Honors Physics Chapter 10 - Rotation

Rotational Variables

• Radians (Rad) : $\theta =\dfrac s r$
  • $1\,\text{rev}=360\degree$ rev : revolutions
• Angular displacement :
  • $\Delta \theta = \theta_2-\theta_1$
  • Counterclockwise direction is positive
• Average angular velocity
• Instantaneous angular velocity : $\omega=\lim_{\Delta t\to 0}\dfrac{\Delta \theta}{\Delta t}=\dfrac{d\theta}{dt}$
• Average angular acceleration
• Instantaneous angular acceleration: $\alpha=\lim_{\Delta t\to 0}\dfrac{\Delta \omega}{\Delta t}=\dfrac{d\omega}{dt}$

Angular displacement are not vectors, because the order of rotation matters for rotations around are not the same.

Rotation with Constant Angular Acceleration

• Tangential acceleration : $a_t=\alpha r$
• Centripetal acceleration : $a_n=\dfrac{v^2}r=\omega r$

*Attachment :

$$\vec r(t)=r(t)(\cos\theta(t),\sin\theta(t))=r(t)\cos\theta(t) \hat x+r(t) \sin\theta(t)\hat y \\[10px] \vec v(t)=\dfrac{d\vec r(t)}{dt}=\dot r(t)\cos\theta(t)+r(t)\dot{\theta}(t)(-\sin\theta(t))\\[10px]$$

Kinetic Energy of Rotation

$$K=\sum\dfrac 1 2m_i(\omega r_i)^2=\dfrac 1 2\boxed{\left(\sum m_ir_i^2\right)}\omega^2$$

The boxed area is called rotational inertia, or moment of inertia, $I$.

This is a constant for a rigid object and given rotational axis.

Caution : the axis for $I$ must have to be specified.

To sum up :

$$I=\sum m_ir_i^2$$

So we can rewrite the kinetic energy as :

$$K=\dfrac 1 2I\omega ^2$$

For a discrete particles:

$$I=\int r^2dm$$

Some common:

An example proof :

The Parallel-Axis Theorem

Assuming the rotation axis is deviated :

$$I=I_{com}+Mh^2$$

Note the axis must be parallel.

Torque

$$\tau=F_{\perp}d=(r)(F\sin\phi)\\ \tau = r\times F$$

Torque, $\tau$.

The line is called the line of action of $\vec F$.

The $\vec r=r\sin \phi$ is called the moment arm.

The unit of torque is $\text{N}\cdot\text{m}$, but torques are never expressed in joules, torque is not energy.

Torque is positive if it would cause a counterclockwise.

For several torques, the net torque or resultant torque is the sum of other torques.

Newton’s Second Law for Rotation

So:

$$\tau_{\text{net}} = I\alpha$$

It is torque that causes angular acceleration

$$\alpha = \lim_{\Delta t\to}\frac{\Delta\omega}{\Delta t}=\frac{d\omega}{dt}$$

Work and Rotational Kinetic Energy

The rotational work-kinetic energy theorem states:

$$\Delta K=K_f-Ki=\frac 1 2I\omega_f^2-\frac 1 2I\omega_i^2=W$$

The work done in a rotation about a fixed axis can be calculated by:

$$W=\int_{\theta_i}^{\theta_f}\tau d\theta\\ \tau_{\text{net}}=I\alpha$$

For a constant torque, reduces to:

$$W=\tau(\theta_f-\theta_i)$$

We can relate work to power with equation:

$$P=\frac{dW}{dt}=\tau\omega$$