Chapter 8, Problem 65P

(a)

To determine

The speed of the bucket after it has fallen through a distance of $0.80\text{ m}$ .

Answer to Problem 65P

The speed of the bucket after it has fallen through a distance of $0.80\text{ m}$ is $3.0\text{ m/s}$ .

Explanation of Solution

The diagram is shown in figure 1.

Write the equation for the conservation of energy for the given situation.

$K_{i\text{cylinder}}+K_{\text{ibucket}}+U_i=K_{\text{fcylinder}}+K_{f\text{bucket}}+U_f$ (I)

Here, $K_{i\text{cylinder}}$ is the initial kinetic energy of the cylinder, $K_{\text{ibucket}}$ is the initial kinetic energy of the bucket, $U_i$ is the initial potential energy, $K_{\text{fcylinder}}$ is the final kinetic energy of the cylinder, $K_{f\text{bucket}}$ is the final kinetic energy of the bucket and $U_f$ is the final potential energy

The initial kinetic energies of the cylinder and the bucket are zero.

$\begin{array}{c}{K}_{i\text{cylinder}}=0\\ K_{\text{ibucket}}=0\end{array}$ (II)

Write the expression for $U_i$ .

$U_i=mgh$ (III)

Here, $m$ is the mass of the bucket, $g$ is the acceleration due to gravity and $h$ is the initial height of the bucket

Write the expression for $K_{\text{fcylinder}}$ .

$K_{\text{fcylinder}}=\frac{1}{2}I{\omega }^2$ (IV)

Here, $I$ is the rotational inertia of the cylinder and $\omega$ is its angular speed

Write the expression for the rotational inertia of the cylinder.

$I=\frac{1}{2}M{R}^2$

Here, $M$ is the mass of the cylinder and $R$ is its radius

Write the expression for $\omega$ .

$\omega =\frac{v}{R}$

Here, $v$ is the tangential speed of the circumference of the cylinder

Put the above two equations in equation (IV).

$\begin{array}{c}{K}_{\text{fcylinder}}=\frac{1}{2}\left(\frac{1}{2}M{R}^2\right){\left(\frac{v}{R}\right)}^2\\ =\frac{1}{4}M{v}^2\end{array}$ (V)

The linear speed of the bucket is the same as the tangential speed of the circumference of the cylinder.

Write the equation for $K_{f\text{bucket}}$ .

$K_{f\text{bucket}}=\frac{1}{2}m{v}^2$ (VI)

The final potential energy is zero.

$U_f=0$ (VII)

Put equations (II), (III), (V), (VI) and (VII) in equation (I) and rewrite it for $v$ .

$\begin{array}{c}0+0+mgh=\frac{1}{4}M{v}^2+\frac{1}{2}m{v}^2+0\\ mgh=\frac{v^2}{4}\left(M+2m\right)\\ v^2=\frac{4mgh}{M+2m}\\ v=\sqrt{\frac{4mgh}{M+2m}}\end{array}$ (VIII)

Conclusion:

Given that the mass of the bucket is $2.0\text{ kg}$ , the mass of the cylinder is $3.0\text{ kg}$ and the distance through which the bucket is fallen is $0.80\text{ m}$ . The value of acceleration due to gravity is $9.80{\text{ m/s}}^2$ .

Substitute $2.0\text{ kg}$ for $m$ , $3.0\text{ kg}$ for $M$ , $9.80{\text{ m/s}}^2$ for $g$ and $0.80\text{ m}$ for $h$ in equation (VIII) to find the value of $v$ .

$\begin{array}{c}v=\sqrt{\frac{4\left(2.0\text{ kg}\right)\left(9.80{\text{ m/s}}^2\right)\left(0.80\text{ m}\right)}{3.0\text{ kg}+2\left(2.0\text{ kg}\right)}}\\ =3.0\text{ m/s}\end{array}$

Therefore, the speed of the bucket after it has fallen through a distance of $0.80\text{ m}$ is $3.0\text{ m/s}$ .

(b)

To determine

The tension in the rope.

Answer to Problem 65P

The tension in the rope is $8.4\text{ N}$ .

Explanation of Solution

Write the work-energy theorem.

$W_{\text{total}}=\Delta K$ (IX)

Here, $W_{\text{total}}$ is the total work done as the bucket falls and $\Delta K$ is the change in kinetic energy of the bucket

Write the expression for $W_{\text{total}}$ .

$W_{\text{total}}=W_{\text{rope}}+W_{\text{grav}}$ (X)

Here, $W_{\text{rope}}$ is the work done by the rope and $W_{\text{grav}}$ is the work done by gravity

Write the expression for $W_{\text{rope}}$ .

$W_{\text{rope}}=-Th$

Here, $T$ is the tension in the rope

Write the expression for $W_{\text{grav}}$ .

$W_{\text{grav}}=mgh$

Put the above two equations in equation (X).

$W_{\text{total}}=-Th+mgh$ (XI)

Write the expression for $\Delta K$ .

$\begin{array}{c}\Delta K=\frac{1}{2}m{v}^2-0\\ =\frac{1}{2}m{v}^2\end{array}$ (XII)

Put equations (XI) and (XII) in equation (IX) and rewrite it for $T$ .

$\begin{array}{c}-Th+mgh=\frac{1}{2}m{v}^2\\ T=mg-\frac{m{v}^2}{2h}\\ =m\left(g-\frac{v^2}{2h}\right)\end{array}$ (XIII)

Conclusion:

Substitute $2.0\text{ kg}$ for $m$ , $9.80{\text{ m/s}}^2$ for $g$ , $3.0\text{ m/s}$ for $v$ and $0.80\text{ m}$ for $h$ in equation (XIII) to find the value of $T$ .

$\begin{array}{c}T=\left(2.0\text{ kg}\right)\left(9.80{\text{ m/s}}^2-\frac{{\left(3.0\text{ m/s}\right)}^2}{2\left(0.80\text{ m}\right)}\right)\\ =8.4\text{ N}\end{array}$

Therefore, the tension in the rope is $8.4\text{ N}$ .

(c)

To determine

The acceleration of the bucket.

Answer to Problem 65P

The acceleration of the bucket is $5.6{\text{ m/s}}^2\text{ down}$ .

Explanation of Solution

Write the expression for the Newton’s second law.

$\Sigma F_y=m{a}_y$ (XIV)

Here, $\Sigma F_y$ is the net force acting on the bucket as it falls and $a_y$ is the acceleration of the bucket

Write the equation for the net force acting on the bucket as it falls.

$\Sigma F_y=T-mg$ (XV)

Equate equations (XIV) (XV) and rewrite it for $a_y$ .

$\begin{array}{c}m{a}_y=T-mg\\ a_y=\frac{T}{m}-g\end{array}$ (XVI)

Conclusion:

Substitute $8.4\text{ N}$ for $T$ , $2.0\text{ kg}$ for $m$ and $9.80{\text{ m/s}}^2$ for $g$ in equation (XVI) to find the value of $a_y$ .

$\begin{array}{c}{a}_y=\frac{8.4\text{ N}}{2.0\text{ kg}}-9.80{\text{ m/s}}^2\\ =-5.6{\text{ m/s}}^2\end{array}$

The negative sign indicates that the acceleration acts downward.

Therefore, the acceleration of the bucket is $5.6{\text{ m/s}}^2\text{ down}$ .