = \tau \Delta \theta = I \alpha \Delta \theta = \frac{1}{2} I \omega^2 - \frac{1}{2} I \omega_0^2
\]

Where:
- \( W_R \) is the rotational work done,
- \( \tau \) is the torque,
- \( \Delta \theta \) is the angular displacement,
- \( I \) is the moment of inertia,
- \( \alpha \) is the angular acceleration,
- \( \omega \) is the final angular velocity,
- \( \omega_0 \) is the initial angular velocity.

This equation expresses that the work done by the torque on a rotating object equals the change in its rotational kinetic energy.$$

$$The formula shown in the image is:

\[ W_R = \Delta KE_R \]

This represents that the work done by a resultant force \( W_R \) is equal to the change in kinetic energy \( \Delta KE_R \).$$

which is the rotational analog of the work-kinetic energy theorem.

## Power

The power is the rate at which the work is done in rotating the object and is given by

$$The formula shown is:

\[
\overline{P} = \frac{W_R}{\Delta t} = \frac{\tau \, d\theta}{\Delta t} = \tau \, \omega
\]

This represents the average power \(\overline{P}\) in rotational motion, where:
- \(W_R\) is the work done by the torque,
- \(\Delta t\) is the time interval,
- \(\tau\) is the torque,
- \(d\theta\) is the angular displacement,
- \(\omega\) is the angular velocity.

The formula shows that average power is the rate of work done by torque, which can also be expressed as the product of torque and angular velocity.$$

Where ¯ P indicates the average power.

or

## Angular Momentum

The angular momentum L of particle of mass m rotating with momentum p around an axis of rotation at distance r is defined as

$$The formula shown is the vector form of angular momentum in physics:

\[
\vec{L} = \vec{r} \times \vec{p}
\]

where:
- \(\vec{L}\) is the angular momentum vector,
- \(\vec{r}\) is the position vector (from the origin to the point where the particle is located),
- \(\vec{p}\) is the linear momentum vector,
- \(\times\) denotes the cross product.

This equation expresses that the angular momentum is the cross product of the position vector and the linear momentum vector.$$

The magnitude of L is

$$The formula \( L = mvr \sin \phi \) represents the angular momentum \(L\) of a particle moving in a plane, where:

- \(m\) is the mass of the particle,
- \(v\) is the velocity of the particle,
- \(r\) is the distance from the axis of rotation (or the point about which angular momentum is calculated),
- \(\phi\) is the angle between the velocity vector \(\vec{v}\) and the position vector \(\vec{r}\).

This formula calculates the magnitude of the angular momentum as the product of mass, speed, radius, and the sine of the angle between velocity and radius vectors.$$

L depends on the distance r between the line of motion of the particle and the axis about which L is defined.

## Angular Momentum and Angular Velocity

For a rigid body rotating about a fixed axis, L is the sum of the angular momentum of all particles making up the body with φ = 90°

$$The formula shown is:

\[
L = \sum L_i = \sum m_i v_i r_i = \sum m_i (\omega r_i) r_i = \left( \sum m_i r_i^2 \right) \omega
\]

This represents the angular momentum \(L\) of a system of particles, where:

- \(L_i\) is the angular momentum of the \(i\)-th particle,
- \(m_i\) is the mass of the \(i\)-th particle,
- \(v_i\) is the tangential velocity of the \(i\)-th particle,
- \(r_i\) is the distance of the \(i\)-th particle from the axis of rotation,
- \(\omega\) is the angular velocity (assumed the same for all particles).

The formula shows that the total angular momentum \(L\) is the sum of the angular momenta of individual particles, which can be expressed in terms of the moment of inertia \(I = \sum m_i r_i^2\) and the angular velocity \(\omega\):

\[
L = I \omega
\]

This is a fundamental relation in rotational dynamics.$$

$$The formula shown is:

\[
\vec{L} = I \vec{\omega}
\]

This represents the angular momentum \(\vec{L}\) of a rigid body, where:
- \(I\) is the moment of inertia of the rigid body,
- \(\vec{\omega}\) is the angular velocity vector.

This equation states that the angular momentum of a rigid body is the product of its moment of inertia and its angular velocity.$$

Engineering drawing

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The angular momentum of an object remains constant unless there is a torque,

$$The formula shown is the time derivative of angular momentum \(\mathbf{L}\) in vector form:

\[
\frac{\Delta \mathbf{L}}{\Delta t} = \frac{\Delta (\mathbf{r} \times \mathbf{p})}{\Delta t} = \frac{\Delta \mathbf{r}}{\Delta t} \times \mathbf{p} + \mathbf{r} \times \frac{\Delta \mathbf{p}}{\Delta t}
\]

This expresses the rate of change of angular momentum \(\mathbf{L}\) as the sum of two terms: the cross product of the rate of change of position \(\mathbf{r}\) with momentum \(\mathbf{p}\), and the cross product of position \(\mathbf{r}\) with the rate of change of momentum \(\mathbf{p}\).

If you want, I can help explain the physical meaning or derive further from this formula.$$

Since Δ r Δ t × p = v × mv = 0   (For any vector, A x A = 0.)

$$The formula shown is:

\[
\frac{\Delta L}{\Delta t} = r \times \frac{\Delta p}{\Delta t} = r \times F
\]

This equation relates the rate of change of angular momentum \( \frac{\Delta L}{\Delta t} \) to the torque \( r \times F \), where \( r \) is the position vector and \( F \) is the force. The symbol \( \times \) denotes the cross product. The equation expresses that the torque is equal to the rate of change of angular momentum.$$

$$The formula shown in the image is:

\[
\vec{\tau} = \frac{\Delta \vec{L}}{\Delta t}
\]

This represents the relationship between torque (\(\vec{\tau}\)) and the rate of change of angular momentum (\(\vec{L}\)) with respect to time (\(t\)).$$

or,

Engineering drawing

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This is analogous to Newton's 2 nd law relating the force to the time rate of change of linear momentum, ⃗ F = Δ⃗ p Δ t .

The above equation can be extended to include the total external torque (internal torques involve action-reaction pairs and cancel) and the total angular momentum of a system of particles.  This leads to the law of conservation of angular momentum , which states that the total angular momentum of a system of particles is constant if there is no net external torque .

If τ = 0 then ΔL = 0 or

$$The formula shown is:

\[
\vec{L}_i = \vec{L}_f
\]

This represents the conservation of angular momentum, which states that the initial angular momentum (\(\vec{L}_i\)) of a system is equal to the final angular momentum (\(\vec{L}_f\)) of the system, assuming no external torque acts on it.$$

## Examples

An object in empty space which rotates around some axis will keeps rotating, without the need of any external force or propulsion.

The reason why as you pedal and keep your bicycle in motion, the bicycle stays in vertical position.

## Example

A figure skater goes into a spin with her arms outstretched.  She then pulls her arms in against her body.  If she is initially rotating at 2 rev/s, what is her final rotational speed?  Assume that by pulling in her arms she reduces her rotational inertia from 2 kg-m 2 to 1.5 kg-m 2 .

$$The formula shown is related to the conservation of angular momentum.

- \( L_f = L_i \) states that the final angular momentum \( L_f \) is equal to the initial angular momentum \( L_i \).
- \( I_f \omega_f = I_i \omega_i \) expresses the relationship between the moment of inertia \( I \) and angular velocity \( \omega \) before and after a change.
- The angular velocity after the change \( \omega_f \) is calculated as:

\[
\omega_f = \frac{I_i \omega_i}{I_f} = \frac{(2 \, kg \cdot m^2)(2 \, rev/s)}{1.5 \, kg \cdot m^2} = 2.67 \, rev/s
\]

This means that when the moment of inertia decreases from \( 2 \, kg \cdot m^2 \) to \( 1.5 \, kg \cdot m^2 \), the angular velocity increases from \( 2 \, rev/s \) to \( 2.67 \, rev/s \), conserving angular momentum.$$

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How much kinetic energy was gained or lost when she pulled in her arms?

$$Let's break down the formulas and calculations shown in the image:

1. Final kinetic energy \( KE_f \):
\[
KE_f = \frac{1}{2} I_f \omega_f^2 = \frac{1}{2} (1.5 \, kg \cdot m^2) \left( (2\pi)(2.67 \, rad/s) \right)^2 = 210.5 \, J
\]

2. Initial kinetic energy \( KE_i \):
\[
KE_i = \frac{1}{2} I_f \omega_i^2 = \frac{1}{2} (2 \, kg \cdot m^2) \left( (2\pi)(2) \, rad/s \right)^2 = 157.9 \, J
\]

3. Change in kinetic energy \( \Delta KE \):
\[
\Delta KE = KE_f - KE_i = 210.5 \, J - 157.9 \, J = 52.6 \, J
\]

This shows the calculation of kinetic energy before and after some event, and the difference between them. If you need further explanation or help with these calculations, feel free to ask!$$

Thus, there was a gain in KE .  Note that we had to convert angular velocity to rad/s when calculating KE .

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Summary of the relations and analogies of the linear and angular quantities:

| Linear Quantity   | Angular Quantity   |
|-------------------|--------------------|
| Δ l, v, a         | Δ θ, ω, α          |
| m                 | I                  |
| p = m v           | L = r × p, L = I ω |
| F = m a           | τ = r × F, τ = I α |
| KE L = ½ mv 2     | KE R = ½ I ω 2     |
| W L = ΔKE L       | W R = ΔKE R        |
| P AVE = Fv        | P AVE = τ ω        |