Chapter 14, Problem 75P

(a)

To determine

The distance $d$ for the pendulum that gives time period of $2.50\text{s}$ .

Explanation of Solution

Given:

The length of rod is $2.00\text{m}$ .

The mass of rod is $0.800\text{kg}$ .

The mass of uniform disk is $1.20\text{kg}$ .

The radius of uniform disk is $0.150\text{m}$ .

The perfect time period of pendulum is $3.50\text{s}$ .

The actual time period of pendulum is $2.50\text{s}$ .

Formula used:

Write the expression for moment of inertia of rod passing through center of mass.

  $I_{\text{rod}}=\frac{1}{3}m{L}^2$

Here, $m$ is mass of rod, $L$ is length of rod and $I_{\text{rod}}$ is moment of inertia of rod.

Write the expression for moment of inertia of disk passing through center of mass.

  $I_{\text{disk}}=\frac{1}{2}M{r}^2$

Here, $M$ is mass of disk, $r$ is radius of disc and $I_{\text{disk}}$ is moment of inertia of disk.

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

  $I=I_{\text{cm}}+M{d}^2$ …… (1)

Here, $I$ is the inertia of system, $d$ is the distance between contact point and center of mass and $I_{\text{cm}}$ is the inertia of rod and disc passing through center of mass.

The moment of inertia of system passing through center of mass is the sum of moment of inertia of rod and moment of inertia disk passing through center of mass.

Substitute $\frac{1}{3}m{L}^2+\frac{1}{2}M{r}^2$ for $I_{\text{cm}}$ in above expression.

  $I=\frac{1}{3}m{L}^2+\frac{1}{2}M{r}^2+M{d}^2$ …… (1)

Write the expression for center of mass of system.

  $x_{\text{cm}}=\frac{m{x}_1+Md}{m+M}$ …… (2)

Here, $x_1$ is center of mass of rod and $x_{\text{cm}}$ is center of mass of system.

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

  $T=2\pi \sqrt{\frac{I}{m_{\text{T}}g{x}_{\text{cm}}}}$

Here, $m_{\text{T}}$ is mass of system, $g$ is the gravitational acceleration and $T$ is the time period of object.

Rearrange the above expression in term of $I/x_{\text{cm}}$ .

  $\frac{I}{x_{\text{cm}}}=\frac{T^{2}g{m}_{\text{T}}}{4{\pi }^2}$ …… (3)

Calculation:

Substitute $0.800\text{kg}$ for $m$ , $1.20\text{kg}$ for $M$ , $2.00\text{m}$ for $L$ and $0.150\text{m}$ for $r$ in equation (1).

  $\begin{array}{c}I=\frac{1}{3}\left(0.800\text{kg}\right){\left(2.00\text{m}\right)}^2+\frac{1}{2}\left(1.20\text{kg}\right){\left(0.150\text{m}\right)}^2+\left(1.20\text{kg}\right)d^2\\ =1.0802\text{kg}\cdot {\text{m}}^2+\left(1.20\text{kg}\right)d^2\end{array}$

Substitute $0.800\text{kg}$ for $m$ , $1.20\text{kg}$ for $M$ and $1.00\text{m}$ for $x_1$ in equation (2).

  $\begin{array}{c}{x}_{\text{cm}}=\frac{\left(0.800\text{kg}\right)\left(1.00\text{m}\right)+\left(1.20\text{kg}\right)d}{0.800\text{kg}+1.20\text{kg}}\\ =0.400\text{m}+0.600d\end{array}$

Substitute $9.81\text{m}/{\text{s}}^2$ for $g$ , $2.00\text{kg}$ for $m_{\text{T}}$ , $2.50\text{s}$ for $T$ , $1.0802\text{kg}\cdot {\text{m}}^2+\left(1.20\text{kg}\right)d^2$ for $I$ and $0.400\text{m}+0.600d$ for $x_{\text{cm}}$ in equation (3) and solve for $d$ .

  $\begin{array}{c}\frac{1.0802\text{kg}\cdot {\text{m}}^2+\left(1.20\text{kg}\right)d^2}{0.400\text{m}+0.600d}=\frac{{\left(2.50\text{s}\right)}^2\left(9.81\text{m}/{\text{s}}^2\right)\left(2.00\text{kg}\right)}{4{\pi }^2}\\ d=1.63574\text{m}\end{array}$

Conclusion:

Thus, the distance $d$ for the pendulum is $1.63574\text{m}$ .

(b)

To determine

The change in distance and direction of disk to keep the clock at perfect time period.

Explanation of Solution

Given:

The length of rod is $2.00\text{m}$ .

The mass of rod is $0.800\text{kg}$ .

The mass of uniform disk is $1.20\text{kg}$ .

The radius of uniform disk is $0.150\text{m}$ .

The perfect time period of pendulum is $3.50\text{s}$ .

The delay in time period of clock is $5.00\text{min}/\text{d}$ .

Formula used:

Write the expression for moment of inertia of rod passing through center of mass.

  $I_{\text{rod}}=\frac{1}{3}m{L}^2$

Here, $m$ is mass of rod, $L$ is length of rod and $I_{\text{rod}}$ is moment of inertia of rod.

Write the expression for moment of inertia of disk passing through center of mass.

  $I_{\text{disk}}=\frac{1}{2}M{r}^2$

Here, $M$ is mass of disk, $r$ is radius of disc and $I_{\text{disk}}$ is moment of inertia of disk.

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

  $I=I_{\text{cm}}+M{d}^2$ …… (1)

Here, $I$ is the inertia of system, $d$ is the distance between contact point and center of mass and $I_{\text{cm}}$ is the inertia of rod and disc passing through center of mass.

The moment of inertia of system passing through center of mass is the sum of moment of inertia of rod and moment of inertia disk passing through center of mass.

Substitute $\frac{1}{3}m{L}^2+\frac{1}{2}M{r}^2$ for $I_{\text{cm}}$ in above expression.

  $I=\frac{1}{3}m{L}^2+\frac{1}{2}M{r}^2+M{d}^2$ …… (1)

Write the expression for center of mass of system.

  $x_{\text{cm}}=\frac{m{x}_1+Md}{m+M}$ …… (2)

Here, $x_1$ is center of mass of rod and $x_{\text{cm}}$ is center of mass of system.

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

  $T=2\pi \sqrt{\frac{I}{m_{\text{T}}g{x}_{\text{cm}}}}$

Here, $m_{\text{T}}$ is mass of system, $g$ is the gravitational acceleration and $T$ is the time period of object.

Rearrange the above expression in term of $I/x_{\text{cm}}$ .

  $\frac{I}{x_{\text{cm}}}=\frac{T^{2}g{m}_{\text{T}}}{4{\pi }^2}$ …… (3)

Calculation:

Substitute $0.800\text{kg}$ for $m$ , $1.20\text{kg}$ for $M$ , $2.00\text{m}$ for $L$ and $0.150\text{m}$ for $r$ in equation (1).

  $\begin{array}{c}I=\frac{1}{3}\left(0.800\text{kg}\right){\left(2.00\text{m}\right)}^2+\frac{1}{2}\left(1.20\text{kg}\right){\left(0.150\text{m}\right)}^2+\left(1.20\text{kg}\right)d^2\\ =1.0802\text{kg}\cdot {\text{m}}^2+\left(1.20\text{kg}\right)d^2\end{array}$

Substitute $0.800\text{kg}$ for $m$ , $1.20\text{kg}$ for $M$ and $1.00\text{m}$ for $x_1$ in equation (2).

  $\begin{array}{c}{x}_{\text{cm}}=\frac{\left(0.800\text{kg}\right)\left(1.00\text{m}\right)+\left(1.20\text{kg}\right)d}{0.800\text{kg}+1.20\text{kg}}\\ =0.400\text{m}+0.600d\end{array}$

Substitute $9.81\text{m}/{\text{s}}^2$ for $g$ , $2.00\text{kg}$ for $m_{\text{T}}$ , $3.50\text{s}$ for $T$ , $1.0802\text{kg}\cdot {\text{m}}^2+\left(1.20\text{kg}\right)d^2$ for $I$ and $0.400\text{m}+0.600d$ for $x_{\text{cm}}$ in equation (3) and solve for $d$ .

  $\begin{array}{c}\frac{1.0802\text{kg}\cdot {\text{m}}^2+\left(1.20\text{kg}\right)d^2}{0.400\text{m}+0.600d}=\frac{{\left(3.50\text{s}\right)}^2\left(9.81\text{m}/{\text{s}}^2\right)\left(2.00\text{kg}\right)}{4{\pi }^2}\\ d=3.37826\text{m}\end{array}$

The clock loses $5.00\text{min}/\text{d}$ means that total number of minute count is $5.00\text{min}$ less than $24\text{hr}$ for clock. There are total $1440\text{min}$ in a day; so the time count for the clock would be $1435\text{min}$ .

Calculate the time period of clock.

  $\begin{array}{c}{T}^{\prime }=\frac{1440\text{min}}{1435\text{min}}\left(3.50\text{s}\right)\\ =3.51220\text{s}\end{array}$

Substitute $9.81\text{m}/{\text{s}}^2$ for $g$ , $2.00\text{kg}$ for $m_{\text{T}}$ , $3.51220\text{s}$ for $T^{\prime }$ , $1.0802\text{kg}\cdot {\text{m}}^2+\left(1.20\text{kg}\right){\left(d^{\prime }\right)}^2$ for $I$ and $0.400\text{m}+0.600d^{\prime }$ for $x_{\text{cm}}$ in equation (3) and solve for $d^{\prime }$ .

  $\begin{array}{c}\frac{1.0802\text{kg}\cdot {\text{m}}^2+\left(1.20\text{kg}\right){\left(d^{\prime }\right)}^2}{0.400\text{m}+0.600d^{\prime }}=\frac{{\left(3.50\text{s}\right)}^2\left(9.81\text{m}/{\text{s}}^2\right)\left(2.00\text{kg}\right)}{4{\pi }^2}\\ d^{\prime }=3.40140\text{m}\end{array}$

Calculate the change in distance of disk.

  $\begin{array}{c}\Delta d=d^{\prime }-d\\ =3.40140\text{m}-3.37826\text{m}\\ =0.02314\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\\ =2.31\text{cm}\end{array}$

Conclusion:

Thus, the disk will be moved upward by $2.31\text{cm}$ to keep the clock at perfect time period.