9781118230725.pdf

10-1 ROTATIONAL VARIABLES 263

t3.0 s to t6.0 s. Sketch the disk and indicate the direc-

tion of turning and the sign of vatt2.0 s, 4.0 s, and tmin.

KEY IDEA

From Eq. 10-6, the angular velocity vis equal to du/dtas

given in Eq. 10-10. So, we have

v0.6000.500t. (10-11)

The graph of this function v(t) is shown in Fig. 10-5c.

Because the function is linear, the plot is a straight line. The

slope is 0.500 rad/s^2 and the intercept with the vertical axis

(not shown) is0.600 rad/s.

Calculations: To sketch the disk at t2.0 s, we substitute

that value into Eq. 10-11, obtaining

v1.6 rad/s. (Answer)

The minus sign here tells us that at t2.0 s, the disk is

turning clockwise (as indicated by the left-hand sketch in

Fig. 10-5c).

Substitutingt4.0 s into Eq. 10-11 gives us

v1.4 rad/s. (Answer)

The implied plus sign tells us that now the disk is turning

counterclockwise (the right-hand sketch in Fig. 10-5c).

For tmin, we already know that du/dt0. So, we must

also have v0. That is, the disk momentarily stops when

the reference line reaches the minimum value of uin

Fig. 10-5b, as suggested by the center sketch in Fig. 10-5c.On

the graph of v versustin Fig. 10-5c, this momentary stop is

the zero point where the plot changes from the negative

clockwise motion to the positive counterclockwise motion.

(d) Use the results in parts (a) through (c) to describe the

motion of the disk from t3.0 s to t6.0 s.

Description: When we first observe the disk at t3.0 s, it

has a positive angular position and is turning clockwise but

slowing. It stops at angular position u1.36 rad and then

begins to turn counterclockwise, with its angular position

eventually becoming positive again.

KEY IDEA

To find the extreme value (here the minimum) of a function,
we take the first derivative of the function and set the result
to zero.

Calculations: The first derivative of u(t) is

(10-10)

Setting this to zero and solving for tgive us the time at
whichu(t) is minimum:

tmin1.20 s. (Answer)

To get the minimum value of u, we next substitute tmininto
Eq. 10-9, finding

u1.36 rad77.9. (Answer)

This minimumofu(t) (the bottom of the curve in Fig. 10-5b)
corresponds to the maximum clockwiserotation of the disk
from the zero angular position, somewhat more than is
shown in sketch 3.

(c) Graph the angular velocity vof the disk versus time from

du

dt

0.6000.500t.

(c)

2

0

–2

–2 0 2 4 6

ω(rad/s)

t(s)

negativeω zeroω positiveω

This is a plot of the angular

velocity of the disk versus time.

The angular velocity is

initially negative and slowing,

then momentarily zero during

reversal, and then positive and

increasing.

Additional examples, video, and

practice available at WileyPLUS