Chapter 5, Problem 17P

To determine

To Derive:An algebraic expression for the tension in each rod.

Answer to Problem 17P

Solution:

The tension in rod connecting block A and B $=4{\pi }^2{f}^2{m}_B{r}_B$

The tension in rod connecting A and post $=4{\pi }^2{f}^2\left(m_B{r}_B+m_A{r}_A\right)$

Explanation of Solution

Given:

Mass of block A $=m_A$

Mass of block B $=m_B$

The radius of circular path in which block A moves $=r_A$

The radius of circular path in which block B moves= $r_B$

Angular frequency = f

Formula used:

Centripetal acceleration $={\omega }^{2}r$

Centripetal force acting on a block moves in a circular path radius r angular speed $\omega$ , and mass m is $m{\omega }^{2}r.$

Angular speed

  $\omega =2\pi f$

Where, f is the frequency.

Calculation:

  

  $N_A$ : Normal reaction by the smooth surface on block A

  $N_B$ : Normal reaction by the smooth surface on block B

  $m_{A.g}$ : Weight of block B

  $T_1:$ The tension in rod connected A and B block

  $T_2:$ The tension in rod connection the post and block A

For Block B:

The tension in rod provide centripetal acceleration $={\omega }^2{r}_B$

Apply Newton's second law of motion:

  $\begin{array}{l}{\displaystyle \sum fy=0}\\ N_B-m_{B}g=0\\ N_B=m_{B}g\\ T_1=m_B.a_c\\ T_1=m_B{\omega }^2{r}_B\\ T_1\text{= }{m}_B{\left(2\pi f\right)}^2.r_B\\ T_1=4{\pi }^2{m}_B.f^2{r}_B\end{array}$

Here, $a_c$ is the centripetal acceleration.

The tension in rod connecting A and B is $4{\pi }^2{m}_B.f^2{r}_B$ .

For Block A:

Apply Newton's second law of motion:

  $\begin{array}{l}{\displaystyle \sum fy}=0\\ N_A-m_{B}g=0\\ N_A=m_{B}g\\ T_1-T_2=m_{Ä}{a}_c\\ T_2-T_1=m_A\left({\omega }^2{r}_A\right)\end{array}$

Here, $a_c$ is the centripetal acceleration.

Substitute the value of $\omega =2\pi f$

  $\begin{array}{l}{T}_2-T_1=m_A{\left(2\pi f\right)}^2.r_A\\ T_2-T_1=4{\pi }^2{m}_A.f^2{r}_A\\ T_2=T_1+4{\pi }^2{m}_A.f^2{r}_A\end{array}$

Substitute the value of $T_1$ :

  $\begin{array}{l}{T}_2=4{\pi }^2{m}_B{f}^2{r}_B+4{\pi }^2{m}_A.f^2{r}_A\\ T_2=4{\pi }^2{f}^2\left(m_B.r_B+m_A.r_A\right)\end{array}$

Hence, tension in rod connecting block A and the post is $4{\pi }^2{f}^2\left(m_B.r_B+m_A.r_A\right)$ .

Conclusion:

Rod in tension connecting block A and B is smaller than tension in rod connecting the post and block B and this tension is equal to $4{\pi }^2{f}^2\left(m_B.r_B+m_A.r_A\right)$ .