PHY PHYSICS 231 Lecture 18: equilibrium and more rotations Remco Zegers Walk-in hour: Tuesday 4:00-5:00 pm Helproom

PHY gravitation Only if an object is near the surface of earth one can use: F gravity =mg with g=9.81 m/s 2 In all other cases: F gravity = GM object M planet /r 2 with G=6.67E-11 Nm 2 /kg 2 This will lead to F=mg but g not equal to 9.8 m/s 2 (see Previous lecture!) If an object is orbiting the planet: F gravity =ma c =mv 2 /r=m  2 r with v: linear velocity  =angular vel. So: GM object M planet /r 2 = mv 2 /r=m  2 r Kepler’s 3 rd law: T 2 =K s r 3 K s =2.97E-19 s 2 /m 3 r: radius of planet T: period(time to make one rotation) of planet Our solar system!

PHY A child of 40 kg is sitting in a Ferris wheel, rotating with an angular velocity of 0.4 rad/s. The radius of the wheel is 9 m. What is the force exerted by the seat on the child at the top and at the bottom.

PHY Previously 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

PHY examples: A lot more in the book! Where is the center of gravity?

PHY Weight of board: w What is the tension in each of the wires (in terms of w)? w T1T1 T2T2 0

PHY  s =0.5 coef of friction between the wall and the 4.0 meter bar (weight w). What is the minimum x where you can hang a weight w for which the bar does not slide?

PHY another example 12.5N 30N What is the tension in the tendon?

PHY Demo: fighting sticks UNTIL HERE FOR MIDTERM II !!!! (until 8.4 in the book)

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 ) compare with: F=ma The moment of inertia in rotations is similar to the mass in Newton’s 2 nd law.

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 2 =0.5 kgm 2 I=(  m i r i 2 ) =0.2*0.+0.3* *0+0.3*0.5 2 =0.3 kgm 2