Chapter 14, Problem 74P

(a)

To determine

The period of small oscillations.

Explanation of Solution

Given:

The distance between center of sphere and point of support is $L$ .

The period of simple pendulum is $T_0=2\pi \sqrt{\frac{L}{g}}$ .

Introduction:

Time period is the time taken to cover one oscillation. The body resists any angular acceleration due to inertia present it which is known as moment of inertia.

The moment of inertia of sphere is $\frac{2}{5}m{r}^2$ .

Write the expression for moment of inertia using parallel axis theorem.

  $I=I_{\text{cm}}+m{L}^2$

Here, $I$ is the inertia of bob, $m$ is mass of bob, $L$ is the distance between center of sphere and point of support and $I_{\text{cm}}$ is the inertia of bob passing through center of mass.

Substitute $$ I_{\text{cm}}$$

25mr2 for in above expression.

  $I=\frac{2}{5}m{r}^2+m{L}^2$

Write the expression for object in term of period of pendulum.

  $T=2\pi \sqrt{\frac{I}{mgh}}$

Here, $g$ is the gravitational acceleration and $T$ is the time period of object.

Substitute $I=\frac{2}{5}m{r}^2+m{L}^2$ for $I$ in above expression.

  $\begin{array}{c}T=2\pi \sqrt{\frac{\frac{2}{5}m{r}^2+m{L}^2}{mgL}}\\ =2\pi \sqrt{\frac{L}{g}}\sqrt{1+\frac{2r^2}{5L^2}}\\ =T_0\sqrt{1+\frac{2r^2}{5L^2}}\end{array}$

Here, $r$ is the radius of bob.

Conclusion:

Thus, the period of small oscillations is $T=T_0\sqrt{1+\frac{2r^2}{5L^2}}$ .

(b)

To determine

The period of small oscillations when radius is much smaller than length of pendulum.

Explanation of Solution

Given:

The distance between center of sphere and point of support is $L$ .

The period of simple pendulum is $T_0=2\pi \sqrt{\frac{L}{g}}$ .

Introduction:

Time period is the time taken to cover one oscillation. The body resists any angular acceleration due to inertia present it which is known as moment of inertia.

The moment of inertia of sphere is $\frac{2}{5}m{r}^2$ .

Write the expression for moment of inertia using parallel axis theorem.

  $I=I_{\text{cm}}+m{L}^2$

Here, $I$ is the inertia of bob, $m$ is mass of bob, $L$ is the distance between center of sphere and point of support and $I_{\text{cm}}$ is the inertia of bob passing through center of mass.

Substitute $\frac{2}{5}m{r}^2$ for $I_{\text{cm}}$ in above expression.

  $I=\frac{2}{5}m{r}^2+m{L}^2$

Write the expression for object in term of period of pendulum.

  $T=2\pi \sqrt{\frac{I}{mgh}}$

Here, $g$ is the gravitational acceleration and $T$ is the time period of object.

Substitute $I=\frac{2}{5}m{r}^2+m{L}^2$ for $I$ in above expression and simplify.

  $T=T_0{\left(1+\frac{2r^2}{5L^2}\right)}^{1/2}$   ...... (I)

Expand the term ${\left(1+\frac{2r^2}{5L^2}\right)}^{1/2}$ by binomial expansion for $r<<L$ .

  ${\left(1+\frac{2r^2}{5L^2}\right)}^{1/2}=1+\frac{r^2}{5L^2}$

Substitute $1+\frac{r^2}{5L^2}$ for ${\left(1+\frac{2r^2}{5L^2}\right)}^{1/2}$ in equation (I).

  $T=T_0\left(1+\frac{r^2}{5L^2}\right)$

Here, $r$ is the radius of bob.

Conclusion:

Thus, the period of small oscillations when radius is much smaller than length of pendulumis $T=T_0\left(1+\frac{r^2}{5L^2}\right)$ .

(c)

To determine

The error in the time period of oscillations; the value of radius for $1.00$ percent error.

Explanation of Solution

Given:

The distance between center of sphere and point of support is $100\text{cm}$ .

The radius of bob in first case is $2.00\text{cm}$ .

The error for second case is $1.00\%$ .

Introduction:

Time period is the time taken to cover one oscillation. The body resists any angular acceleration due to inertia present it which is known as moment of inertia.

The moment of inertia of sphere is $\frac{2}{5}m{r}^2$ .

Write the expression for moment of inertia using parallel axis theorem.

  $I=I_{\text{cm}}+m{L}^2$

Here, $I$ is the inertia of bob, $m$ is mass of bob, $L$ is the distance between center of sphere and point of support and $I_{\text{cm}}$ is the inertia of bob passing through center of mass.

Substitute $\frac{2}{5}m{r}^2$ for $I_{\text{cm}}$ in above expression.

  $I=\frac{2}{5}m{r}^2+m{L}^2$

Write the expression for object in term of period of pendulum.

  $T=2\pi \sqrt{\frac{I}{mgh}}$

Here, $g$ is the gravitational acceleration and $T$ is the time period of object.

Substitute $I=\frac{2}{5}m{r}^2+m{L}^2$ for $I$ in above expression and simplify.

  $T=T_0{\left(1+\frac{2r^2}{5L^2}\right)}^{1/2}$   ...... (I)

Expand the term ${\left(1+\frac{2r^2}{5L^2}\right)}^{1/2}$ by binomial expansion for $r<<L$ .

  ${\left(1+\frac{2r^2}{5L^2}\right)}^{1/2}=1+\frac{r^2}{5L^2}$

Substitute $1+\frac{r^2}{5L^2}$ for ${\left(1+\frac{2r^2}{5L^2}\right)}^{1/2}$ in equation (I) and rearrange in term of $T/T_0$ .

  $\frac{T}{T_0}=\left(1+\frac{r^2}{5L^2}\right)$

Here, $r$ is the radius of bob.

Calculate the error of time period for approximation of $T=T_0$ .

  $\begin{array}{c}\frac{\Delta T}{T}\times 100\%\approx \left(\frac{T-T_0}{T_0}\right)\times 100\%\\ =\left(\frac{T}{T_0}-1\right)\times 100\%\end{array}$

Substitute $\left(1+\frac{r^2}{5L^2}\right)$ for $\frac{T}{T_0}$ in above expression and simplify.

  $\frac{\Delta T}{T}\times 100\%=\frac{r^2}{5L^2}\times 100\%$   ...... (2)

Conclusion:

For case1:

Substitute $2.00\text{cm}$ for $r$ and $100\text{cm}$ for $L$ in equation (2).

  $\begin{array}{c}\frac{\Delta T}{T}\times 100\%=\frac{{\left(2.00\text{cm}\right)}^2}{5{\left(100\text{cm}\right)}^2}\times 100\%\\ =0.00008\times 100\%\\ =0.00800\%\end{array}$

For case2:

Substitute $1.00\%$ for $\frac{\Delta T}{T}\times 100\%$ and $100\text{cm}$ for $L$ in equation (2).

  $\begin{array}{c}1.00\%=\frac{{\left(r\right)}^2}{5{\left(100\text{cm}\right)}^2}\times 100\%\\ r=\left(100\text{cm}\right)\sqrt{0.0500}\\ =22.4\text{cm}\end{array}$

Thus, the error in the time period of oscillation is $0.008\%$ ; the value of radius for $1.00$ percent error is $22.4\text{cm}$ .