Course Notes 📚
Radian Definition
One radian is the angle subtended by an arc length equal to the radius \(r\) of the circle. Each point on a rotating rigid object undergoes the same angular displacement in any given time interval.
\[ 1 \text{ rad} = 57.3^\circ \]
\[ 180^\circ = \pi \text{ rad} \]
Conversion: To convert degrees to radians, multiply by \( \frac{\pi \text{ rad}}{180^\circ} \).
Angular Displacement (\(\Delta\theta\))
The change in angular position. SI unit: rad.
\[ \Delta\theta = \theta_f - \theta_i \]
Arc Length (s)
Relates angular displacement to linear displacement: \( s = r\theta \)
Angular Velocity (\(\omega\))
The rate of change of angular displacement. SI unit: rad/s.
Average: \( \omega_{av} = \frac{\Delta\theta}{\Delta t} \)
Instantaneous: \( \omega = \lim_{\Delta t \to 0} \frac{\Delta\theta}{\Delta t} \)
Angular Acceleration (\(\alpha\))
The rate of change of angular velocity. SI unit: rad/s\(^2\). When a rigid object rotates about a fixed axis, every portion of the object has the same angular velocity and the same angular acceleration.
Average: \( \alpha_{av} = \frac{\Delta\omega}{\Delta t} \)
Instantaneous: \( \alpha = \lim_{\Delta t \to 0} \frac{\Delta\omega}{\Delta t} \)
Example 7.1: Whirllybirds
A helicopter rotor turns at an angular velocity of \(3.20 \times 10^2\) rev/min.
(a) Express this in rad/s:
\( \omega = 3.20 \times 10^2 \frac{\text{rev}}{\text{min}} \cdot \frac{2\pi \text{ rad}}{1 \text{ rev}} \cdot \frac{1 \text{ min}}{60.0 \text{ s}} = 33.5 \text{ rad/s} \)(b) Arc length traced by a 2.00 m rotor in 300 s:
\( \Delta\theta = \omega t = (33.5)(300) = 1.01 \times 10^4 \text{ rad} \)
\( s = r\Delta\theta = (2.00)(1.01 \times 10^4) = 2.02 \times 10^4 \text{ m} \)
The kinematic equations for rotational motion with constant angular acceleration are analogous to linear motion equations.
Example 7.2: A Rotating Wheel
A wheel with constant \(\alpha = 3.50 \text{ rad/s}^2\) starts with \(\omega_i = 2.00 \text{ rad/s}\).
(a) Angle rotated in 2.00 s:
\( \Delta\theta = \omega_i t + \frac{1}{2} \alpha t^2 = (2)(2) + \frac{1}{2}(3.5)(2)^2 = 11.0 \text{ rad} \)(b) Angular velocity at t = 2.00 s:
\( \omega = \omega_i + \alpha t = 2 + (3.5)(2) = 9.00 \text{ rad/s} \)
Comparison of Linear and Rotational Kinematic Equations
| No. | Linear Kinematic Equation | Rotational Kinematic Equation |
|---|---|---|
| 1 | \(s = \left( \frac{v_i + v_f}{2} \right) t\) | \(\theta = \left( \frac{\omega_i + \omega_f}{2} \right) t\) |
| 2 | \(s = v_i t + \frac{1}{2} a t^2\) | \(\theta = \omega_i t + \frac{1}{2} \alpha t^2\) |
| 3 | \(s = v_f t - \frac{1}{2} a t^2\) | \(\theta = \omega_f t - \frac{1}{2} \alpha t^2\) |
| 4 | \(s_f = v_i t + \frac{1}{2} a t^2 + s_i\) | \(\theta_f = \omega_i t + \frac{1}{2} \alpha t^2 + \theta_i\) |
| 5 | \(v_f = v_i + a t\) | \(\omega_f = \omega_i + \alpha t\) |
| 6 | \(v_f^2 = v_i^2 + 2 a s\) | \(\omega_f^2 = \omega_i^2 + 2 \alpha \theta\) |
| 7 | \(v_i^2 = v_f^2 - 2 a s\) | \(\omega_i^2 = \omega_f^2 - 2 \alpha \theta\) |
| 8 | \(a = \frac{v_f - v_i}{t}\) | \(\alpha = \frac{\omega_f - \omega_i}{t}\) |
Tangential Velocity (\(v_t\)): The linear velocity of a point on a rotating object. \( v_t = r\omega \)
Tangential Acceleration (\(a_t\)): The linear acceleration of a point on a rotating object. \( a_t = r\alpha \)
Centripetal Acceleration (\(a_c\)): The acceleration directed towards the center of the circular path. \( a_c = \frac{v^2}{r} = r\omega^2 \)
Example 7.3: Compact Discs
A CD accelerates from rest to -31.4 rad/s in 0.892 s.
(a) Angular acceleration:
\( \alpha = \frac{\omega}{t} = \frac{-31.4}{0.892} = -35.2 \text{ rad/s}^2 \)(b) Angle turned:
\( \Delta\theta = \frac{1}{2} \alpha t^2 = \frac{1}{2}(-35.2)(0.892)^2 = -14.0 \text{ rad} \)
Centripetal Force (\(F_c\)): The net force acting toward the center of the circular path, causing centripetal acceleration. \( F_c = ma_c = m\frac{v^2}{r} \)
Problem-solving Strategy:
- Draw a free-body diagram.
- Choose a coordinate system.
- Find the net force \(F_c\) toward the center.
- Apply Newton's second law.
- Solve for the unknown.
Example 7.6: Buckle Up For Safety
A car at 13.4 m/s takes a circular turn of radius 50.0 m. Find the minimum \(\mu_s\).
The static friction provides the centripetal force: \( f_s = m\frac{v^2}{r} \).
The maximum static friction is \(f_{s, \text{max}} = \mu_s n = \mu_s mg\).
Setting them equal: \( \mu_s mg = m\frac{v^2}{r} \)
\( \mu_s = \frac{v^2}{rg} = \frac{(13.4)^2}{(50.0)(9.80)} = 0.366 \)
The gravitational force between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force exerted by a uniform sphere on a particle outside is as if the sphere's mass were concentrated at its center.
\[ F = G\frac{m_1 m_2}{r^2} \]
\(G = 6.673 \times 10^{-11} \frac{\text{N} \cdot \text{m}^2}{\text{kg}^2}\)
Gravitational Potential Energy
\[ PE = -G\frac{m_1 m_2}{r} \]
Example 7.11: A Near-Earth Asteroid
An asteroid (\(m = 10^9\) kg) falls from infinity toward Earth.
(a) Change in PE at \(4 \times 10^8\) m from Earth's center:
\( \Delta PE = -G\frac{M_E m}{r} = -(6.67 \times 10^{-11})\frac{(5.98 \times 10^{24})(10^9)}{4 \times 10^8} = -9.97 \times 10^{14} \text{ J} \)(b) Asteroid's speed at that point:
From conservation of energy (\(\Delta KE + \Delta PE = 0\)):
\( \frac{1}{2}mv^2 - 9.97 \times 10^{14} = 0 \)
\( v = \sqrt{\frac{2(9.97 \times 10^{14})}{10^9}} = 1.41 \times 10^3 \text{ m/s} \)