Question 8.14: THE FALLING BUCKET GOAL Combine Newton's second law with its...

(a) Find the tension in the cord and the acceleration of the bucket.

Apply Newton’s second law to the bucket in Figure 8.28b. There are two forces: the tension $\overrightarrow{\mathbf{T}}$ acting upward and gravity $m \overrightarrow{\mathbf{g}}$ acting downward.

(1) $m a=-m g+T$

Apply $\tau=I \alpha$ to the cylinder in Figure $8.28 \mathrm{c}$ :

$\sum \tau=I \alpha=\frac{1}{2} M R^{2} \alpha \quad \text { (solid cylinder) }$

Notice the angular acceleration is clockwise, so the torque is negative. The normal and gravity forces have zero moment arm and don’t contribute any torque.

(2) $-T R=\frac{1}{2} M R^{2} \alpha$

Solve for $T$ and substitute $\alpha=a / R$ (notice that both $\alpha$ and $a$ are negative):

(3) $T=-\frac{1}{2} M R \alpha=-\frac{1}{2} M a$

Substitute the expression for $T$ in Equation (3) into Equation (1), and solve for the acceleration:

$m a=-m g \quad \frac{1}{2} M a \quad \rightarrow \quad a=-\frac{m g}{m+{ }_{2}^{1} M}$

Substitute the values for $m, M$, and $g$, getting $a$, then substitute $a$ into Equation (3) to get $T$ :

$a=-5.60 \mathrm{~m} / \mathrm{s}^{2} \quad T=8.40 \mathrm{~N}$

(b) Find the distance the bucket falls in $3.00 \mathrm{~s}$.

Apply the displacement kinematic equation for constant acceleration, with $t=3.00 \mathrm{~s}$ and $v_{0}=0$ :

$\Delta y=v_{0} t+\frac{1}{2} a t^{2}=-\frac{1}{2}\left(5.60 \mathrm{~m} / \mathrm{s}^{2}\right)(3.00 \mathrm{~s})^{2}=-25.2 \mathrm{~m}$

REMARKS Proper handling of signs is very important in these problems. All such signs should be chosen initially and checked

mathematically and physically. In this problem, for example, both the angular acceleration $\alpha$ and the acceleration $a$ are negative, so $\alpha=a / R$ applies. If the rope had been wound the other way on the cylinder, causing counterclockwise rotation, the torque would have been positive, and the relationship would have been $\alpha=-a / R$, with the double negative making the right-hand side positive, just like the left-hand side.