Chapter 15, Problem 59AP

(a)

To determine

Tension in the rod at the pivot point.

Answer to Problem 59AP

Tension in the rod at the pivot point is $\underset{\_}{2Mg}$.

Explanation of Solution

Given that the mass of rod and ball is equal to $M$.

The tension at the pivot point is equal to the sum of the force of gravity on the rod and ball.

Write the expression for the force of gravity on the rod.

    $F_r=Mg$                                                                                                                     (I)

Here, $F_r$ is the gravitational force on the rod, $M$ is the mass of the rod, and $g$ is the acceleration due to gravity.

Write the expression for the force of gravity on the ball.

    $F_b=Mg$                                                                                                                   (II)

Here, $F_b$ is the force of gravity on the ball, and $M$ is the mass of ball.

Write the expression for the tension in the rod.

    $F=F_r+F_b$                                                                                                             (III)

Here, $F$ is the tension in the rod.

Conclusion:

Use expressions (I) and (II) in expression (III) to find $F$.

    $\begin{array}{c}F=Mg+Mg\\ =2Mg\end{array}$

Therefore, tension in the rod at the pivot point is $\underset{\_}{2Mg}$.

(b)

To determine

The tension in the rod at the point $P$ when the system of rod and ball is at rest.

Answer to Problem 59AP

The tension in the rod at the point $P$ when the system of rod and ball is at rest is $\underset{\_}{Mg\left(1+\frac{y}{L}\right)}$.

Explanation of Solution

At point $P$ the whole weight of rod is not to be considered. At this point only $\frac{y}{L}$ portion of the rod only contributes. Therefore the tension at point $P$ is equal to the sum of weight of rod due to $\frac{y}{L}$ portion of rod and the weight of the ball.

Force of gravity on the ball is equal to $Mg$. Write the expression for the force of gravity on $\frac{y}{L}$ portion of rod.

    $F_p=Mg\left(\frac{y}{L}\right)$                                                                                                          (IV)

Write the expression for the tension.

    $F=F_p+F_b$                                                                                                              (V)

Conclusion:

Substitute $Mg$ for $F_b$, and $Mg\left(\frac{y}{L}\right)$ for $F_p$ in equation (V) to find $F$.

    $\begin{array}{c}F=Mg\left(\frac{y}{L}\right)+Mg\\ =Mg\left(1+\frac{y}{L}\right)\end{array}$

Therefore, the tension in the rod at the point $P$ when the system of rod and ball is at rest is $\underset{\_}{Mg\left(1+\frac{y}{L}\right)}$.

(c)

To determine

The period of oscillation for small displacements.

Answer to Problem 59AP

The period of oscillation for small displacements is $\underset{\_}{\frac{4\pi }{3}\sqrt{\frac{2L}{g}}}$.

Explanation of Solution

Write the expression for moment of inertia of the ball with respect to the point of pivot.

    $I_{\text{ball}}=M{L}^2$                                                                                                               (VI)

Here, $I_{\text{ball}}$ is the moment of inertia of the ball, $M$ is the mass of the ball, and $L$ is the distance between ball and pivot.

Write the expression for the moment of inertia of the rod rotating about one end.

    $I_{\text{rod}}=\frac{1}{3}M{L}^2$                                                                                                           (VII)

Here, $I_{\text{rod}}$ is the moment of inertia of the rod rotating about one end.

Write the expression for the total moment of inertia of system.

    $I=I_{\text{rod}}+I_{\text{ball}}$                                                                                                         (VIII)

Use expression (VI) and (VII) in expression (VIII) to find $I$.

    $\begin{array}{c}I=\frac{1}{3}M{L}^2+M{L}^2\\ =\frac{4}{3}M{L}^2\end{array}$                                                                                                     (IX)

Write the expression for the time period of the physical pendulum.

    $T=2\pi \sqrt{\frac{I}{mgd}}$                                                                                                          (X)

Here, $T$ is the time period of the physical pendulum, $I$ is the moment of inertia of the pendulum, $m$ is the mass of pendulum, and $d$ is the distance from pivot to center of mass of rod and ball combination.

Write the expression to find the distance from pivot to center of mass of rod and ball combination.

    $d=\frac{m_1{d}_1+m_2{d}_2}{m_1+m_2}$                                                                                                     (XI)

Here, $m_1$ is the mass of rod, $m_2$ is the mass of ball, $d_1$ is the distance from pivot to center of mass of rod, and $d_2$ is the distance between pivot to ball.

Conclusion:

Substitute $M$ for $m_1$ and $m_2$, $\frac{L}{2}$ for $d_1$, and $L$ for $d_2$ in expression (XI) to find $d$.

    $\begin{array}{c}d=\frac{M\left(\frac{L}{2}\right)+ML}{M+M}\\ =\frac{3L}{4}\end{array}$

Substitute $\frac{4}{3}M{L}^2$ for $I$, $2M$ for $m$, and $\frac{3L}{4}$ for $d$ in expression (X) to find $T$.

    $\begin{array}{c}T=2\pi \sqrt{\frac{\frac{4}{3}M{L}^2}{\left(2M\right)g\left(\frac{3L}{4}\right)}}\\ =\frac{4\pi }{3}\sqrt{\frac{2L}{g}}\end{array}$

Therefore, the period of oscillation for small displacements is $\underset{\_}{\frac{4\pi }{3}\sqrt{\frac{2L}{g}}}$.

(d)

To determine

The time period of oscillation if the length of the pendulum is $2.00\text{m}$.

Answer to Problem 59AP

The time period of oscillation if the length of the pendulum is $2.00\text{m}$ is $\underset{\_}{2.68\text{s}}$.

Explanation of Solution

Write the expression for time period of oscillation of the pendulum as found in part (c).

    $T=\frac{4\pi }{3}\sqrt{\frac{2L}{g}}$                                                                                                          (XII)

Conclusion:

Substitute $2.00\text{m}$ for $L$, and $9.80{\text{m/s}}^2$ for $g$ in equation (XII) to find $T$.

    $\begin{array}{c}T=\frac{4\pi }{3}\sqrt{\frac{2\left(2.00\text{m}\right)}{\left(9.80{\text{m/s}}^2\right)}}\\ =2.68\text{s}\end{array}$

Therefore, the time period of oscillation if the length of the pendulum is $2.00\text{m}$ is $\underset{\_}{2.68\text{s}}$.