AP Physics C Mechanics- 5.5 Rotational Equilibrium and Newton’s First Law in Rotational Form- Study Notes- New Syllabus
AP Physics C Mechanics- 5.5 Rotational Equilibrium and Newton’s First Law in Rotational Form – Study Notes
AP Physics C Mechanics- 5.5 Rotational Equilibrium and Newton’s First Law in Rotational Form – Study Notes – per latest Syllabus.
Key Concepts:
- Rotational Equilibrium and Newton’s First Law for Rotation
- Rotational Corollary to Newton’s Second Law
Rotational and Translational Equilibrium
A system may exhibit rotational equilibrium (constant angular velocity) without being in translational equilibrium, and vice versa. These are two distinct types of equilibrium:
- Translational equilibrium: Occurs when the net force on the system is zero (\( \mathrm{\sum \vec{F} = 0} \)), meaning the center of mass has no linear acceleration.
- Rotational equilibrium: Occurs when the net torque on the system is zero (\( \mathrm{\sum \tau = 0} \)), meaning the system’s angular velocity is constant.
Key Idea: An object can rotate steadily (constant \( \mathrm{\omega} \)) while still accelerating linearly, or vice versa. Both forms of equilibrium are independent of one another.
Example
A wheel mounted on a frictionless axle spins at a constant angular velocity while being pulled horizontally by a rope with tension \( \mathrm{T} \). Explain whether the wheel is in translational and/or rotational equilibrium.
▶️ Answer / Explanation
Step 1: The wheel spins with constant angular velocity, so \( \mathrm{\alpha = 0} \) and \( \mathrm{\sum \tau = 0} \). It is in rotational equilibrium.
Step 2: However, the tension force \( \mathrm{T} \) on the wheel causes a net horizontal acceleration of its center of mass, so \( \mathrm{\sum \vec{F} \neq 0} \). Therefore, it is not in translational equilibrium.
Result: The wheel is in rotational equilibrium but not in translational equilibrium.
Force and Torque Diagrams
Free-body diagrams and force diagrams are used to represent the magnitudes, directions, and points of application of forces on an object. When analyzing rotational motion:
- Each force is represented as a vector applied at its point of contact on the object.
- The line of action of each force helps determine its contribution to torque about the chosen axis.
- The direction of each torque (clockwise or counterclockwise) is labeled according to a sign convention.
Key Idea: These diagrams allow for clear visualization of both the translational and rotational effects of forces acting on a rigid body.
Example
A uniform meter stick of weight \( \mathrm{10\,N} \) is balanced horizontally on a fulcrum placed at its center. A \( \mathrm{2\,N} \) weight is hung at the left end. Where must a \( \mathrm{3\,N} \) weight be hung on the right side to keep the stick balanced?
▶️ Answer / Explanation
Step 1: Choose the fulcrum as the axis of rotation. In equilibrium, \( \mathrm{\sum \tau = 0} \).
Step 2: Let \( \mathrm{x} \) be the distance of the \( \mathrm{3\,N} \) weight from the center.
\( \mathrm{(2\,N)(0.50\,m) = (3\,N)(x)} \)
Step 3: Solve for \( \mathrm{x} \):
\( \mathrm{x = \dfrac{(2)(0.50)}{3} = 0.33\,m} \)
Result: The \( \mathrm{3\,N} \) weight must be placed \( \mathrm{0.33\,m} \) from the fulcrum on the right side for equilibrium.
Condition for Rotational Equilibrium
Rotational equilibrium occurs when all torques acting on a system balance so that the net torque is zero:
\( \mathrm{\sum \tau_i = 0} \)
- Clockwise torques are taken as negative, counterclockwise torques as positive (or vice versa, depending on convention).
- If \( \mathrm{\sum \tau_i = 0} \), the system has no angular acceleration (\( \mathrm{\alpha = 0} \)).
Key Idea: Rotational equilibrium means there is no change in rotational motion — angular velocity remains constant (which may be zero or nonzero).
Rotational Analog of Newton’s First Law
Newton’s First Law of Motion states that an object maintains constant velocity unless acted upon by a net external force. The rotational analog of this law is:
A system will have a constant angular velocity only if the net torque exerted on it is zero.
Equation Form:
\( \mathrm{\sum \tau = 0 \Rightarrow \alpha = 0} \)
Key Idea: Just as forces change linear motion, torques change rotational motion. If no net torque acts, angular velocity remains constant — including the special case of being at rest.
Example
A uniform beam of length \( \mathrm{3.0\,m} \) and weight \( \mathrm{60\,N} \) is supported horizontally by a hinge at one end and a vertical cable at the other. Find the tension in the cable if the beam is in rotational equilibrium.
▶️ Answer / Explanation
Step 1: Choose the hinge as the axis of rotation. The beam is in rotational equilibrium, so \( \mathrm{\sum \tau = 0} \).
Step 2: Identify torques:
- The tension \( \mathrm{T} \) at the far end of the beam creates a counterclockwise torque: \( \mathrm{\tau_T = T(3.0)} \).
- The beam’s weight \( \mathrm{W = 60\,N} \) acts at its center, \( \mathrm{1.5\,m} \) from the hinge, producing a clockwise torque: \( \mathrm{\tau_W = W(1.5)} \).
Step 3: Apply rotational equilibrium condition:
\( \mathrm{\sum \tau = 0 \Rightarrow T(3.0) – 60(1.5) = 0} \)
Step 4: Solve for \( \mathrm{T} \):
\( \mathrm{T = \dfrac{60(1.5)}{3.0} = 30\,N} \)
Result: The tension in the cable is \( \mathrm{30\,N} \). The system is in rotational equilibrium because torques balance each other.
Rotational Corollary to Newton’s Second Law
The rotational corollary to Newton’s Second Law states that when the torques exerted on a rigid system are not balanced, the system experiences a change in angular velocity — that is, it undergoes an angular acceleration.
Mathematically, the net torque acting on a rigid body is proportional to its angular acceleration and is given by:
\( \mathrm{\sum \tau = I\alpha} \)
- \( \mathrm{\sum \tau} \): net torque acting on the system (in \( \mathrm{N·m} \))
- \( \mathrm{I} \): rotational inertia (in \( \mathrm{kg·m^2} \))
- \( \mathrm{\alpha} \): angular acceleration (in \( \mathrm{rad/s^2} \))
Vector Form:
\( \mathrm{\sum \vec{\tau} = I\vec{\alpha}} \)
The direction of \( \mathrm{\vec{\alpha}} \) (angular acceleration) is along the axis of rotation, determined by the right-hand rule, and it matches the direction of the net torque.
Key Idea:
- A nonzero net torque causes a change in angular velocity (rotational acceleration or deceleration).
- If torques are balanced (\( \mathrm{\sum \tau = 0} \)), the system rotates at a constant angular velocity or remains at rest.
- This is the rotational analog of Newton’s Second Law for linear motion, \( \mathrm{\sum F = ma} \).
Analogous Relationship:
| Linear Motion | Rotational Motion |
|---|---|
| \( \mathrm{\sum F = ma} \) | \( \mathrm{\sum \tau = I\alpha} \) |
| Force causes linear acceleration | Torque causes angular acceleration |
| Mass resists changes in linear motion | Moment of inertia resists changes in rotation |
Example
A solid disk of mass \( \mathrm{2.0\,kg} \) and radius \( \mathrm{0.40\,m} \) is free to rotate about its center. A constant tangential force of \( \mathrm{3.0\,N} \) is applied at its rim. Find the angular acceleration of the disk.
▶️ Answer / Explanation
Step 1: Identify known quantities:
- \( \mathrm{F = 3.0\,N} \)
- \( \mathrm{r = 0.40\,m} \)
- \( \mathrm{I = \tfrac{1}{2}MR^2 = \tfrac{1}{2}(2.0)(0.40)^2 = 0.16\,kg·m^2} \)
Step 2: Calculate the torque:
\( \mathrm{\tau = rF = (0.40)(3.0) = 1.2\,N·m} \)
Step 3: Apply Newton’s second law for rotation:
\( \mathrm{\sum \tau = I\alpha \Rightarrow \alpha = \dfrac{\tau}{I}} \)
\( \mathrm{\alpha = \dfrac{1.2}{0.16} = 7.5\,rad/s^2} \)
Step 4: Interpret the result:
The positive \( \mathrm{\alpha} \) indicates that the angular velocity increases in the direction of the applied torque.
Result: The disk’s angular acceleration is \( \mathrm{7.5\,rad/s^2} \).