Rotational Kinematics and Dynamics

Angular velocity, acceleration, and torque with calculus

Rotational Kinematics and Dynamics

Angular Quantities

Angular position: $\theta$ (radians)

Angular velocity: $\omega = \frac{d\theta}{dt}$

Angular acceleration: $\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}$

Relationship to Linear Quantities

For point at distance $r$ from axis:

Arc length: $s = r\theta$

Linear velocity: $v = r\omega$

Tangential acceleration: $a_t = r\alpha$

Centripetal acceleration: $a_c = \frac{v^2}{r} = r\omega^2$

Rotational Kinematics

For constant angular acceleration $\alpha$ :

$\omega = \omega_0 + \alpha t$

$\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2$

$\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$

(Analogous to linear kinematics)

Variable Angular Acceleration

When $\alpha(t)$ is given:

$\omega(t) = \omega_0 + \int_0^t \alpha(t') \, dt'$

$\theta(t) = \theta_0 + \int_0^t \omega(t') \, dt'$

When $\alpha(\theta)$ is given, use: $\alpha = \frac{d\omega}{dt} = \frac{d\omega}{d\theta}\frac{d\theta}{dt} = \omega\frac{d\omega}{d\theta}$

Torque

Definition: $\vec{\tau} = \vec{r} \times \vec{F}$

Magnitude: $\tau = rF\sin\phi$

where $\phi$ is angle between $\vec{r}$ and $\vec{F}$ .

Perpendicular distance form: $\tau = r_{\perp}F = rF_{\perp}$

Rotational Dynamics

Newton's second law for rotation: $\tau_{net} = I\alpha$

where $I$ is moment of inertia.

Differential form: $\tau = I\frac{d\omega}{dt}$

Work and Power in Rotation

Work by torque: $W = \int_{\theta_1}^{\theta_2} \tau \, d\theta$

For constant torque: $W = \tau\Delta\theta$

Rotational power: $P = \tau\omega$

(Analogous to $P = Fv$ for linear motion)

Rotational Kinetic Energy

$KE_{rot} = \frac{1}{2}I\omega^2$

Work-energy theorem: $W_{net} = \Delta KE_{rot} = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2$

Combined Translation and Rotation

For rolling object:

Total kinetic energy: $KE = KE_{trans} + KE_{rot} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$

Rolling without slipping: $v_{cm} = R\omega$

where $R$ is radius.

Example: Rolling Down Incline

Energy conservation: $Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$

Using $v = R\omega$ and $I = \beta MR^2$ :

$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}\beta MR^2\frac{v^2}{R^2}$

$gh = \frac{1}{2}v^2(1 + \beta)$

$v = \sqrt{\frac{2gh}{1 + \beta}}$

where $\beta$ depends on shape:

Solid cylinder: $\beta = 1/2$
Solid sphere: $\beta = 2/5$
Hoop: $\beta = 1$

Angular Impulse

$\int_{t_1}^{t_2} \tau \, dt = \Delta L = I\Delta\omega$

where $L = I\omega$ is angular momentum.

📚 Practice Problems

1Problem 1easy

❓ Question:

A wheel starts from rest and accelerates uniformly at α = 2.0 rad/s² for 5.0 seconds. Find: (a) the final angular velocity, (b) the total angle rotated, and (c) the number of complete revolutions.

💡 Show Solution

Given:

ω₀ = 0
α = 2.0 rad/s²
t = 5.0 s

(a) Final angular velocity:

$\omega = ω_0 + αt = 0 + (2.0)(5.0)$

$\boxed{\omega = 10 \text{ rad/s}}$

(b) Total angle rotated:

$\theta = ω_0 t + \frac{1}{2}αt^2 = 0 + \frac{1}{2}(2.0)(5.0)^2$

$\theta = (1.0)(25)$

$\boxed{\theta = 25 \text{ rad}}$

(c) Number of revolutions:

$N = \frac{\theta}{2\pi} = \frac{25}{2\pi}$

$\boxed{N = 3.98 \text{ rev} \approx 4 \text{ revolutions}}$

2Problem 2easy

❓ Question:

A wheel starts from rest and accelerates uniformly at α = 2.0 rad/s² for 5.0 seconds. Find: (a) the final angular velocity, (b) the total angle rotated, and (c) the number of complete revolutions.

💡 Show Solution

Given:

ω₀ = 0
α = 2.0 rad/s²
t = 5.0 s

(a) Final angular velocity:

$\omega = ω_0 + αt = 0 + (2.0)(5.0)$

$\boxed{\omega = 10 \text{ rad/s}}$

(b) Total angle rotated:

$\theta = ω_0 t + \frac{1}{2}αt^2 = 0 + \frac{1}{2}(2.0)(5.0)^2$

$\theta = (1.0)(25)$

$\boxed{\theta = 25 \text{ rad}}$

(c) Number of revolutions:

$N = \frac{\theta}{2\pi} = \frac{25}{2\pi}$

$\boxed{N = 3.98 \text{ rev} \approx 4 \text{ revolutions}}$

3Problem 3medium

❓ Question:

A solid disk (mass M = 5.0 kg, radius R = 0.3 m) rotates about its central axis. A tangential force F = 15 N is applied at the rim. Find: (a) the torque, (b) the angular acceleration, and (c) the angular velocity after 3.0 seconds (starting from rest).

💡 Show Solution

Given:

M = 5.0 kg
R = 0.3 m
F = 15 N (tangential at rim)
ω₀ = 0

(a) Torque:

$\tau = rF = RF = (0.3)(15)$

$\boxed{\tau = 4.5 \text{ N·m}}$

(b) Angular acceleration:

Moment of inertia for solid disk: $I = \frac{1}{2}MR^2 = \frac{1}{2}(5.0)(0.3)^2 = 0.225 \text{ kg·m}^2$

$\alpha = \frac{\tau}{I} = \frac{4.5}{0.225}$

$\boxed{\alpha = 20 \text{ rad/s}^2}$

(c) Angular velocity after 3.0 s:

$\omega = ω_0 + αt = 0 + (20)(3.0)$

$\boxed{\omega = 60 \text{ rad/s}}$

Linear speed at rim: $v = Rω = (0.3)(60) = 18 \text{ m/s}$

4Problem 4medium

❓ Question:

💡 Show Solution

Given:

M = 5.0 kg
R = 0.3 m
F = 15 N (tangential at rim)
ω₀ = 0

(a) Torque:

$\tau = rF = RF = (0.3)(15)$

$\boxed{\tau = 4.5 \text{ N·m}}$

(b) Angular acceleration:

Moment of inertia for solid disk: $I = \frac{1}{2}MR^2 = \frac{1}{2}(5.0)(0.3)^2 = 0.225 \text{ kg·m}^2$

$\alpha = \frac{\tau}{I} = \frac{4.5}{0.225}$

$\boxed{\alpha = 20 \text{ rad/s}^2}$

(c) Angular velocity after 3.0 s:

$\omega = ω_0 + αt = 0 + (20)(3.0)$

$\boxed{\omega = 60 \text{ rad/s}}$

Linear speed at rim: $v = Rω = (0.3)(60) = 18 \text{ m/s}$

5Problem 5hard

❓ Question:

A uniform rod (length L = 1.0 m, mass M = 2.0 kg) is hinged at one end and held horizontal. It is released from rest. Find: (a) the initial angular acceleration, (b) the angular velocity when the rod is vertical, and (c) the linear speed of the free end when vertical.

💡 Show Solution

Given:

L = 1.0 m
M = 2.0 kg
Rod hinged at one end, released from horizontal

(a) Initial angular acceleration:

Moment of inertia about end: $I = \frac{1}{3}ML^2 = \frac{1}{3}(2.0)(1.0)^2 = 0.667 \text{ kg·m}^2$

Torque due to gravity (acts at CM, distance L/2 from hinge): $\tau = Mg\frac{L}{2} = (2.0)(9.8)\frac{1.0}{2} = 9.8 \text{ N·m}$

$\alpha_0 = \frac{\tau}{I} = \frac{9.8}{0.667}$

$\boxed{\alpha_0 = 14.7 \text{ rad/s}^2}$

(b) Angular velocity when vertical:

Using energy conservation:

Initial PE (relative to final): $U_i = Mg\frac{L}{2}$ (CM drops by L/2)

Final KE: $K_f = \frac{1}{2}Iω^2$

$Mg\frac{L}{2} = \frac{1}{2}Iω^2$

$ω^2 = \frac{MgL}{I} = \frac{(2.0)(9.8)(1.0)}{0.667}$

$ω^2 = 29.4$

$\boxed{\omega = 5.42 \text{ rad/s}}$

(c) Linear speed of free end:

$v = Lω = (1.0)(5.42)$

$\boxed{v = 5.42 \text{ m/s}}$

Check with energy: $\frac{1}{2}Mv_{cm}^2 = Mg\frac{L}{2}$

$v_{cm} = \sqrt{gL} = \sqrt{9.8} = 3.13 \text{ m/s}$

Since $v_{end} = 2v_{cm}$ for rotation about end: $v_{end} \approx 5.4$ m/s ✓

6Problem 6hard

❓ Question:

💡 Show Solution

Given:

L = 1.0 m
M = 2.0 kg
Rod hinged at one end, released from horizontal

(a) Initial angular acceleration:

Moment of inertia about end: $I = \frac{1}{3}ML^2 = \frac{1}{3}(2.0)(1.0)^2 = 0.667 \text{ kg·m}^2$

Torque due to gravity (acts at CM, distance L/2 from hinge): $\tau = Mg\frac{L}{2} = (2.0)(9.8)\frac{1.0}{2} = 9.8 \text{ N·m}$

$\alpha_0 = \frac{\tau}{I} = \frac{9.8}{0.667}$

$\boxed{\alpha_0 = 14.7 \text{ rad/s}^2}$

(b) Angular velocity when vertical:

Using energy conservation:

Initial PE (relative to final): $U_i = Mg\frac{L}{2}$ (CM drops by L/2)

Final KE: $K_f = \frac{1}{2}Iω^2$

$Mg\frac{L}{2} = \frac{1}{2}Iω^2$

$ω^2 = \frac{MgL}{I} = \frac{(2.0)(9.8)(1.0)}{0.667}$

$ω^2 = 29.4$

$\boxed{\omega = 5.42 \text{ rad/s}}$

(c) Linear speed of free end:

$v = Lω = (1.0)(5.42)$

$\boxed{v = 5.42 \text{ m/s}}$

Check with energy: $\frac{1}{2}Mv_{cm}^2 = Mg\frac{L}{2}$

$v_{cm} = \sqrt{gL} = \sqrt{9.8} = 3.13 \text{ m/s}$

Since $v_{end} = 2v_{cm}$ for rotation about end: $v_{end} \approx 5.4$ m/s ✓

🎴

Practice with Flashcards

Review key concepts with our flashcard system

📖

Browse All Topics

Explore other calculus topics