Question 2.SP.27: Determine the linear velocity and total acceleration of a po......
Determine the linear velocity and total acceleration of a point on the rim of the flywheel in Problem 2.25, 0.6 s after it has started from rest.
The velocity of a point on the rim is found by multiplying the radius by the angular velocity. The angular velocity is
\omega=\omega_{0}+\alpha t=0+\left(10.5 \mathrm{rad} / \mathrm{s}^{2}\right)(0.6 \mathrm{~s})=6.30 \mathrm{rad} / \mathrm{s}
The magnitude of the linear velocity of a point on the rim when t=0.6 \mathrm{~s} is
v=r \omega=(0.6 \mathrm{~m})(6.30 \mathrm{rad} / \mathrm{s})=\underline{3.78 \mathrm{~m} / \mathrm{s}} \quad(\text { tangent to the rim })
To determine the acceleration completely, use the normal and tangential components. The tangential component a_{t} is
a_{t}=r \alpha=(0.6 \mathrm{~m})\left(10.5 \mathrm{rad} / \mathrm{s}^{2}\right)=6.3 \mathrm{~m} / \mathrm{s}^{2}
The normal component a_{n} is
a_{n}=r \omega^{2}=(0.6 \mathrm{~m})(6.3 \mathrm{rad} / \mathrm{s})^{2}=23.8 \mathrm{~m} / \mathrm{s}^{2}
Figure 2-21 illustrates these components for any point P on the rim.
The total acceleration a is the vector sum of the two components a_{t} and a_{n}. Let \phi be the angle between the total acceleration and the radius. The normal acceleration a_{n} is directed toward the center of the circle:
\begin{array}{l}a=\sqrt{a_{t}^{2}+a_{n}^{2}}=\sqrt{(6.3)^{2}+(23.8)^{2}}=\underline{24.6 \mathrm{~m} / \mathrm{s}^{2}} \\ \\\phi=\tan ^{-1} \frac{a_{t}}{a_{n}}=\tan ^{-1} \frac{6.3}{23.8}=0.259 \mathrm{rad}=\underline{14.8^{\circ}}\end{array}
where the angle \phi is shown in Fig. 2-21.
