Rotation - Unit 5

Rotation - Unit 5 Rotational motion is based on completely different quantities for certain things. We use 𝜃 as the rotational equivalent for position (in radians), 𝜔 as the rotational equivalent for velocity (angular frequency/velocity in radians per second) and 𝛼 as the rotational equivalent for acceleration. If you have a spinning object and its tangential velocity is v, then the angular velocity is related to the cross product with position vector r. 𝑣 =𝜔×𝑟 There is also the scalar form. 𝑣 = 𝑟𝜔 Similarly, we can also find the relation between angular acceleration and tangential acceleration. 𝑎𝜃 = 𝑟𝛼 Relations Between Each Quantity Each quantity is related in very similar ways to how rectilinear quantities (position, velocity and acceleration) are related. 𝛼 = 𝜔̇ = 𝜃̈ Rotational Kinematics With constant 𝛼, just like with rectilinear motion, we can have kinematic equations that are basically the same. 𝜃 1 = 𝜃0 + 𝜔0 𝑡 + 𝛼𝑡2 2 𝜃 𝜔 2 = 𝜃0 + 𝜔0 + 𝜔 𝑡 2 = 𝜔02 + 2𝛼 (𝜃 − 𝜃0 ) Remember: this is only for constant 𝛼. Without a constant angular acceleration, you need to integrate. However, where it is constant, you’re golden! For example: an object moves in a circle with angular acceleration of 8.0 rad/s2. It starts from rest. How long does it take for the object to go in a circle? We can just use a kinematics equation and substitute in what we know. 𝜃 𝜃0 1 = 𝜃0 + 𝜔0 𝑡 + 𝛼𝑡2 2 = 0, 𝜔0 = 0, 𝜃 = 𝜋, 𝛼 = 4.0𝜋 𝑡=√ 2𝜃 1 = 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 𝛼 2 Torque Torque is the tendency for a force to change the rotational motion of a body. A force acting at distance r from a fulcrum or pivot that applies at least some force tangentially will provide torque. When applying a force Fa at a distance r and at an angle 𝜙 to the body, then we can find the torque it provides. 𝜏 = 𝑟𝐹𝑎 𝑠𝑖𝑛 𝑠𝑖𝑛 𝜙 There is also the vector form of the equation. 𝜏 = 𝑟 × 𝐹𝑎 Note: - A force applied only towards the pivot (no tangential force) will not provide a torque. - A force applied at the pivot (r = 0) will not provide any torque, either. It shows in the equation: if r = 0, then the torque will always be zero. We can also relate torque to angular acceleration. Just like F = ma, we have a relation for the rotational equivalent. 𝜏 = 𝐼𝛼 I is the rotational inertia of the body. Don’t worry, we cover this soon. Rotational Inertia/Moment of Inertia The moment of inertia of a body is basically the rotational equivalent of mass. If you have a single point of mass at a distance r from a pivot, then the rotational inertia is given by: 𝐼 = 𝑟2𝑚 For a collection of particles, you just add them up. 𝐼=∑ 𝑟𝑖2 𝑚𝑖 For a continuous distribution of mass, this extends to an integral. 𝐼=∫ 𝑟 2 𝑑𝑚 It’s used absolutely everywhere when dealing with rotational problems. You can see how it acts as an equivalent to mass. For example: 𝐹 = 𝑚𝑎 ⟶ 𝜏 = 𝐼𝛼 To calculate moment of inertia for continuous distributions of mass, you need to set up integrals for the mass and shape. However, usually, if the College Board doesn’t want you to specifically calculate the moment of inertia, they’ll just give it to you. Rotational Momentum Rotational momentum just describes the overall rotational state of a system, just like how linear momentum describes the overall translational state of the system. Just like linear momentum, rotational momentum is related to rotational inertia and angular velocity, and its rate of change is torque. Angular momentum is described most basically as the cross product of the radius and momentum. 𝐿 =𝑟×𝑝 Let’s derive the relationship with angular velocity! Given a particle of mass m, travelling at speed v at a distance r from a center: 𝐿 = 𝑟 × (𝑚𝑣) = 𝑚𝑟 × (𝜔 × 𝑟) = 𝑚𝑟 2 𝜔 = 𝐼𝜔 We can really see the equivalence with linear concepts here. 𝑝 = 𝑚𝑣 ⟶ 𝐿 = 𝐼𝜔 𝐹= 𝑑𝑝 𝑑𝐿 ⟶𝜏= 𝑑𝑡 𝑑𝑡 Rotational Kinetic Energy Rotational kinetic energy is derived either by summing (or integrating) the kinetic energy of the individual particles in the rotating physical system. You get the expression for rotational kinetic energy. 1 𝐾 = 𝐼𝜔2 2 We also have expressions for work done by torque. For constant torque: 𝑊 = 𝐹𝑑 ⟶ 𝑊 = 𝜏𝜃 For torque that isn’t constant: 𝑊=∫ 𝐹𝑑𝑥 ⟶ 𝑊 = ∫ 𝜏𝑑𝜃 The total amount of kinetic energy in a rotating system of particles is given by the sum of the rotational kinetic energy about the system’s center of mass and any translational (linear) kinetic energy. Rolling Rolling without slipping is a very special type of rolling: it means that the point of contact of the rolling object with the ground isn’t moving. Generally, it means that there is static friction between the object and the ground. This means that 𝑣 = 𝑟𝜔, and that 𝑎 = 𝑟𝛼. You can use this to your advantage. For example: If you have a cylinder that rolls without slipping down a slope with a known change in elevation, you can find both the translational and rotational velocity using the mechanical energy: you know the exact relation between the linear and rotational velocity, so you can make an expression for the energy. 1 1 𝑣 2 2 𝑚𝑔ℎ = 𝑚𝑣 + 𝐼 ( ) 2 2 𝑟 However, do note: not all rolling is rolling without slipping. You can ONLY consider something to be rolling without slipping if the problem says that it is rolling without slipping, or you can determine through other means that 𝑣 = 𝑟𝜔. Otherwise, you can not rely on this relationship to determine angular velocity from linear velocity, and must rely on angular acceleration, torque, etc.