Chapter 15, Problem 15.41P

(a)

To determine

The maximum speed of the bob.

Answer to Problem 15.41P

The maximum speed of the bob is $0.820\text{m}/\text{s}$.

Explanation of Solution

The formula to calculate amplitude is,

    $A=L\theta$

Here, $L$ is the length of pendulum and $\theta$ is the displaced angle.

Substitute $1.00\text{m}$ for $L$ and $15.0°$ for $\theta$ in above equation to find $A$.

    $\begin{array}{c}A=\left(1.00\text{m}\right)\left(15.0°\left(\frac{\text{π}}{180°}\right)\right)\\ =0.262\text{m}\end{array}$

The formula to calculate angular frequency is,

    $\omega =\sqrt{\frac{g}{L}}$

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

Substitute $1.00\text{m}$ for $L$ and $9.8\text{m}/{\text{s}}^2$ for $g$ in above equation to find $\omega$.

    $\begin{array}{c}\omega =\sqrt{\frac{9.8\text{m}/{\text{s}}^2}{1.00\text{m}}}\\ =3.13\text{rad}/\text{s}\end{array}$

The formula to calculate maximum speed is,

    $v_{\text{max}}=A\omega$

Substitute $0.262\text{m}$ for $A$ and $3.13\text{rad}/\text{s}$ for $\omega$ in above equation to find $v_{\max }$.

    $\begin{array}{c}{v}_{\text{max}}=\left(0.262\text{m}\right)\left(3.13\text{rad}/\text{s}\right)\\ =0.820\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the maximum speed of the bob is $0.820\text{m}/\text{s}$.

(b)

To determine

The maximum acceleration of the bob.

Answer to Problem 15.41P

The maximum acceleration of the bob is $2.57\text{rad}/{\text{s}}^2$.

Explanation of Solution

The formula to calculate maximum acceleration of the bob is,

    $a_{\text{max}}=A{\omega }^2$

Substitute $0.262\text{m}$ for $A$ and $3.13\text{rad}/\text{s}$ for $\omega$ in above equation to find $a_{\max }$.

    $\begin{array}{c}{a}_{\text{max}}=\left(0.262\text{ m}\right){\left(3.13\text{rad}/\text{s}\right)}^2\\ =2.57\text{rad}/{\text{s}}^2\end{array}$

Conclusion:

Therefore, the maximum acceleration of the bob is $2.57\text{rad}/{\text{s}}^2$.

(c)

To determine

The maximum restoring force of the bob.

Answer to Problem 15.41P

The maximum restoring force of the bob is $0.641\text{N}$.

Explanation of Solution

The formula to calculate maximum restoring force of the bob is,

    $F=m{a}_{\text{max}}$

Here, $m$ is the mass of the pendulum.

Substitute $A{\omega }^2$ for $a_{\text{max}}$ in above equation.

    $F=mA{\omega }^2$

Substitute $0.250\text{kg}$ for $m$, $0.262\text{ m}$ for $A$ and $3.13\text{rad}/\text{s}$ for $\omega$ in above equation to find $F$.

    $\begin{array}{c}F=\left(0.250\text{kg}\right)\left(0.262\text{m}\right){\left(3.13\text{rad}/\text{s}\right)}^2\\ =0.641\text{N}\end{array}$

Conclusion:

Therefore, the maximum restoring force of the bob is $0.641\text{N}$.

(d)

To determine

The maximum speed, angular acceleration and restoring force of the bob using the model introduced earlier chapter.

Answer to Problem 15.41P

The maximum speed of the bob is $0.817\text{m}/\text{s}$, the angular acceleration of the bob is $2.54\text{rad}/{\text{s}}^2$ and the restoring force of the bob is $0.634\text{N}$.

Explanation of Solution

Consider the figure given below.

In triangle $ABC$,

    $\begin{array}{c}\cos \theta =\frac{AC}{AB}\\ AC=AB\cos \theta \\ =L\cos \theta \end{array}$

The height of the bob is,

    $\begin{array}{c}h=AD-AC\\ =L-L\cos \theta \\ =L\left(1-\cos \theta \right)\end{array}$

The law of conservation of energy is,

    $mgh=\frac{1}{2}m{v}_{\text{max}}^2$

Substitute $L\left(1-\cos \theta \right)$ for $h$ in above expression.

    $\begin{array}{c}mgL\left(1-\cos \theta \right)=\frac{1}{2}m{v}_{\text{max}}^2\\ gL\left(1-\cos \theta \right)=\frac{1}{2}{v}_{\text{max}}^2\end{array}$

Substitute $15.0°$ for $\theta$, $1.00\text{m}$ for $L$ and $9.8\text{m}/{\text{s}}^2$ for $g$ in above equation to find $v_{\text{max}}$.

    $\begin{array}{c}\left(9.8\text{m}/{\text{s}}^2\right)\left(1.00\text{m}\right)\left(1-\cos \left(15.0°\right)\right)=\frac{1}{2}{v}_{\text{max}}^2\\ v_{\text{max}}^2=2\left(0.333\right){\text{m}}^2/{\text{s}}^2\\ v_{\text{max}}=\sqrt{2\left(0.333\right){\text{m}}^2/{\text{s}}^2}\\ =0.817\text{m}/\text{s}\end{array}$

The formula for the moment of inertia of the pendulum is,

    $I=m{L}^2$

The equation for the conservation of energy is,

    $I\alpha =mgL\sin \theta$

Here, $\alpha$ is the angular acceleration.

Substitute $m{L}^2$ for $I$ in above expression and rearrange for $\alpha$.

    $\begin{array}{c}m{L}^2\alpha =mgL\sin \theta \\ \alpha =\frac{mgL\sin \theta }{m{L}^2}\\ =\frac{g\sin \theta }{L}\end{array}$

Substitute $9.8\text{m}/{\text{s}}^2$ for $g$, $1.00\text{m}$ for $L$ and $15.0°$ for $\theta$ in above equation to find $\alpha$.

    $\begin{array}{c}\alpha =\frac{\left(9.8\text{m}/{\text{s}}^2\right)\sin \left(15.0°\right)}{1.00\text{m}}\\ =2.54\text{rad}/{\text{s}}^2\end{array}$

The force is maximum, when the angle is maximum.

The restoring force is calculated as,

    $F=mg\sin \theta$

Substitute $15.0°$ for $\theta$, $0.250\text{kg}$ for $m$ and $9.8\text{m}/{\text{s}}^2$ for $g$ in above equation to find $F$.

    $\begin{array}{c}F=\left(0.250\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)\sin \left(15.0°\right)\\ =0.634\text{N}\end{array}$

Conclusion:

Therefore, the maximum speed of the bob is $0.817\text{m}/\text{s}$, the angular acceleration of the bob is $2.54\text{rad}/{\text{s}}^2$ and the restoring force of the bob is $0.634\text{N}$.

(e)

To determine

The answers of part (a), part (c) and part (d).

Explanation of Solution

The restoring force is defined as the force or torque that tends to restore a system to equilibrium after displacement.

The answers are closest but not exactly the same. The angular amplitude of $15.0°$ is not small, so the simple harmonic oscillation is not accurate. The answers computed from conservation of the energy and from Newton’s second law are more accurate.

Conclusion:

Therefore, the answers are closest but not exactly the same.