Rotational motion

• If the object has a regular (symmetrical) shape, then the axis will be halfway through the object, and the centre of mass will be at the halfway point of that line. If the object has a weird shape, the axis and centre of mass might not be exactly in the middle, but usually close by.

  
• Special terms (translational = linear)

  
• Equations:

  -> ω =Δθ/Δt

  -> α = Δω/Δt

  
  

  Δω = ωf - ωi
• When a whole object is rotating, each of the points on the object have circular, i.e. translational, motion. The individual points in a rotating object travel in circles with linear speed v = rω. (merry go round, whole thing has rotational, individual people have circular motion)

  
  

  All points in a rotating object have the same ω, but their v depends on distance from the centre, r

  v = rω
• Rolling objects have both linear and rotational velocity.
•  link linear and rotational acceleration:

  a = rα

  (a = acceleration, α = angular acceleration)
• rotational kinematics

  ωf = ωi + α x t

  ωf$^2$ = ωi$^2$ + 2α x θ

  θ = (ωi + ωf /2)× t

  θ = ωi x t + 1/2 x α x t$^2$
• Torque

  Think of torque as a twisting force. Torque represented by tau (τ) is what happens when force x by distance between it and pivot point (τ = Fd) in Nm.

  
  

  Equations:

  F = ma

  τ = Iα (I = rotational inertia)
• Rotational inertia


  Think of rotational inertia (I) as the rotational version of mass. It is measured in kgm$^2$. The harder something is to start rotating (or stop once it’s going), the larger its rotational inertia.

  
  

  Two things increase rotational inertia: more mass, and that mass being further away from the centre of mass. 

  
  

  More mass further away from the rotation axis (COM) means it is harder to get rotating. The rotational inertia of an object is always I=kmr$^2$, m is the mass, r is radius, k is a constant.
• Rotational inertia question Part 1

  A cord is wrapped around the edge of a wheel of radius 0.4 m. A 0.5 kg mass is attached to the end of the cord, as shown. The mass has a downward acceleration of 0.2 ms$^-$$^2$. Find the rotational inertia of the wheel.

  
  

  Normally object accelerates down at 9.81ms$^-$$^2$, but this is slower so upwards force is slowing it down.

  
• Rotational inertia question part 2

  Tension force upwards:

  Fnet = Fg - Ftension => (0.5 x 9.81) - (0.5 x 0.2) = 4.805N

  
  

  (Newtons law, every force as an equal and opposite reaction) so tension force also applied in opposite direction causing torque:

  τ = Fr = 4.805 × 0.4 = 1.922 Nm

  
  

  need angular acceleration to find rotational inertia:

  a = αr = 0.2 /0.4 = 0.5 rad s−$^2$
• Rotational inertia question part 3


  Find rotational inertia:

  I =τ / α = 1.922 / 0.5 = 3.84 kgm$^2$
• Angular momentum


  Has symbol L and unit kgm$^2$s$^-$$^1$

  
  

  Equations to find angular momentum:

  L = I x ω (p = m x v)

  L = m x v x r (mv from linear multiplied by r to get angular)
• Angular momentum

  (linear momentum is conserved when there's no external forces.)

  
  

  Torque is the rotational version of force, so angular momentum is conserved as long as there are no external torques, i.e. the net torque on the object/system is zero.

  
  

  This means that total angular momentum will stay the same over time, i.e. Li = Lf
• Rotational kinetic energy formula:

  EK(ROT) = 1/2 x I x ω$^2$

   

  -> It takes more energy to spin up objects that

  have high rotational inertia.

  -> It also takes more energy to make an object spin faster. 
• Rolling objects have both linear and rotational kinetic energy use v = r x ω
• Example question part 1

  A ball has mass 7.0 kg and radius 10 cm. It rolls down a 50 cm ramp, converting all potential energy into kinetic energy. It has a speed of 2.5 ms$^-$$^1$ at the bottom of the ramp. Find its rotational inertia.

  
  

  First find potential energy: Ep = mgh = 7 x 9.81 x 0.5 = 34.335 J
• Example question part 2

  Find linear component at bottom and use to find rotational component:

  Ek(LIN) = 1/2mv$^2$ = 1/2 x 7 x 2.5$^2$ = 21.875J

  Ek(ROT) = Etotal - Ek(LIN) = 34.335 - 21.875 = 12.46J
• Example question part 3

  To find rotational inertia, rearrange rotational kinetic energy equation after finding angular velocity (ω):

  v = rω -> ω = v/r = 2.5/0.1 = 25 rad s$^-$$^1$

  Ek(ROT) = 1/2Iω$^2$

  I = 2Ek(ROT)/ω$^2$ = 2 x 12.46/25$^2$ = 0.040 kmg$^2$
• Objects with the same mass but difference rotational inertias will roll differently.

  Disc has lots of mass away from middle axis so it has a larger rotational inertia. Ball has less rotational inertia than disc but more than rod.

  All the potential energy gets converted into a mixture of rotational and linear kinetic energy. The more energy needed for rotation, the less energy available for linear velocity. Therefore, the rod would have more EK(LIN) and therefore be moving

  faster. The disc would be the slowest.