PHY 231 1 PHYSICS 231 Lecture 18: equilibrium and more rotations Remco Zegers Walk-in hour: Tuesday 4:00-5:00 pm Helproom.

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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 T: period(time to make one rotation) of planet r: distance object to planet Our solar system!

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 question m 20N 40N A wooden bar is initially balanced. Suddenly, 3 forces are applied, as shown in the figure. Assuming that the bar can only rotate, what will happen (what is the sum of torques)? a)the bar will remain in balance b)the bar will rotate counterclockwise c)the bar will rotate clockwise Torque:  =Fd

PHY Weight of board: w What is the tension in each of the wires (in terms of w)? w T1T1 T2T2 0 Translational equilibrium  F=ma=0 T 1 +T 2 -w=0 so T 1 =w-T 2 Rotational equilibrium   =0 T *w+0.75*T 2 =0 T 2 =0.5/0.75*w=2/3w T 1 =1/3w T 2 =2/3w

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? w snsn n w T TyTy TxTx (x=0,y=0) Translational equilibrium (Hor.)  F x =ma=0 n-T x =n-Tcos37 o =0 so n=Tcos37 o Translational equilibruim (vert.)  F y =ma=0  s n-w-w+T y =0  s n-2w+Tsin37 o =0  s Tcos w+Tsin37 0 =0 1.00T=2w Rotational equilibrium:   =0 xw+2w-4Tsin37 0 =0 so w(x+2-4.8)=0 x=2.8 m

PHY another example 12.5N 30N 0.2L 0.5L L Does not move! What is the tension in the tendon? Rotational equilibrium:  T =0.2LTsin(155 o )=0.085LT  w =0.5L*30sin(40 o )=-9.64L  F =L*12.5sin(40 0 )=-8.03L  =-17.7L+0.085LT=0 T=208 N

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