PHY PHYSICS 231 Lecture 19: More about rotations Remco Zegers Walk-in hour: Thursday 11:30-13:30 am Helproom Demo: fighting sticks

PHY What we did so far Translational equilibrium:  F=ma=0 The center of gravity does not move! Rotational equilibrium:   =0 The object does not rotate Mechanical equilibrium:  F=ma=0 &   =0 No movement! Torque:  =Fd Center of Gravity: Demo: Leaning tower Demo: turning screws

PHY r F t =ma t Torque and angular acceleration m F Newton 2nd law: F=ma F t r=mra t F t r=mr 2  Used a t =r   =mr 2  Used  =F t r The angular acceleration goes linear with the torque. Mr 2 =moment of inertia

PHY Two masses r m F m r  =mr 2   =(m 1 r 1 2 +m 2 r 2 2 )  If m 1 =m 2 and r 1 =r 2  = 2mr 2  Compared to the case with only one mass, the angular acceleration will be twice smaller when applying the same torque, if the mass increases by a factor of two. The moment of inertia has increased by a factor of 2.

PHY Two masses at different radii r m F m r  =mr 2   =(m 1 r 1 2 +m 2 r 2 2 )  If m 1 =m 2 and r 2 =2r 1  = 5mr 2  When increasing the distance between a mass and the rotation axis, the moment of inertia increases quadraticly. So, for the same torque, you will get a much smaller angular acceleration.

PHY A homogeneous stick Rotation point m m m m m m m m m m F  =mr 2   =(m 1 r 1 2 +m 2 r 2 2 +…+m n r n 2 )   =(  m i r i 2 )   =I  Moment of inertia I: I=(  m i r i 2 )

PHY Two inhomogeneous sticks m m m m m m m m 5m F m m m m m m m m F 18m  =(  m i r i 2 )   118mr 2   =(  m i r i 2 )   310mr 2  r Easy to rotate!Difficult to rotate

PHY More general.  =I  Moment of inertia I: I=(  m i r i 2 )

PHY A simple example A and B have the same total mass. If the same torque is applied, which one accelerates faster? F F r r Answer: A  =I  Moment of inertia I: I=(  m i r i 2 )

PHY The rotation axis matters! I=(  m i r i 2 ) =0.2* * * *0.5 =0.5 kgm 2 I=(  m i r i 2 ) =0.2*0.+0.3* *0+0.3*0.5 =0.3 kgm 2

PHY Extended objects (like the stick) I=(  m i r i 2 ) =(m 1 +m 2 +…+m n )R 2 =MR 2 M

PHY Some common cases

PHY Example A monocycle (bicycle with one wheel) has a wheel that has a diameter of 1 meter. The mass of the wheel is 5 kg (assume all mass is sitting at the outside of the wheel). The friction force from the road is 25 N. If the cycle is accelerating with 0.3 m/s, what is the force applied on each of the paddles if the paddles are 30 cm from the center of the wheel? 25N 0.5m 0.3m F  =I   =a/r so  =0.3/0.5=0.6 rad/s I=(  m i r i 2 )=MR 2 =5(0.5 2 )=1.25 kgm 2  friction =-25*0.5=-12.5  paddles =F*0.3+F*0.3=0.6F 0.6F-12.5=1.25*0.6, so F=22.1 N

PHY Lon-capa You can now do all problems up to and including Look at the lecture sheets of last 2 lectures and this one for help!!

PHY Rotational kinetic energy Consider a object rotating with constant velocity. Each point moves with velocity v i. The total kinetic energy is: KE r =½I  2 Conservation of energy for rotating object: [PE+KE t +KE r ] initial = [PE+KE t +KE r ] final

PHY Example. 1m Consider a ball and a block going down the same 1m-high slope. The ball rolls and both objects do not feel friction. If both have mass 1kg, what are their velocities at the bottom (I.e. which one arrives first?). The diameter of the ball is 0.4 m. Block: [½mv 2 +mgh] initial = [½mv 2 +mgh] final 1*9.8*1 = 0.5*1*v 2 so v=4.4 m/s Ball: [½mv 2 +mgh+½I  2 ] initial = [½mv 2 +mgh+½I  2 ] final I=0.4*MR 2 =0.064 kgm 2 and  =v/R=2.5v 1*9.8*1 = 0.5*1*v *0.064*(2.5v) 2 so v=3.7 m/s Part of the energy goes to the rotation: slower!

PHY Lon-capa You can now do all problems up to and including Next lecture: Last part of chapter 8 and examples.