Lecture 15 Physics 111 Results from Exam #2 Chapter 8: Rotational Equilibrium and Rotational Dynamics Monday, October 26, 1998 Physics 111

The Physics 111 Help Session None This Week Mondays 5:00 - 6:30 pm 8:00 - 9:00 pm NSC Room 118

Exam 2 Generally, most of you did better this time! The grading has been completed. However, there are still a couple of students who have not completed the exam yet. So, I will return them on Wednesday. If you’re dying to know how you did, stop by my office--I have scores for those of you who took the exam.

D C B A If I had to put letter grades on this exam...

Chapter 8 Rotational Dynamics We’ve already started down this path, looking at the motion of objects that rotate. We’re now going to formalize and generalize the types of motion we consider using a few new concepts.

Let’s look at a balance scale... X? 10 g 50 g If I hang a 10 g mass at the end of the left side of the balance... …where should I hang the 50 g mass to get the scale to balance?

How did we arrive at such an answer? 10 g L 50 g L/5 How did we arrive at such an answer? Perhaps we used one of these formulae:

Let’s restate our balance condition so that we 10 g L 50 g L/5 Let’s restate our balance condition so that we get all the subscript 1’s on the same side and all the subscript 2’s on the same side...

98 L = 98 L L/5 L So in the case of this balance, we have... 10 g L 50 g L/5 So in the case of this balance, we have... 98 L = 98 L Good! The balance is in balance!

The scale rotates around the pivot point. 5 g L 50 g L/5 What happens to our balance if I change the mass on the left side from 10 g to 5 g? 5 g L 50 g L/5 The scale rotates around the pivot point.

X = L/2 Let’s look at one more case with our balance... L L X? 10 g L L 5 g 10 g X? Now where should I hang the 10 g mass to get the scale to balance? How did we arrive at this answer? X = L/2

First, we examined the left side of the balance: 10 g L X? 5 g First, we examined the left side of the balance: Then, we examined the right side:

If the scale is balanced, the two sides should equal one another: 10 g L X? 5 g If the scale is balanced, the two sides should equal one another:

In our scale experiments, we’ve been playing around with a physical quantity that we’ve not yet encountered formally. Our experiments show us that this quantity is related to a force applied at a distance. The farther away from the pivot point the force is applied, the greater our new quantity is. When present and unbalanced, our new physical quantity cause the scale to rotate in the direction of the unbalanced force.

We’ve also seen that the quantity is additive. That is, if we have two forces on one side of the pivot at two different distances, the resulting physical quantity is simply the sum of the two force * distance products. Before we name our new physical quantity, let’s examine one more thought experiment.

Let’s put a mass on one side of our balance and let go. What will eventually happen to this system? (Assume there is SOME friction in the pivot, but that the pivot is free to rotate 360o.)

The mass eventually comes to rest with the balance aligned vertically. We still have a mass of 50 g at a distance L from the pivot point, so why has the motion stopped? What has happened to the value of our new physical quantity?

that the gravitational force on the mass is acting along For this case, we notice that the gravitational force on the mass is acting along the length of the balance. 50 g L In other words, the force is parallel to the radius of length L around the pivot point. In this case, the magnitude of our new physical quantity must be

Torque Let’s finally give a name to this new physical quantity. How about... Any suggestions? Torque Where d is the distance from the pivot point (the point of rotation) to the point at which the force is applied and is the component of the applied force that is perpendicular to the line joining the pivot point to the point at which the force is applied.

Units! Although this has the same units as energy (J), we generally denote torques as forces acting at a distance, and therefore leave the units in the form N m.

0.5 m What is the magnitude of the torque experienced pivot What is the magnitude of the torque experienced by this unbalanced balance? Fg Here, the force of gravity acts perpendicularly to the radius joining the two blue points above.

f = 30o 0.5 m In this case, the force of gravity is NOT pivot 50 g 0.5 m In this case, the force of gravity is NOT perpendicular to the line joining the two blue points. Fg f = 30o Only the component of Fg that acts perpendicularly to this line will cause torque. Fg

f = 30o pivot 50 g 0.5 m Fg

Concept Quiz! Torque