0.2 8.3 - revisit kinematics of rotational motion  (Page 3/4)

Calculating the slow acceleration of trains and their wheels

Large freight trains accelerate very slowly. Suppose one such train accelerates from rest, giving its 0.350-m-radius wheels an angular acceleration of $0\text{.}\text{250}{\text{rad/s}}^2$ . After the wheels have made 200 revolutions (assume no slippage): (a) How far has the train moved down the track? (b) What are the final angular velocity of the wheels and the linear velocity of the train?

Strategy

In part (a), we are asked to find $x$ , and in (b) we are asked to find $\omega$ and $v$ . We are given the number of revolutions $\theta$ , the radius of the wheels $r$ , and the angular acceleration $\alpha$ .

Solution for (a)

The distance $x$ is very easily found from the relationship between distance and rotation angle:

Solving this equation for $x$ yields

Before using this equation, we must convert the number of revolutions into radians, because we are dealing with a relationship between linear and rotational quantities:

Now we can substitute the known values into $x=\mathrm{r\theta }$ to find the distance the train moved down the track:

Solution for (b)

We cannot use any equation that incorporates $t$ to find $\omega$ , because the equation would have at least two unknown values. The equation ${\omega }^2={{\omega }_0}^2+2\text{αθ}$ will work, because we know the values for all variables except $\omega$ :

Taking the square root of this equation and entering the known values gives

We can find the linear velocity of the train, $v$ , through its relationship to $\omega$ :

Discussion

The distance traveled is fairly large and the final velocity is fairly slow (just under 32 km/h).

Calculating the distance traveled by a fly on the edge of a microwave oven plate

A person decides to use a microwave oven to reheat some lunch. In the process, a fly accidentally flies into the microwave and lands on the outer edge of the rotating plate and remains there. If the plate has a radius of 0.15 m and rotates at 6.0 rpm, calculate the total distance traveled by the fly during a 2.0-min cooking period. (Ignore the start-up and slow-down times.)

Strategy

First, find the total number of revolutions $\theta$ , and then the linear distance $x$ traveled. $\theta =\overline{\omega }t$ can be used to find $\theta$ because $\stackrel{-}{\omega }$ is given to be 6.0 rpm.

Solution

Entering known values into $\theta =\overline{\omega }t$ gives

As always, it is necessary to convert revolutions to radians before calculating a linear quantity like $x$ from an angular quantity like $\theta$ :

Now, using the relationship between $x$ and $\theta$ , we can determine the distance traveled:

Discussion

Quite a trip (if it survives)! Note that this distance is the total distance traveled by the fly. Displacement is actually zero for complete revolutions because they bring the fly back to its original position. The distinction between total distance traveled and displacement was first noted in One-Dimensional Kinematics .