Chapter 4, Problem 84P

(a)

To determine

The expression for $g$ in terms of $L$ , $t$ , $m_1$ and $m_2$ .

Answer to Problem 84P

The expression for $g$ in terms of $L$ , $t$ , $m_1$ and $m_2$ is $\left(\frac{m_1+m_2}{m_1-m_2}\right)\frac{2L}{t^2}$ .

Explanation of Solution

Formula used:

The expression for the acceleration of the wood system is given by,

  $a=\frac{\left(m_1-m_2\right)g}{m_1+m_2}$ ............ (1)

The expression for the distance of the system is given by,

  $L=v_{0}t+\frac{1}{2}a{t}^2$

Calculation:

The block is start from the rest position which means that the initial velocity of the block is, $v_0=0.0\text{m}/\text{s}$

The acceleration from Kinematics equation is calculated as,

  $\begin{array}{c}L=v_{0}t+\frac{1}{2}a{t}^2\\ L=0.0\text{m}/\text{s}\left(t\right)+\frac{1}{2}a{\left(t\right)}^2\\ L=\frac{1}{2}a{\left(t\right)}^2\\ a=\frac{2L}{t^2}\end{array}$ ............ (2)

From equation (1) and equation (2), the expression of the gravity is written as,

  $\begin{array}{c}a=\frac{\left(m_1-m_2\right)}{m_1+m_2}g\\ \frac{2L}{t^2}=\frac{\left(m_1-m_2\right)}{m_1+m_2}g\\ g=\left(\frac{m_1+m_2}{m_1-m_2}\right)\frac{2L}{t^2}\end{array}$

Conclusion:

Therefore, the expression for $g$ in terms of $L$ , $t$ , $m_1$ and $m_2$ is $\left(\frac{m_1+m_2}{m_1-m_2}\right)\frac{2L}{t^2}$ .

(b)

To determine

The proof that the small error in time measurement lead to the error in $g$ such that $dg/g=-2dt/t$

Answer to Problem 84P

The value of $dg/g$ is $-2dt/t$ for the small error.

Explanation of Solution

Formula used:

The expression for $g$ in terms of $L$ , $t$ , $m_1$ and $m_2$ is given by,

  $g=\left(\frac{m_1+m_2}{m_1-m_2}\right)\frac{2L}{t^2}$

Calculation:

Differentiate the expression of $g$ with respect to $t$ .

  $\begin{array}{l}g=\left(\frac{m_1+m_2}{m_1-m_2}\right)\frac{2L}{t^2}\\ dg=\left(\frac{m_1+m_2}{m_1-m_2}\right)2L\times d\left(\frac{1}{t^2}\right)dt\\ dg=\left(\frac{m_1+m_2}{m_1-m_2}\right)2L\times \left(\frac{-2}{t^3}\right)dt\\ dg=\left(\frac{m_1+m_2}{m_1-m_2}\right)2\left(\frac{a{t}^2}{2}\right)\times \left(\frac{-2}{t^3}\right)dt\end{array}$

Solve further as,

  $\begin{array}{l}dg=\left(\frac{m_1+m_2}{m_1-m_2}\right)a\times \left(\frac{-2}{t}\right)dt\\ dg=g\times \left(\frac{-2dt}{t}\right)\\ \frac{dg}{g}=\frac{-2dt}{t}\end{array}$

Conclusion:

Therefore, the value of $dg/g$ is $-2dt/t$ for the small error.

(c)

To determine

The value of $m_2$ .

Answer to Problem 84P

The value of $m_2$ is $0.997\text{kg}$ .

Explanation of Solution

Given:

The value of the $dg/g$ is, $dg/g=0.05$

The value of the $dt$ is, $dt=0.1\text{s}$

The value of the $L$ is, $L=3\text{m}$

The value of the mass is, $m_1=1\text{kg}$

Formula used:

The expression for $\frac{dg}{g}$ is given by,

  $\frac{dg}{g}=\frac{-2dt}{t}$

The expression for $g$ in terms of $L$ , $t$ , $m_1$ and $m_2$ is given by,

  $g=\left(\frac{m_1+m_2}{m_1-m_2}\right)\frac{2L}{t^2}$

Calculation:

The value of the time from the expression of $\frac{dg}{g}$ is calculated as,

  $\begin{array}{c}\frac{dg}{g}=\frac{-2dt}{t}\\ 0.05=\frac{-2\left(0.1\right)}{t}\\ t=-20\text{s}\end{array}$

The value of the mass $m_2$ from the expression of $g$ is calculated as,

  $\begin{array}{c}g=\left(\frac{m_1+m_2}{m_1-m_2}\right)\frac{2L}{t^2}\\ 9.8\text{m}/{\text{s}}^{\text{2}}=\left(\frac{1\text{kg}+m_2}{1\text{kg}-m_2}\right)\frac{2\left(3\text{m}\right)}{{\left(-20\text{s}\right)}^2}\\ 653.3=\left(\frac{1+m_2}{1-m_2}\right)\\ 653.3-653.3m_2=1+m_2\end{array}$

Solve further as,

  $\begin{array}{l}{m}_2+653.3m_2=653.3-1\\ 654.3m_2=652.3\\ m_2=\frac{652.3}{654.3}\\ m_2=0.997\text{kg}\end{array}$

Conclusion:

Therefore, the value of $m_2$ is $0.997\text{kg}$ .