Kinematics of rotational motion IB DP Physics Study Notes - 2025 Syllabus
Kinematics of rotational motion IB DP Physics Study Notes
Kinematics of rotational motion IB DP Physics Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand
- the torque $\tau$ of a force about an axis as given by $\tau = Fr \sin \theta$
- that bodies in rotational equilibrium have a resultant torque of zero
- that an unbalanced torque applied to an extended, rigid body will cause angular acceleration
- that the rotation of a body can be described in terms of angular displacement, angular velocity and angular acceleration
- that equations of motion for uniform angular acceleration can be used to predict the body’s angular position $\theta$, angular displacement $\Delta \theta$, angular speed $\omega$ and angular acceleration $\alpha$, as given by
\( \Delta \theta = \frac{\omega_f + \omega_i}{2} t \)
\( \omega_f = \omega_i + \alpha t \)
\( \Delta \theta = \omega_i t + \frac{1}{2} \alpha t^2 \)
\( \omega_f^2 = \omega_i^2 + 2 \alpha \Delta \theta \)
Standard level and higher level: There is no standard level content in A.4
Additional higher level: 7 hours
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
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Torque of a Force About an Axis
The torque \( \tau \) (also called the moment of a force) is a measure of how effectively a force causes a rotation about a pivot or axis.
It is given by the formula:
\( \tau = Fr \sin \theta \)
Where:
- \( F \) = applied force (in N)
- \( r \) = perpendicular distance from the axis to the line of action of the force (in m)
- \( \theta \) = angle between the force vector and the lever arm
Torque is a vector, and its direction is determined by the right-hand rule. SI unit: \( \text{N·m} \).
Example:
A 30 N force is applied at the end of a 0.5 m wrench at an angle of 60° to the handle. Calculate the torque about the pivot.
▶️ Answer/Explanation
Use the torque formula:
\( \tau = Fr \sin \theta = 30 \times 0.5 \times \sin(60^\circ) \)
\( = 15 \times 0.866 = \boxed{12.99 \, \text{N·m}} \)
Rotational Equilibrium
A body is in rotational equilibrium when it is either:
- Not rotating at all, or
- Rotating with a constant angular velocity (no angular acceleration)
The condition for rotational equilibrium is:
\( \sum \tau = 0 \)
This means the net torque acting on the body is zero—clockwise and anticlockwise torques balance out.
This is analogous to translational equilibrium where the net force is zero.
Example:
A horizontal beam is balanced on a fulcrum placed 2 m from the left end. A 20 N weight is suspended 1 m from the left end. Where should a 30 N weight be hung (on the right side of the fulcrum) to maintain equilibrium?
▶️ Answer/Explanation
Step 1: Set clockwise and anticlockwise torques equal
Anticlockwise torque = \( 20 \times 1 = 20 \, \text{N·m} \)
Let the 30 N weight be placed \( x \) meters from the fulcrum (on the right). Then:
Clockwise torque = \( 30x \)
Step 2: Equilibrium condition
\( 30x = 20 \Rightarrow x = \boxed{0.67 \, \text{m}} \)
The 30 N weight should be placed 0.67 m to the right of the fulcrum.
Unbalanced Torque and Angular Acceleration
When a net (unbalanced) torque acts on an object, it causes the object to experience angular acceleration.
This is the rotational analogue of Newton’s Second Law:
\( \tau_{\text{net}} = I \alpha \)
Where:
- \( \tau_{\text{net}} \) = net torque acting on the object (in N·m)
- \( I \) = moment of inertia of the object (in kg·m²)
- \( \alpha \) = angular acceleration (in rad/s²)
The larger the moment of inertia, the harder it is to rotate the object. The angular acceleration is directly proportional to the net torque and inversely proportional to the moment of inertia.
Example:
A constant torque of \( 6.0 \, \text{N·m} \) is applied to a wheel with moment of inertia \( 2.0 \, \text{kg·m}^2 \). Calculate the angular acceleration produced.
▶️ Answer/Explanation
Use the formula:
\( \tau = I \alpha \Rightarrow \alpha = \frac{\tau}{I} \)
\( \alpha = \frac{6.0}{2.0} = \boxed{3.0 \, \text{rad/s}^2} \)
The wheel experiences an angular acceleration of 3.0 rad/s².
Rotational Quantities: Angular Displacement, Angular Velocity, and Angular Acceleration
Rotational motion is described using angular analogues of linear quantities:
- Angular Displacement \( \theta \): The angle (in radians) through which a point or line has been rotated around a fixed axis.
- Angular Velocity \( \omega \): The rate of change of angular displacement with respect to time.
\( \omega = \frac{d\theta}{dt} \)
- Angular Acceleration \( \alpha \): The rate of change of angular velocity with respect to time.
\( \alpha = \frac{d\omega}{dt} \)
These quantities describe the rotational equivalent of displacement, velocity, and acceleration in linear motion. Their units are:
- \( \theta \): radians (rad)
- \( \omega \): radians per second (rad/s)
- \( \alpha \): radians per second squared (rad/s²)
These quantities are vectors and follow the right-hand rule for direction.
Example:
A rotating disk starts from rest and reaches an angular velocity of \( 12 \, \text{rad/s} \) in 4 seconds with constant angular acceleration. Find the angular acceleration and angular displacement during this time.
▶️ Answer/Explanation
Step 1: Use \( \omega = \omega_0 + \alpha t \)
\( \omega_0 = 0, \omega = 12, t = 4 \)
\( 12 = 0 + \alpha \cdot 4 \Rightarrow \alpha = \boxed{3 \, \text{rad/s}^2} \)
Step 2: Use \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \)
\( \theta = 0 + \frac{1}{2}(3)(4^2) = \boxed{24 \, \text{rad}} \)
Equations of Motion for Uniform Angular Acceleration
When a rigid body rotates with constant angular acceleration \( \alpha \), we can use angular kinematic equations analogous to linear motion. These equations allow prediction of:
- Angular displacement \( \theta \)
- Angular velocity \( \omega \)
- Angular acceleration \( \alpha \)
Assuming motion starts at \( t = 0 \), with initial angular velocity \( \omega_0 \), and constant \( \alpha \), the following equations hold:
- \( \omega = \omega_0 + \alpha t \)
- \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \)
- \( \omega^2 = \omega_0^2 + 2\alpha\theta \)
- \( \theta = \frac{\omega + \omega_0}{2} \cdot t \)
All quantities are rotational analogues:
- Displacement \( s \rightarrow \theta \)
- Velocity \( v \rightarrow \omega \)
- Acceleration \( a \rightarrow \alpha \)
Units:
- \( \theta \): radians
- \( \omega \): rad/s
- \( \alpha \): rad/s²
Example:
A fan blade starts from rest and makes 5 complete revolutions in 2 seconds with constant angular acceleration. Find (a) angular acceleration, and (b) final angular velocity.
▶️ Answer/Explanation
Step 1: Convert revolutions to radians
\( \theta = 5 \times 2\pi = 10\pi \, \text{rad} \)
Step 2: Use \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \)
\( \omega_0 = 0 \Rightarrow 10\pi = \frac{1}{2} \alpha (2)^2 \Rightarrow 10\pi = 2\alpha \Rightarrow \alpha = \boxed{5\pi \, \text{rad/s}^2} \)
Step 3: Use \( \omega = \omega_0 + \alpha t = 0 + 5\pi \cdot 2 = \boxed{10\pi \, \text{rad/s}} \)