Notes on Newton's Second Law for Rotation

Newton’s Second Law for Rotation Notes

1. Review of Newton’s Second Law

• Newton’s second law can be expressed in two forms: momentum and acceleration.

  • Momentum form: \vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}

  • Acceleration form: \vec{F}_{\text{net}} = m\vec{a}

• The law states that the net force acting on an object is equal to the mass of the object times its acceleration.

• This principle is fundamental for developing dynamic equations of motion in linear physics.

• Example with Hooke’s Law for an ideal spring:

  • Given in the x direction: \vec{F}_{\text{net}} = < -kx, 0, 0 >

  • Substituting into Newton’s law results in:

  • \vec{F}_{\text{net}} = m\vec{a} or

  • m\vec{a} - < -kx, 0, 0 > = 0

  • Therefore, focusing on the x component: \text{max} = -kx (3)

• This equation describes the dynamic behavior of an ideal spring.

2. Torque

• Torque is a crucial concept before deriving dynamic equations of motion in rotational dynamics.

2.1. Torque Vector Direction

• Torque is defined as the cross product of two vectors:
  \vec{\tau} = \vec{r} \times \vec{F}

• In rotation about the x-y plane, the torque vector must be in the z direction.

  • For rotating on the x-y plane, set:

  • \vec{r} = < rx, ry, 0 >

  • \vec{F} = < Fx, Fy, 0 >

• Calculating torque:

  • Resulting torque vector:

  • < \tau{x}, \tau{y}, \tau{z} >= < 0, 0, rx Fy - ry F_x > (15)

  • Non-zero component for z-direction:

  • \tau_{z}=rxFy-ryF_{x} (16)

• Magnitude of torque when rotating in the x-y plane:

  • \tau = |\tau| = |\tauz| = |rx Fy - ry F_x| \

2.2. Sin Definition of Cross Product

• Torque magnitude defined with sine of the angle between the two vectors:

  • |\vec{\tau}| = |\vec{r} \times \vec{F}| = |\vec{r}| |\vec{F}| \sin(\phi) (18)

• Individual magnitudes:

  • For the vectors:

  • r = |\vec{r}| = \sqrt{rx^2 + ry^2 + r_z^2} (22)

  • F = |\vec{F}| = \sqrt{Fx^2 + Fy^2 + F_z^2} (23)

2.3. Maximum Torque

• The maximum torque occurs when \phi = 90^{\circ} leading to:

  • \tau_{\text{max}} = r F (28)

• Maximum torque occurs when force is applied perpendicularly.

2.4. Minimum Torque

• Minimum torque (zero) occurs when the applied force is radial:

  • For \phi = 0^{\circ} , thus:

  • \tau = r F \sin(0) = 0 (33)

2.5. Example Problem

• For example, if we apply a force of 10N at a distance of 0.2m from the rotation axis:

  • Represent vectors as:

  • \vec{r} = <0.2, 0, 0>

  • \vec{F} = <0, 10, 0>

  • Using the torque definition:

  • \vec{\tau} = \vec{r} \times \vec{F} = <0 - 0, 0 - 0, (0.2)(10) - 0> = <0, 0, 2> (37)

  • Magnitude of torque is:

  • \tau = |\vec{\tau}| = \sqrt{0^2 + 0^2 + 2^2} = 2 \text{ Nm} (38)

  • Confirm using sin definition:

  • \tau = r F \sin(90^{\circ}) = (0.2)(10)(1) = 2 \text{ Nm} (39)

3. Newton’s Second Law for Rotation

• Analogous to the linear scenario, for rotational motion, the Newton's law introduces net torque relating to angular momentum:

  • \vec{\tau}_{\text{net}} = \frac{d \vec{L}}{dt} (40)

• For 2D rotation specifically on the x-y plane, the simplified form is:

  • \tau{\text{net} z} = I \alpha{z} (41)

• This relates net torque to the moment of inertia I and the angular acceleration \alpha_z .

4. Applying N2 for Rotation

• Similar to applying Newton’s second law in linear scenarios, we can apply it to rotational problems:

  • Knowing the net force will allow for calculation of net torque hence angular acceleration:

  • Example: Suppose a force is applied on a disk at its center:

  • From the cross product definition, \vec{\tau} = \vec{r} \times \vec{F}

  • Evaluating:

  • \tau{\text{net z}} = rF{ ext{net}}

  • Rearranging provides:

  • \alphaz = \frac{r F{\text{net}}}{I} (47)

• Using moment of inertia for a solid disk, $I = \frac{1}{2}mr^2$ , leads to:

  • \alphaz = \frac{2 F{\text{net}}}{mr} (49)

5. Static Equilibrium

• Definition: Static equilibrium occurs when both net force and net torque on an object are zero:

  • \vec{F}_{\text{net}} = < 0, 0, 0 > (51)

  • \vec{\tau}_{\text{net}} = < 0, 0, 0 > (52)

• Essential for structural integrity in construction, termed as 'statics'.

5.1 Example - Teeter Totter Problem

• Utilizing concepts of static equilibrium:

• Problem involves balancing two masses, with variables altering their respective positions and weights.

5.1.1 Example Problem Part 1

• Given:

  • Rigid plank 5m long, mass of plank: 15kg

  • Child's mass: 20kg (4m from pivot)

  • Adult's unknown mass (1m from pivot)

• Static conditions:

  • Net force and torque equations yield:

  • F{ ext{net y}} = FN - mc g - mp g - m_a g (57)

  • \tau{\text{net z}} = 4 mc g + 1.5 mp g - ma g (62)

• Solving leads to required adult mass:

  • ma = 4mc + 1.5m_p = (4)(20) + (1.5)(15) = 102.5 kg (66)

5.1.2 Example Problem Part 2

• If the child moves 1m closer (3m from the pivot), determine adult's position.

• Develop new torque equation:

  • For the updated positions:

  • \tau{\text{net z}} = 3 mc g + 1.5 mp g - ra m_a g (70)

• This results in the equation for adult’s position (ra):

  • ra = \frac{1.5 mp + 3 mc}{ma}

  • After substituting known quantities: r_a = 0.8049 m (72)

5.2 Important Note on Reference Point for Static Equilibrium Torque Calculations

• Selection of reference point for torque calculations is crucial yet flexible in static equilibrium scenarios.

• While the pivot is commonly chosen, any point can suffice, ensuring conditions remain valid:

  • \vec{F}{\text{net}} = <0, 0, 0> and \vec{\tau}{\text{net}} = <0, 0, 0> .