Rotational Motion I The radian The radian Angular Velocity Translation vs. Rotation

Rotational Motion I The radian The radian Angular Velocity Translation vs. Rotation

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Angular Velocity Since velocity is defined as the rate of change of displacement. ANGULAR VELOCITY is defined as the rate of change of ANGULAR DISPLACEMENT. NOTE: Translational motion tells you THREE THINGS magnitude of the motion in addition to the units Axis the motion occurs on direction on the given axis Example: v =3i This tells us that the magnitude is 3 m/s, the axis is the “x” axis in addition to the direction is in the “positive sense”. Translation vs. Rotation Translational motion tells you THREE THINGS magnitude of the motion in addition to the units Axis the motion occurs on direction on the given axis Example: v =3i This tells us that the magnitude is 3 m/s, the axis is the “x” axis in addition to the direction is in the “positive sense”. Rotational motion tells you THREE THINGS: magnitude of the motion in addition to the units the PLANE in which the object rotates in the directional sense ( counterclockwise or clockwise) Counterclockwise rotations are defined as having a direction of POSITVE K motion on the “z” axis Rotation Example: Unscrewing a screw or bolt = 5 rad/sec k Clockwise rotations are defined as having a direction of NEGATIVE K motion on the “z” axis Example: Tightening a screw or bolt = -5 rad/sec k

Angular Acceleration Once again, following the same lines of logic. Since acceleration is defined as the rate of change of velocity. We can say the ANGULAR ACCELERATION is defined as the rate of change of the angular velocity. Also, we can say that the ANGULAR ACCELERATION is the TIME DERIVATIVE OF THE ANGULAR VELOCITY. All the rules as long as integration apply as well. Combining motions Tangential velocity First we take our equation as long as the radian measure in addition to divide BOTH sides by a change in time. The left side is simply the equation as long as LINEAR velocity. BUT in this case the velocity is TANGENT to the circle (according to Newton’s first law). There as long as e we call it TANGENTIAL VELOCITY. Inspecting the right side we discover the as long as mula as long as ANGULAR VELOCITY. There as long as e, substituting the appropriate symbols we have a as long as mula that relates Translational velocity to Rotational velocity. Tangential acceleration in addition to rotational kinematics Using the same kind of mathematical reasoning we can also define Linear tangential acceleration. Inspecting each equation we discover that there is a DIRECT relationship between the Translational quantities in addition to the Rotational quantities. We can there as long as e RE-WRITE each translational kinematic equation in addition to turn it into a rotational kinematic equation.

Example A turntable capable of angularly accelerating at 12 rad/s2 needs to be given an initial angular velocity if it is to rotate through a net 400 radians in 6 seconds. What must its initial angular velocity be 30.7 rad/s Rotational Kinetic Energy in addition to Inertia Just like massive bodies tend to resist changes in their motion ( AKA – “Inertia”) . Rotating bodies also tend to resist changes in their motion. We call this ROTATIONAL INERTIA. We can determine its expression by looking at Kinetic Energy. We now have an expression as long as the rotation of a mass in terms of the radius of rotation. We call this quantity the MOMENT OF INERTIA (I) with units kgm2 Moment of Inertia, I m1 m2 Consider 2 masses, m1 & m2, rigidly connected to a bar of negligible mass. The system rotates around its CM. This is what we would see if m1 = m2. Suppose m1>m2. m1 m2 r1 r2 Since it is a rigid body, the have the SAME angular velocity, w. The velocity of the center, vcm of mass is zero since it is rotating around it. We soon see that the TANGENTIAL SPEEDS are NOT EQUAL due to different radii.

Moment of Inertia, I Since both masses are moving they have kinetic energy or rotational kinetic in this case. m1 m2 r1 r2 So this example clearly illustrates the idea behind the SUMMATION in the moment of inertia equation. Example A very common problem is to find the velocity of a ball rolling down an inclined plane. It is important to realize that you cannot work out this problem they way you used to. In the past, everything was SLIDING. Now the object is rolling in addition to thus has MORE energy than normal. So lets assume the ball is like a thin spherical shell in addition to was released from a position 5 m above the ground. Calculate the velocity at the bottom of the incline. 7.67 m/s If you HAD NOT included the rotational kinetic energy, you see the answer is very much different. Example: Moment of Inertia, I m1 m2 Let’s use this equation to analyze the motion of a 4-m long bar with negligible mass in addition to two equal masses(3-kg) on the end rotating around a specified axis. EXAMPLE 1 -The moment of Inertia when they are rotating around the center of their rod. EXAMPLE 2-The moment if Inertia rotating at one end of the rod m1 m2 24 kgm2 48 kgm2

Example cont Now lets calculate the moment of Inertia rotating at a point 2 meters from one end of the rod. m1 m2 2m 120 kgm2 As you can see, the FARTHER the axis of rotation is from the center of mass, the moment of inertia increases. We need an expression that will help us determine the moment of inertia when this situation arises. Parallel Axis Theorem This theorem will allow us to calculate the moment of inertia of any rotating body around any axis, provided we know the moment of inertia about the center of mass. It basically states that the Moment of Inertia ( Ip) around any axis “P” is equal to the known moment of inertia (Icm) about some center of mass plus M ( the total mass of the system) times the square of “d” ( the distance between the two parallel axes) Using the prior example lets use the parallel axis theorem to calculate the moment of inertia when it is rotating around one end in addition to 2m from a fixed axis. Exam  Parallel Axis Theorem m1 m2 d = 2m m1 m2 4m (not drawn to scale) 48 kgm2 120 kgm2

Continuous Masses The earlier equation, I =Smr2, worked fine as long as what is called POINT masses. But what about more continuous masses like disks, rods, or sphere where the mass is extended over a volume or area. In this case, calculus is needed. This suggests that we will take small discrete amounts of mass in addition to add them up over a set of limits. Indeed, that is what we will do. Lets look at a few example we MIGHT encounter. Consider a solid rod rotating about its CM. Will, I =Smr2, be the equation as long as a rod The rod We begin by using the same technique used to derive the center of mass of a continuous body. dr The CM acts as the origin in the case of determining the limits. Your turn What if the rod were rotating on one of its ENDS dr As you can see you get a completely different expression depending on HOW the body is rotating.

The disk R r dx dx 2pr The bottom line Will you be asked to derive the moment of inertia of an object Possibly! Fortunately, most of the time the moment of inertia is given within the free response question. Consult the file ( on the notes page) called Moments of Inertia to view common expressions as long as I as long as various shapes in addition to rotational situations.

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