Question 8.14: THE FALLING BUCKET GOAL Combine Newton's second law with its...
THE FALLING BUCKET
GOAL Combine Newton’s second law with its rotational analog.
PROBLEM A solid, uniform, frictionless cylindrical reel of mass M=3.00 \mathrm{~kg} and radius R=0.400 \mathrm{~m} is used to draw water from a well (Fig. 8.28a). A bucket of mass m=2.00 \mathrm{~kg} is attached to a cord that is wrapped around the cylinder. (a) Find the tension T in the cord and acceleration a of the bucket. (b) If the bucket starts from rest at the top of the well and falls for 3.00 \mathrm{~s} before hitting the water, how far does it fall?
STRATEGY This problem involves three equations and three unknowns. The three equations are Newton’s second law applied to the bucket, m a=\Sigma F_{i}; the rotational version of the second law applied to the cylinder, I \alpha=\sum \tau_{i}; and the relationship between linear and angular acceleration, a=r \alpha, which connects the dynamics of the bucket and cylinder. The three unknowns are the acceleration a of the bucket, the angular acceleration a of the cylinder, and the tension T in the rope. Assemble the terms of the three equations and solve for the three unknowns by substitution. Part (b) is a review of kinematics.
(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.