When objects accelerate in the sense that we can describe it in terms of a vector [math]\vec{a}[/math] and relate that to a velocity and a displacement, it is called[b] translational motion[/b]. In other words, nearly everything we've discussed this semester so far has dealt with translational motion. [br][br]Now we turn to [b]rotational motion[/b]. As the name suggests, this motion implies not that the object is going anywhere, but that it is spinning. For a rigid body (one which does not have movable parts), these are the only two types of motion it can undergo. In other words, every ball that flies or wheel that rolls or rock that gets tossed can be described by a combination of translational and rotational motion. The focus in this chapter is the rotational part of that discussion.

Recall that given a circular arc segment of radius r that subtends an angle [math]\phi[/math], that the arc length is [math]s=r\phi.[/math]  Taking a time derivative of both sides gives us our definition of angular velocity: [math]\frac{ds}{dt}=\frac{d}{dt}(r\phi)=r\frac{d\phi}{dt} = r\omega.[/math]  The [math]\omega[/math] (omega) is NOT a w (double u)!  Note that [math]\frac{ds}{dt}=v_{tan}[/math] represents the tangential velocity.[br][br]Taking one more time derivative gives us an expression for angular acceleration:  [math]\frac{d^2s}{dt^2}=r\frac{d\omega}{dt} = r\alpha.[/math] The tangential acceleration is [math]a_{tan}=\frac{d^2s}{dt^2}.[/math][br][br]So in short: [br][center][math]s=\phi r \\[br]v_{tan}=\omega r \\[br]a_{tan}=\alpha r. [/math][/center][br][color=#1e84cc]EXAMPLE: Suppose a CD is rotating at 500rpm and has a diameter of 12cm. What is the speed of a point on the edge of the CD?[br]SOLUTION: [center][math]v_{tan}=\omega r \\[br]\omega = \tfrac{500 rot}{1 min}\cdot\tfrac{1 min}{60 s}\cdot\tfrac{2\pi rad}{1 rot}=52.4\tfrac{rad}{s} \\[br]v_{tan}=\omega r = 52.4\tfrac{rad}{s}(6\times 10^{-2}m) = 3.14 \tfrac{m}{s}.[br][/math][/center][br][br]EXAMPLE: How fast is a person traveling with respect to earth's center while standing on the surface at latitude [math]\theta=34^o[/math]? [br]SOLUTION: Recall that the radius of the circular motion is earth's radius times the cosine of the latitude, and that [math]\omega[/math] must be expressed in radians per second. The earth rotates at roughly [math]\omega = \frac{2\pi \text{rad}}{86400s}.[/math] The earth's radius is [math]6.37\times 10^6m.[/math] Putting this together gives [math]v_{tan}=\omega r\cos\theta = 384m/s.[/math][br][/color]

Notice that the units on these expressions can look odd.  The radians on the right side terms just vanishes.  This is ok since a radian is not a unit, but a ratio of arc length to radius in a circle.  Some people prefer to call it a unit, but in any case [b]a radian is dimensionless[/b]. This disappearance of units should not happen with other units that you encounter, but it happens all the time when converting from translational terms to rotational terms.

If you take what you know of ordinary kinematics and apply it to angular kinematics, all of it still holds. Recall that our old definitions were:[br][br][center][math]\Delta\vec{r}=\int\vec{v}\;dt \\[br]\Delta\vec{v}=\int\vec{a}\;dt \\[br]\text{Now we have } \\[br]\Delta\vec{\phi}=\int\vec{\omega}\;dt \\[br]\Delta\vec{\omega}=\int\vec{\alpha}\;dt [/math][/center][br][br][color=#1e84cc]EXAMPLE: A merry-go-round starts from rest and accelerates at [math]\vec{\alpha} = 0.10\frac{rad}{s^2}\hat{k}[/math] (assuming the z-axis is the upward direction) until reaching an angular velocity of [math]\vec{\omega}=1.5\frac{rad}{s}\hat{k}.[/math]  How long did this take, how many times around did the children travel while this was taking place, and what was the total angular displacement (vector)?[br][br]SOLUTION: Using [math]\Delta\vec{\omega} = \vec{\omega}_f - \vec{\omega}_i = \vec{\omega}_f - \vec{0}= \int_0^t\vec{\alpha} \;dt= \vec{\alpha} (t-0).[/math] This means [math]\vec{\omega}(t)=\vec{\alpha}t,[/math] and we set that equal [math]1.5\frac{rad}{s}\hat{k}[/math] to find the time it took to get up to speed. It is [math]t=15s.[/math] To find the angular displacement, use [math]\Delta\vec{\theta} = \int\vec{\omega}(t) \;dt =\int_0^t \vec{\alpha} t\;dt = \frac{\vec{\alpha} t^2}{2},[/math]  gives [math]\Delta\vec{\theta} = 11.25 rad\; \hat{k},[/math] or 1.8 rotations.  [br][br]EXAMPLE: A child sits 1.5m from the axis of rotation of the merry-go-round from the previous example. What is the child's speed as a function of time? What is the child's total acceleration as a vector? Please use [math]\hat{r}[/math] and [math]\hat{\theta}[/math] to describe the vector. Assume the polar coordinates are in the x-y plane.[br][br]SOLUTION: We use [math]v_{tan}=r\omega[/math] to find the speed, which only has a component tangent to the path. This gives [math]v_{tan}=r \alpha t = (1.5m)0.10\frac{rad}{s^2}t=0.15\tfrac{m}{s^2}t.[/math] The total acceleration vector is made of two orthogonal components - the tangential acceleration in the [math]\hat{\theta}[/math] direction and the inward [math]-\hat{r}[/math] centripetal acceleration. Recall that the centripetal acceleration may be written [math]\vec{a}_{c}=-r\omega^2\hat{r} = r\alpha^2t^2\hat{r},[/math] and that as a vector we may write [math]\vec{a}_{tan}=r\alpha\hat{\theta}.[/math] Put together this gives the total acceleration of the child: [math]\vec{a}= -r\alpha^2t^2\hat{r}+r\alpha\hat{\theta}.[/math][/color]