1. Equilibrium and Dynamics
PHYSICS 220
Lecture 14
Equilibrium and Dynamics
Lecture 14
Purdue University, Physics 220

2. Purdue University, Physics 220
Lecture 14
Purdue University, Physics 220

3. Purdue University, Physics 220
Find Center of Mass
Center of mass
pivot d
W=mg
Torque about pivot  0
Not in equilibrium
Center of mass
pivot
Torque about pivot = 0
Equilibrium
A method to find center of mass of an irregular object
Lecture 14
Purdue University, Physics 220

4. Purdue University, Physics 220
Exercise
A 50 kg diver stands at the end of a 4.6 m diving board. Neglecting the weight of the board, what is the force on the pivot 1.2 meters from the end?
1) Draw FBD
2) Choose Axis of rotation
3) t = Rotational EQ
F1 (1.2) – mg (4.6) = 0
F1 = 4.6 (50 *9.8) / 1.2
F1 = 1880 N
4) F = Translational EQ
F1 – F2 – mg = 0
F2 = F1 – mg = 1390 N mg F1 F2
Lecture 14
Purdue University, Physics 220

5. Purdue University, Physics 220
Exercise
A 75 kg painter stands at the center of a 50 kg 3 meter plank. The supports are 1 meter in from each edge. Calculate the force on support A.
1 meter
1 meter A B FA FB Mg mg
1) Draw FBD
2) F = 0
3) Choose pivot
4)  = 0
1 meter
0.5meter
Painter homework problem. Get demo here have man stand at several places calculate force on each support. Right above support, middle
FA + FB – mg – Mg = 0
-FA (1) sin(90)+ FB (0) sin(90) + mg (0.5)sin(90) + Mg(0.5) sin(90) = 0
FA = 0.5 mg Mg = N
Lecture 14
Purdue University, Physics 220

6. Purdue University, Physics 220
iClicker
If the painter moves to the right, the force exerted by support A
A) Increases B) Unchanged C) Decreases
1 meter
1 meter
Painter homework problem. Get demo here have man stand at several places calculate force on each support. Right above support, middle A B
Lecture 14
Purdue University, Physics 220

7. Purdue University, Physics 220
Question
How far to the right of support B can the painter stand before the plank tips?
1 meter
1 meter A B
Just before board tips, force from A becomes zero FB Mg mg
1) Draw FBD
2) F = 0
3) Choose pivot
4)  = 0
Painter homework problem. Get demo here have man stand at several places calculate force on each support. Right above support, middle
FB – mg – Mg = 0
0.5meter x
FB (0) sin(90) + mg (0.5)sin(90) – Mg(x) sin(90) = 0
0.5 m = x M -> x = 0.3 meter
Lecture 14
Purdue University, Physics 220

8. Spool on a Rough Surface
Consider all of the forces acting: tension T and friction f.
Using FNET = 0 in the x direction: a b  T f y x
Using NET = 0 about the CM axis:
Solving:
Lecture 14
Purdue University, Physics 220

9. Spool on a Rough Surface
There is another (slick) way to see this:
Consider the torque about the point of contact between the spool and the ground. We know the net torque about this (or any other) point is zero.
Since both Mg and f act through this point, they do not contribute to the net toque.
Therefore the torque due to T must also be zero.
Therefore T must act along a line that passes through this point! T  a y x b Mg f
Lecture 14
Purdue University, Physics 220

10. Spool on a Rough Surface
So we can use geometry to get the same result T  a  b
Lecture 14
Purdue University, Physics 220

11. Purdue University, Physics 220
Rotational Dynamics
t = I a
Torque is amount of twist provide by a force
Signs: positive = CCW
Moment of Inertial like mass. Large I means hard to start or stop from spinning.
Angular acceleration is defined as the rate at which the angular velocity changes.
Problems Solving:
Draw free body diagram
Pick an axis
Compute torque due to each force
Take Eric’s version
Lecture 14
Purdue University, Physics 220

12. Falling Weight & Pulley
A mass m is hung by a string that is wrapped around a pulley of radius R attached to a heavy flywheel. The moment of inertia of the pulley + flywheel is I. The string does not slip on the pulley. Starting at rest, how long does it take for the mass to fall a distance L.  I R T m a mg L
Lecture 14
Purdue University, Physics 220

13. Falling Weight & Pulley
Using 1-D kinematics we can solve for the time required for the weight to fall a distance L:  I R T
Careful to point out the different R’s that exist. R for pulley where rope is attached, and r for wheel, which is hidden in I. m a mg L
Lecture 14
Purdue University, Physics 220

14. Falling Weight & Pulley
For the hanging mass use F = ma
mg - T = ma
For the flywheel use t = I
 = TR sin(90) = I
Realize that a = R
Now solve for a using the above equations.  I R T m
Transparency? a mg L
Lecture 14
Purdue University, Physics 220

15. Purdue University, Physics 220
Rolling
An object with mass M, radius R, and moment of inertia I rolls without slipping down a plane inclined at an angle  with respect to horizontal. What is its acceleration?
Consider CM motion and rotation about the CM separately when solving this problem I M R 
Lecture 14
Purdue University, Physics 220

16. Purdue University, Physics 220
Rolling
Static friction f causes rolling. It is an unknown, so we must solve for it.
First consider the free body diagram of the object and use FNET = Macm :
In the x direction Mg sin  - f = Macm
Now consider rotation about the CM and use  = I realizing that
 = Rf and acm = R M y x f Mg  R 
Lecture 14
Purdue University, Physics 220

17. Purdue University, Physics 220
Rolling
We have two equations: Mg sin  - f = Ma
We can combine these to eliminate f: I
For a sphere: a M R 
Lecture 14
Purdue University, Physics 220

18. Purdue University, Physics 220
Energy Conservation
Friction causes an object to roll, but if it rolls w/o slipping friction does NO work!
No dissipated work, energy is conserved
Need to include both translation and rotation kinetic energy.
KE = ½ m vcm2 + ½ I 2
Lecture 14
Purdue University, Physics 220

19. Translational + Rotational KE
Consider a cylinder with radius R and mass M, rolling w/o slipping down a ramp. Determine the ratio of the translational to rotational KE.
Translational: KET = ½ M Vcm2
Rotational: KER = ½ I w2
use
and
KER = ½ (½ M R2) (Vcm/R)2
= ¼ M Vcm2
= ½ KET
Finish up H
Lecture 14
Purdue University, Physics 220

20. Purdue University, Physics 220
iClicker
A hoop and a cylinder of equal mass roll down a ramp with height h. Which has greater kinetic energy at bottom?
A) Hoop B) Same C) Cylinder
Lecture 14
Purdue University, Physics 220

21. Rolling Race (Hoop vs Cylinder)
A hoop and a cylinder of equal mass roll down a ramp with height h. Which has greater speed at the bottom of the ramp?
A) Hoop B) Same C) Cylinder
I = MR2
I = ½ MR2
Cylinder will get to the bottom first because inertia for a cylinder is less than that for a hoop type object.
Lecture 14
Purdue University, Physics 220