Chapter 8 Rotational Motion Some definitions Rigid body

Chapter 8 Rotational Motion

Some definitions • Rigid body: any object with a definite shape so that all particles stay in a fixed position with regards to each other. • Pure rotational motion: all points in the body move in a circle. The absence of translational motion (the center of mass does not move). • Axis of rotation: the line on which all the centers of all the circles lie. This will be the center of mass.

Angular Quantities • When dealing with rotational motion, it is often useful to work with angular quantities such as angular velocity and angular acceleration. • These are analogous to linear velocity and linear acceleration, which means we must now define angular position.

Angular Position • As a body rotates, a point on the circle moves from point P to P’ along the circumference of its circle. • It will sweep out an angle, θ, and an arc length, l. • We usually measure θ in degrees, but for rotational motion it is usually beneficial to use radians.

Radians • Radian(rad): one rad is defined as the angle subtended by an arc whose arc length (l) equals its radius (r) • If r = l, then θ = 1 rad • Typically, θ = l/r • Because there are 360 o in a circle and l would equal the circumference, 2πr, θ = l/r = 2πr / r = 2π • So 360 o = 2π radians or 1 radian = 360/ 2π, which is approximately 57. 3 o.

Example • A particular bird’s eye can distinguish objects that subtend an angle no smaller than 3 E-4 rad. • A) how many degrees is this? • B) how small an object can the bird see from 100 m in the air?

Solution • A. (3 E-4 rad)(360 degrees / 2π rad) = 0. 017 degrees. • B. from θ = l/r we get l = θr note: for small θ (less than 15 degrees), θ is very close to the cord length. l = 100 m(3 E-4 rad) = 3 E-2 m or 3 cm • Radians are unit less because it is the ratio of two lengths (units cancel).

Angular Velocity • Velocity is displacement over time this is true in the angular sense as well. • Displacement: if we begin at some initial θ 0 and ending at θ, we can find our angular displacement, Δθ, by Δθ = θ – θ 0 • Angular velocity (ω, omega) = Δθ/Δt

Angular Acceleration • Acceleration is velocity/time, so • Angular acceleration (α, alpha) = Δω/Δt

Linear Velocity • Imagine a line of points along a radius of a rotation circle. • All of the points would have the same ω, but the points further from the center would have more distance to cover in the same amount of time, meaning its has a greater linear, or tangential, velocity, v. • v = rω

Linear Acceleration • The tangential acceleration is given by atan = Δv/Δt = r Δω/Δt = rα • But wait! We already covered in chapter 5 that objects in circular motion experience a centripetal acceleration a. R. • So the total linear acceleration of a rotating point is the vector sum of atan and a. R, where a. R can be rewritten as a. R = v 2/r = (ωr)2/r = ω2 r

Frequency • f can be written in terms of ω easily, f = ω/2π • Or ω = 2πf

Example • What is the linear velocity of a child sitting 1. 2 m from the center of a merry-go-round that makes one complete revolution on 4. 0 s? • What is the child’s acceleration?

Solution • • • First find f: f = 1/T = 1 rev/4. 0 s = 0. 25 rev/s ω = 2πf = (2π rad/rev)(0. 25 rev/s) = 1. 6 rad/s v = rω = 1. 2 m(1. 6 rad/s) = 1. 9 m/s As for the acceleration: atan = 0 because the rotation is constant a. R = ω2 r = (1. 6 rad/s)2(1. 2 m) = 3. 1 m/s 2

Rotational Kinematics • The rotational kinematic equations are simply the 4 linear kinematic equations from chapter 2 with the following substitutions: • x becomes θ • v becomes ω • a becomes α

The “New” 4 Kinematic Equations • ω = ω0 + αt • θ = ω0 t + ½αt 2 • ω2 = ω02 + 2αθ •

Rolling Motion • Rolling without slipping combines rotational and translational motion. • 1 revolution of the tires corresponds to a translational distance of 2πr • Also, remember that v = ωr

Rotational Dynamics • Obviously rotations require forces and the stronger the force the higher the acceleration just like in the linear case. • BUT… there’s more to it than that. • Imagine opening a swing door. – You apply a force, F 1, perpendicular to the door at the far end vs. – You apply a force, F 2, that is equal in magnitude to F 1, you apply it at the middle of the door.

Lever arm • What we see is the magnitude of the force alone is not enough. • It also matters where the force is applied. • The angular acceleration is proportional to the magnitude of the force and the perpendicular distance from the axis of rotation to the line along which the force acts. • This distance is called the lever arm or moment arm, which is given the variable r.

Torque • The acceleration is proportional to the product of the force times the lever arm, r. F. • This product is called torque, τ (tau: t + ”ow”) • There is a word we have been glossing over thus far, perpendicular. • Let’s go back to our door example: – Picture another force, F 3, that is equal in magnitude to F 1, acts on the same spot as F 1, but acts at an angle to the door.

Torque continued • So just like in the linear case we only care about the force component that causes the motion. • HOWEVER, unlike the linear case, we want the perpendicular component. (the linear case wanted the parallel component) • Therefore, τ = r. Fperpendicular = r. Fsinθ

Units of Torque • Torque is a distance (the lever arm) times a force (the perpendicular component); therefore, the units of torque are m. N • NOTE: we write this as m. N and not Nm to help you distinguish torque from energy. Energy is a scalar quantity and torque is a vector quantity. • Nm is called joules, m. N is NEVER referred to as joules.

Example • The biceps muscle exerts a vertical force on the lower arm. The point of attachment for the biceps muscle is 5. 0 cm from the elbow (the axis of rotation) and can generate a force of 700 N. Calculate the torque when the arm is at A. ) a 90 degrees angle to the upper arm and B. ) an angle 30 degrees below perpendicular.

Solution • A. F = 700 N and the Fperp = 0. 050 m, so • τ = (0. 050 m)(700 N) = 35 m. N • B. The angle of the arm shortens the lever arm down to 0. 050 sin 60, so • τ = (0. 050 m)(sin 60)(700 N) = 30 m. N

It’s how you use it • Adult chimpanzees only have about 1/3 the muscle mass of an adult human male; however, they have been observed to be twice as strong as human for certain movements. How is this possible?

Return of the sigma • If more than one torque is acting on a body, the acceleration α is proportional to the net torque. • Remember the vector nature of torque: clockwise subtracts from counterclockwise and vice versa. • It is accepted by the scientific community that counterclockwise is positive and clockwise is negative.

Similar to chapter 4 • Let us begin with F = ma, • Remember, a = rα so F = mrα • We no longer have force in chapter 8, we have torque (Fr). • So, let’s multiply both sides by r so that we can morph F into τ: Fr = mr 2α or τ = mr 2α

Rotational Inertia • So we finally know exactly how τ relates to α, but it depends on this quantity mr 2. • We give a name to this quantity, we call it the moment of inertia.

I • Now picture a rotating rigid body like a wheel. We can think of the wheel as having many particles located at various distances from the axis of rotation. • The sum of the various torques is given by • Σmr 2 = m 1 r 12 + m 2 r 22 +…+mnrn 2 = I = the moment of inertia of the body.

Torque and Inertia • Στ = Iα • This is the rotational equivalent of Newton’s second law. • Rotational inertia plays the same roll in rotations as regular inertia does in translational motion: a resistance to changing velocity.

Location • I depends not only on mass but on how that mass is distributed about the axis. • UNLIKE for gravity, the mass CANNOT be considered to be concentrated at its center of mass here. • Consider two round objects of equal mass. One is a solid plate and the other is a thin ring. They will rotate differently.

Example • A 5. 0 kg mass and a 7. 0 kg mass are mounted 4. 0 m apart on a light rod (considered massless). Calculate the moment of inertia of the system when: • A. ) it is rotated about an axis halfway between the masses. B. ) it is rotated about an axis 0. 50 m to the left of the 5. 0 kg mass.

Solution • A. ) I = m 1 r 12 + m 2 r 22 = (5. 0 kg)(2. 0 m)2 + (7. 0 kg)(2. 0 m)2 = 20 kgm 2 + 28 kgm 2 = 48 kgm 2 • B. ) I = m 1 r 12 + m 2 r 22 = (5. 0 kg)(0. 50 m)2 + (7. 0 kg)(3. 5 m)2 = 1. 3 kgm 2 + 142 kgm 2 = 143 kgm 2

Analysis • Obviously, the moment of inertia is different for different axes of rotation. • Secondly, masses very close to the axis of rotation contribute very little to the inertia.