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

2. PHY 231 2 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!

3. PHY 231 3 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

4. PHY 231 4 examples: A lot more in the book! Where is the center of gravity?

5. PHY 231 5 question 0-212 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

6. PHY 231 6 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 1 0-0.5*w+0.75*T 2 =0 T 2 =0.5/0.75*w=2/3w T 1 =1/3w T 2 =2/3w

7. PHY 231 7  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 Tcos37 0 -2w+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

8. PHY 231 8 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

9. PHY 231 9 Demo: fighting sticks UNTIL HERE FOR MIDTERM II !!!! (until 8.4 in the book)

10. PHY 231 10 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

11. PHY 231 11 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.

12. PHY 231 12 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.

13. PHY 231 13 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 )

14. PHY 231 14 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

15. PHY 231 15 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.

16. PHY 231 16 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 )

17. PHY 231 17 The rotation axis matters! I=(  m i r i 2 ) =0.2*0.5 2 +0.3*0.5 2 + 0.2*0.5 2 +0.3*0.5 2 =0.5 kgm 2 I=(  m i r i 2 ) =0.2*0.+0.3*0.5 2 + 0.2*0+0.3*0.5 2 =0.3 kgm 2