Question

. II A wheel initially rotating at experiences the angular acceleration shown in FIGURE EX4.39. What is the wheel's angular velocity, in rpm, at ?

FIGURE EX4.39

Short answer

The angular velocity of the wheel at $t=3.0\text{s}$is $98.2\text{rpm}$.$98.2\text{rpm}$

Step-by-step solution

Step 1. Given information

The wheel is initially rotating at $60\text{rpm}$.

The following figure shows the graph of angular acceleration versus time for a rotating wheel.

From the show figure-(I) it can be seen that for the interval of 0< t < 1 the angular acceleration zero. Thus the angular acceleration in this period is constant and that is equal to 60 rpm.

Step 2. Explanation

From the show figure-(I) it can be seen that for the interval of $$" localid="1650816406559" width="9">0<t<1the angular acceleration zero. Thus the angular acceleration in this period is constant and that is equal to $60\text{rpm}$.

The angular velocity of the of the wheel at $t=3\text{s}$from the shown figure-(I) is,

${\omega }_3-{\omega }_2=\frac{1}{2}\left(\Delta t\right)\left(\Delta h\right)$

Here, ${\omega }_2$is the angular velocity of the wheel at $t=2\text{s}$, $\Delta t$is the change in time from $t=1\text{s}$to ,

is the angular velocity of the wheel at , and is the change in angular acceleration at the time interval.

${\omega }_3-\left(60\text{rpm}\right)=\frac{1}{2}(2\text{s})\left(4\text{rad}/{\text{s}}^2\times \frac{1\text{rev}}{2\pi \text{rad}}\times \frac{60\text{s}}{1\text{min}}\right)$

$\begin{array}{l}{\omega }_3=38.2\text{rpm}+60\text{rpm}\\ {\omega }_3=98.2\text{rpm}\end{array}$

Thus, the angular velocity of the wheel at $t=3.0\text{s}$is $98.2\text{\hspace{0.17em}rpm}$.