Chapter 14, Problem 66P

(a)

To determine

The speed of the bob.

Explanation of Solution

Introduction:

In the simple pendulum, there is the bob that has mass and is hanged from the string that has the certain length. When the bob is displaced from the equilibrium position the string follows forth and the back motion which is known as the periodic motion.

Write the expression to relate the speed of the bob with the angular speed.

  $v=L\frac{d\varphi }{dt}$ …… (1)

Here, $v$ is the speed of the bob, $L$ is the length of the pendulum and $\varphi$ is the angle.

Write the expression for the angular position with respect to time.

  $\varphi ={\varphi }_0\cos \omega t$

Here, ${\varphi }_0$ is the angle displaced by the bob and $\omega$ is the angular speed.

Differentiate the above equation with respect to time.

  $\frac{d\varphi }{dt}=-{\varphi }_0\omega \sin \omega t$

Substitute $-{\varphi }_0\omega \sin \omega t$ for $\frac{d\varphi }{dt}$ in equation (1).

  $v=-L{\varphi }_0\omega \sin \omega t$ …… (2)

Write the expression for the velocity.

  $v=-v_m\sin \omega t$ …… (3)

Here, is the maximum velocity.

Equate equation (2) and (3) for $v$ .

  $L{\varphi }_0\omega \sin \omega t=v_m\sin \omega t$

Solve the above equation for $v_m$ .

  $v_m=L{\varphi }_0\omega$ …… (4)

Write the expression for the angular velocity.

  $\omega =\sqrt{\frac{g}{L}}$

Here, $g$ is the acceleration due to gravity.

Substitute $\sqrt{\frac{g}{L}}$ for $\omega$ in equation (4).

  $\begin{array}{l}{v}_m=L{\varphi }_0\sqrt{\frac{g}{L}}\\ v_m={\varphi }_0\sqrt{gL}\end{array}$

Conclusion:

Thus, the speed of the bob is ${\varphi }_0\sqrt{gL}$ .

(b)

To determine

The speed of the bob.

Explanation of Solution

Introduction:

In the simple pendulum, there is the bob that has mass and is hanged from the string that has the certain length. When the bob is displaced from the equilibrium position the string follows forth and the back motion which is known as the periodic motion.

Write the expression for the conservation for the energy.

  $\left(K_F-K_I\right)-\left(U_F-U_I\right)=0$ …… (5)

Here, $K_F$ is the final kinetic energy, $K_I$ is the initial kinetic energy, $U_F$ is the final potential energy and $U_I$ is the initial potential energy.

The initial kinetic energy and final potential energy are zero.

Substitute $0$ for $K_I$ and $0$ for $U_F$ in equation (5).

  $K_F-U_I=0$ …… (6)

Write the expression for the kinetic energy.

  $K_F=\frac{1}{2}m{v}^2$

Here, $v$ is the velocity.

Write the expression for the potential energy.

  $U_I=mgh$

Substitute $\frac{1}{2}m{v}^2$ for $K_F$ and $mgh$ for $U_I$ in equation (6).

  $\frac{1}{2}m{v}^2-mgh=0$ ……. (7)

Write the expression for the height.

  $h=L\left(1-\cos {\varphi }_0\right)$

Substitute $L\left(1-\cos {\varphi }_0\right)$ for $h$ in equation (7).

  $\frac{1}{2}m{v}^2-mg\left(L\left(1-\cos {\varphi }_0\right)\right)=0$

Solve the above equation for $v$ .

  $v=\sqrt{2gL\left(1-\cos {\varphi }_0\right)}$

Conclusion:

Thus, the speed of the bob is $\sqrt{2gL\left(1-\cos {\varphi }_0\right)}$ .

(c)

To determine

The speed of the bob.

Explanation of Solution

Introduction:

In the simple pendulum, there is the bob that has mass and is hanged from the string that has the certain length. When the bob is displaced from the equilibrium position the string follows forth and the back motion which is known as the periodic motion.

Write the expression for the velocity.

  $v=\sqrt{2gL\left(1-\cos {\varphi }_0\right)}$

Write the expression for the small angle.

  $1-\cos {\varphi }_0=\frac{1}{2}{\varphi }_0^2$

Substitute $\frac{1}{2}{\varphi }_0^2$ for $1-\cos {\varphi }_0$ in equation (7).

  $v={\varphi }_0\sqrt{gL}$

Conclusion:

Thus, the speed of the bob is $\sqrt{2gL\left(1-\cos {\varphi }_0\right)}$ .

(d)

To determine

The difference in the speeds of the bob.

Explanation of Solution

Given:

The length is $1.0\text{m}$ .

The angle is $0.20\text{rad}$ .

Formula used:

Write the expression for the change in velocity.

  $\Delta v=v_m-v$ …… (8)

Substitute ${\varphi }_0\sqrt{gL}$ for $v_m$ and $\sqrt{2gL\left(1-\cos {\varphi }_0\right)}$ for $v$ in equation (8).

  $\Delta v=\sqrt{gL}\left({\varphi }_0-\sqrt{2\left(1-\cos {\varphi }_0\right)}\right)$ …… (9)

Calculation:

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ , $1.0\text{m}$ for $L$ and $0.20\text{rad}$ for ${\varphi }_0$ in equation (9).

  $\begin{array}{l}\Delta v=\sqrt{9.81\text{m}/{\text{s}}^{\text{2}}\left(1.0\text{m}\right)}\left(0.20\text{rad}-\sqrt{2\left(1-0.20\text{rad}\right)}\right)\\ \Delta v=1\text{mm}/\text{s}\end{array}$

Conclusion:

Thus, the difference in the speed of the bob is $1\text{mm}/\text{s}$ .

(e)

To determine

The difference in the speed of the bob.

Explanation of Solution

Given:

The length is $1.0\text{m}$ .

The angle is $1.20\text{rad}$ .

Formula used:

Write the expression for the change in velocity.

  $\Delta v=v_m-v$

Substitute ${\varphi }_0\sqrt{gL}$ for $v_m$ and $\sqrt{2gL\left(1-\cos {\varphi }_0\right)}$ for $v$ in equation (8).

  $\Delta v=\sqrt{gL}\left({\varphi }_0-\sqrt{2\left(1-\cos {\varphi }_0\right)}\right)$

Calculation:

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ , $1.0\text{m}$ for $L$ and $1.20\text{rad}$ for ${\varphi }_0$ in equation (9).

  $\begin{array}{l}\Delta v=\sqrt{9.81\text{m}/{\text{s}}^{\text{2}}\left(1.0\text{m}\right)}\left(1.20\text{rad}-\sqrt{2\left(1-1.20\text{rad}\right)}\right)\\ \Delta v=0.2\text{mm}/\text{s}\end{array}$

Conclusion:

Thus, the difference in the velocity of the bob is $0.2\text{mm}/\text{s}$ .