Rotational Dynamics

This section is interesting, as all your kinematic equations can be converted into rotational terms.
w = angular velocity.
alpha = angular acceleration.
w = w0 + alpha(t)
w^2 = w0^2 + 2(alpha)(theta) Use when time is not given.
Theta = w0t + 1/2(alpha)t^2 Use when time is given.
2pi(f)=w. F is in revs/second.
I = moment of inertia (equation depends on object shape).
L = angular momentum = Iw.
Torque = r_|_F = I(alpha)
v = rw
a_tan = r(alpha)

1. There is a difference between translation and rotational motion. In translation motion, objects move without rotating. Rotational motion on the other hand, is when all points in the rigid body move in circles.

2.
A radian is defined as the angle subtended by an arc whose length is equal to the radius. In a circle, there are 2pi radians, so 2pi radians is equivalent to 360 degrees. Theta = length of arc/radius.

3.
1 RPM is the same as 1/60 revolutions per second. Since one complete revolution is 2pi rads, then it is 2pi/60 revolutions per second which simplifies to pi/30. Therefore to convert from 1 RPM to rads/sec multiply by pi/30.

4.
The symbol for angular distance is theta, it’s measured in radians. The symbol for angular velocity is omega, which looks like a lowercase w, it is measured in radians per second. Angular acceleration is measured in radians per second squared, it is represented by the symbol alpha.

5.
Average angular velocity is the angular velocity of a body during a time period found by change in theta divided by change in time. Instantaneous angular velocity is the angular velocity at a “point in time,” meaning at a certain moment. This is found by w = lim /\t -> 0, /\ theta / /\ time.

6.
The relationship between linear velocity and angular velocity is that v =rw since v = r* /\ theta / /\ time.

7.
Average angular acceleration is defined as the change in angular velocity divided the change in time. alpha = /\ omega / /\ time. Like instantaneous angular velocity, instantaneous angular acceleration is alpha = lim /\t -> 0, /\ omega / /\ time.

8.
To compute the radial component of angular acceleration, it is a = rw^2. To compute the tangential angular acceleration, it is a = r*alpha. Since these are components, the vector sum is a (linear) = a(tan.) + a(radial)

9.
Kinematics and uniformly accelerated rotational motion can be related simply by letting a = alpha, v = omega, and x = theta. Thus w=w0+alpha(t), theta = wot + ½(alpha)t^2. w^2 =w0^2 + 2(alpha)(theta). ave. w = ( w + wo )/2

10.
Kinematics and uniformly accelerated rotational motion can be related simply by letting a = alpha, v = omega, and x = theta.

11.
The lever arm, or moment arm is the perpendicular distance from the axis of rotation to the line along which the force acts. Moment of the force about the axis defined as the product of force and the lever arm is also known as torque which is abbreviated tau. Torque = r | F.

12.
In order to use Torque = r | F or Torque = rF sin theta, you must first identify the axis of rotation and from that identify the distance that the force is applied through. In order to use net torque = I(alpha), you must identify the correct moment of inertia equation to use for the object which can be located on p 223 and you must know alpha.

13.
Since in rotational dynamics, all the points in the body are being rotated about in a circle, we no longer can just use the mass, we must use the product of the mass and radius squared times a coefficient that corresponds with the object’s shape. This can be used for solving net torque or kinetic energy problems.

14.
The radius of gyration describes the way in which the area of a cross-section is distributed around its central axis. r = sqrt (I/A), where r = radius of gyration, I = moment of inertia, and A = area of cross section.

15.
The relationship between torque, moment of inertia, and angular acceleration is that Net torque is the product of moment of inertia and angular acceleration. Net torque = I(alpha)

16.
The equation for rotational kinetic energy is KE(rotational) = ½ I w^2. The moment of inertia will vary dependent on the object.

17.
To calculate the total energy of a rotating object, you must also consider its translational movement. Thus KE = ½ mv^2 + ½ I w^2. If the object is stationary but rotating then it will have rotational kinetic energy only, since v = 0. If the object is moving linearly but not rotating, then it will only have translational kinetic energy, since w = 0.

18.
Newton’s second law can be stated in terms of angular momentum as well. Torque = change in angular momentum divided by change in time. From this, you can derive the equation net Torque = I(alpha) since angular momentum = I(w).

19.
The law of conservation of angular momentum states that the total angular momentum of a rotating body remains constant if the net torque acting on it is zero. This means as long as no external forces are being applied, the total angular momentum will remain the same.

Problem 1.
A tire is accelerates east from rest at 1.0m/s^2. If the diameter of the wheel is .68m, what is the velocity tangent after 3 seconds at the top of the tire if a point at the bottom of the tire is considered to be at rest?

Calculate angular acceleration.
a = r(alpha).
1.0m/s^2 = .34m(alpha).

Calculate angular velocity.
w=w0+alpha(t)

Now, consider a circle with its center at the bottom of the tire with a radius of the tire's diameter. Since angular velocity is the same no matter what the radius is, it can be used to determine the tangential velocity.
v_tan = rw
v_tan = 6m/s.

Problem 2
A ducky is watching a chicken crawling on the still blade of a ceiling fan when he decides to turn on the fan and watch the chicken go for a ride. If the chicken sits on the fan blade at a distance of 0.80m from the center of the fan, and turns with a frequency of 1.2Hz, what is the linear speed of the chicken?

Calculate the angular velocity of the chicken, I mean, fan. (inside joke)
2pi(f) = w.
2pi(1.2revs/second) = w

Calculate the linear velocity of the chicken.
v = rw
v = 6.03m/s

Problem 3 ( hi rina! )
The moon orbits the earth so that the same side always faces the earth. Determine the ratio of its spin angular momentum to its orbital angular momentum, treating the moon as a particle orbiting the earth.

The moon spins about its axis once every revolution about the earth, this is because the problem states that the same side always faces the earth.
As a result, the angular velocity, omega, is the same for both angular momentums since the time to revolve about the earth is the same as the time to spin about its axis.
Angular momentum is given by L = Iw.

The moon can be considered as a solid sphere, so I = 2/5M_moon*R_moon^2. We use this I in the spin angular momentum, L_spin.
For the orbital angular momentum, we can consider the earth-moon system as a hoop (moon as particle), since the moon revolves about the earth: I = M_moon*R_moon-earth^2 is used in L_orbital.

The ratio is then L_spin/L_orbital = ((2/5)M_moon*R_moon^2)/(M_moon*R_moon-earth^2).
*Note: the omegas canceled.