02_Crossprod

1. Rotational Dynamics: Atwood's Machine Examples

Example 1: Atwood's Machine with Two Pulleys

Scenario: Two masses (\(m_1\) and \(m_2\), where \(m_1 > m_2\)) are connected by a string passing over two identical pulleys (radius \(R\), moment of inertia \(I\)). We want to find the accelerations and the three string tensions (\(T_1, T_2, T_3\)).

• Assumptions: Counter-clockwise rotation is defined as positive (\(\odot\hat{\omega}\) fixed by hand; choosing the opposite direction simply introduces a "-" sign, but remains physically correct).

Equations of Motion:

1. Left Pulley (\(n_1\)): \(T_1 R - T_2 R = I\alpha\)

2. Right Pulley (\(n_2\)): \(T_2 R - T_3 R = I\alpha\)

3. Mass 1 (moving down): \(m_1 g - T_1 = m_1 a\)

4. Mass 2 (moving up): \(T_3 - m_2 g = m_2 a\)

5. Kinematic Constraint: \(a = R\alpha\)

Solution:

Adding equations \((1) + (2) + (3) + (4) + (5)\) to eliminate the tensions:

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

Solving for linear acceleration \(a\) and angular acceleration \(\alpha\):

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

(Note: If \(m_1 = m_2 \implies \alpha = 0\), which is intuitively obvious.)

Example 2: Standard Atwood Machine (Energy Method)

Scenario: A standard one-pulley Atwood machine (pulley \(I, R\)) is released from rest. Find the velocity \(v\) after mass \(m_1\) falls by a displacement \(h\).

Energy Conservation:

The loss in gravitational potential energy equals the gain in translational and rotational kinetic energy:

\[ (m_1 - m_2)gh = \frac{1}{2}m_1 v^2 + \frac{1}{2}m_2 v^2 + \frac{1}{2}I\omega^2 \]

Using \(v = \omega R\):

\[ v^2 = \frac{2(m_1 - m_2)gh}{m_1 + m_2 + I/R^2} \]

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2. The Vector Cross Product: Mathematical Foundations

Definition and Properties

For two vectors \(\vec{A}\) and \(\vec{B}\), the cross product is \(\vec{C} = \vec{A} \times \vec{B}\).

• Orthogonality: \(\vec{C} \perp \vec{A}\) and \(\vec{C} \perp \vec{B}\)

• Magnitude: \(C = AB\sin\theta\)

• Direction: Determined by the Right-Hand Rule.

Key Rules:

1. If \(\vec{A} \parallel \vec{B}\) (\(\theta = 0\) or \(\pi\)), then \(\vec{A} \times \vec{B} = 0\). (Consequently, \(\vec{A} \times \vec{A} = 0\)).

2. If \(\vec{A} \perp \vec{B}\), then \(|\vec{A} \times \vec{B}| = AB\).

3. Non-commutative: \(\vec{A} \times \vec{B} = -\vec{B} \times \vec{A}\)

4. Distributive: \(\vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C}\)

Orthogonal Unit Vectors and Determinant Form

Using Cartesian basis vectors (\(\hat{i} \equiv \hat{e}_x\), \(\hat{j} \equiv \hat{e}_y\), \(\hat{k} \equiv \hat{e}_z\)):

• \(\hat{i} \times \hat{j} = \hat{k}\)

• \(\hat{j} \times \hat{k} = \hat{i}\)

• \(\hat{k} \times \hat{i} = \hat{j}\)

Expanding \(\vec{A} \times \vec{B}\) component-wise yields the determinant form:

\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} \]

Vector Identities and Calculus

• Time Derivative:

  \[ \frac{d}{dt} (\vec{A} \times \vec{B}) = \left(\frac{d\vec{A}}{dt}\right) \times \vec{B} + \vec{A} \times \left(\frac{d\vec{B}}{dt}\right) \]

• Scalar Triple Product:

  \[ (\vec{A} \times \vec{B}) \cdot \vec{C} = \begin{vmatrix} A_x & A_y & A_z \\ B_x & B_y & B_z \\ C_x & C_y & C_z \end{vmatrix} = (\vec{B} \times \vec{C}) \cdot \vec{A} = (\vec{C} \times \vec{A}) \cdot \vec{B} \]

• Vector Triple Product (BAC-CAB Rule):

  \[ \vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C} \]

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3. Kinematics of Rigid Body Rotation

Acceleration in Fixed Axis Rotation

Taking the derivative of linear velocity \(\vec{v} = \vec{\omega} \times \vec{r}\):

\[ \vec{a} = \frac{d}{dt}(\vec{\omega} \times \vec{r}) = \frac{d\vec{\omega}}{dt} \times \vec{r} + \vec{\omega} \times \frac{d\vec{r}}{dt} \]

\[ \vec{a} = \vec{\alpha} \times \vec{r} + \vec{\omega} \times (\vec{\omega} \times \vec{r}) \]

• Tangential Acceleration: \(\vec{a}_t = \vec{\alpha} \times \vec{r}\) (Magnitude: \(a_t = |\vec{\alpha} \times \vec{r}|\))

• Radial/Centripetal Acceleration: \(\vec{a}_r = \vec{\omega} \times (\vec{\omega} \times \vec{r})\) (Magnitude: \(a_r = |\vec{\omega} \times (\vec{\omega} \times \vec{r})|\))

Rotational Kinetic Energy Derivation

Using the scalar triple product rearrangement where \((\vec{A} \times \vec{B}) \cdot \vec{C} = \vec{B} \cdot (\vec{C} \times \vec{A})\):

\[ K = \frac{1}{2} \sum m_i v_i^2 = \frac{1}{2} \sum m_i (\vec{\omega} \times \vec{r}_i) \cdot (\vec{\omega} \times \vec{r}_i) \]

\[ K = \frac{1}{2} \sum m_i \left[ \vec{r}_i \times (\vec{\omega} \times \vec{r}_i) \right] \cdot \vec{\omega} \]

Applying the BAC-CAB rule:

\[ K = \frac{1}{2} \sum m_i \left[ r_i^2 \vec{\omega} - (\vec{r}_i \cdot \vec{\omega})\vec{r}_i \right] \cdot \vec{\omega} = \frac{1}{2} \sum m_i \left[ r_i^2 - (\vec{r}_i \cdot \hat{\omega})^2 \right] \omega^2 \]

\[ K = \frac{1}{2} \left( \sum m_i r_{i\perp}^2 \right) \omega^2 = \frac{1}{2} I \omega^2 \]

(Where \(r_{i\perp}^2 = r_i^2 - (\vec{r}_i \cdot \hat{\omega})^2\) is the perpendicular distance from the axis).

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4. Fictitious Forces and the Coriolis Effect

Inertial vs. Rotating Frames

• Inertial Frame: Follows Newton's Laws. e.g., A mass moving in a circle requires a normal force \(F_N = m\frac{v^2}{r}\).

• Rotating Frame: Non-inertial. Fictitious forces appear to explain motion. e.g., If the mass appears at rest in the rotating frame, \(F_{\text{fictitious}} - F_N = 0 \implies F_{\text{fictitious}} = m\frac{v^2}{r}\).

Transformation Between Frames (The Earth)

Define the position vector \(\vec{r}\) in both the fixed inertial basis \(\{\hat{e}_i\}\) and the rotating basis \(\{\hat{e}_i'\}\):

\[ \vec{r} = \sum r_i \hat{e}_i = \sum r_i' \hat{e}_i' = \vec{r}' \]

Because the basis vectors themselves are rotating with angular velocity \(\vec{\omega}\), their derivative is:

\[ \frac{d\hat{e}_i'}{dt} = \vec{\omega} \times \hat{e}_i' \]

Velocity Transformation:

\[ \frac{d\vec{r}}{dt} = \sum \left( \frac{dr_i'}{dt} \hat{e}_i' + r_i' \frac{d\hat{e}_i'}{dt} \right) = \sum v_i' \hat{e}_i' + \vec{\omega} \times \sum r_i' \hat{e}_i' \]

\[ \vec{v}_{\text{inertial}} = \vec{v}' + \vec{\omega} \times \vec{r}' \]

(The \(\vec{\omega} \times \vec{r}'\) term is the velocity due to rotation).

Acceleration Transformation:

Differentiating again and applying the same logic:

\[ \frac{d^2\vec{r}}{dt^2} = \sum a_i' \hat{e}_i' + 2\vec{\omega} \times \sum v_i' \hat{e}_i' + \vec{\omega} \times \left[ \vec{\omega} \times \sum r_i' \hat{e}_i' \right] \]

\[ \vec{a}_{\text{inertial}} = \vec{a}' + 2\vec{\omega} \times \vec{v}' + \vec{\omega} \times (\vec{\omega} \times \vec{r}') \]

The Fictitious Forces

In the rotating frame, the apparent acceleration is \(\vec{a}'\). The forces that act on an object are the real forces plus two fictitious contributions due to rotation:

\[ \vec{F}_{\text{fictitious}} = - 2m\vec{\omega} \times \vec{v}' - m\vec{\omega} \times (\vec{\omega} \times \vec{r}') \]

• \(- 2m\vec{\omega} \times \vec{v}'\): Coriolis Force (depends on the velocity of the object in the rotating frame).

• \(- m\vec{\omega} \times (\vec{\omega} \times \vec{r}')\): Centrifugal Force (points outward from the rotation axis).

Application: Typhoons (Cyclones)

When air rushes inward toward a central region of low pressure:

• Northern Hemisphere: The Coriolis force deflects the moving air to the right relative to its direction of motion. This forces the air into a counter-clockwise spiral.

• Southern Hemisphere: The Coriolis force deflects the air to the left, creating a clockwise spiral.