Question

current attempt in progress
ball a is attached to one end of a rigid massless rod, while an identical ball b is attached to the center of the rod, as shown in the figure. each ball has a mass of m = 0.410 kg, and the length of each half of the rod is l = 0.390 m. this arrangement is held by the empty end and is whirled around in a horizontal circle at a constant rate, so that each ball is in uniform circular motion. ball a travels at a constant speed of va = 5.30 m/s. find (a) the tension of the part between a and b of the rod and (b) the tension of the part between b and the empty end.
(a) number
(b) number

Explanation:

Step1: Find the angular - speed of the system

Since both balls are part of the same rigid - body rotation, they have the same angular speed $\omega$. For ball A, the relationship between linear speed $v_A$ and angular speed $\omega$ is $v_A = r_A\omega$, where $r_A = 2L$ and $L = 0.390$ m, so $r_A=2\times0.390 = 0.780$ m. Then $\omega=\frac{v_A}{r_A}$.
$\omega=\frac{5.30}{0.780}\text{ rad/s}\approx6.795\text{ rad/s}$

Step2: Analyze the centripetal force on ball A

The centripetal force acting on ball A is provided by the tension $T_A$ in the part of the rod between A and B. The centripetal - force formula is $F_c = m\frac{v^2}{r}=mr\omega^2$. For ball A, $F_{cA}=T_A$, $m = 0.410$ kg, and $r = r_A=0.780$ m.
$T_A=m r_A\omega^2$
$T_A = 0.410\times0.780\times(6.795)^2$
$T_A=0.410\times0.780\times46.17$
$T_A\approx14.8$ N

Step3: Analyze the centripetal force on ball B

The centripetal force acting on ball B is provided by the net force $T_B - T_A$, where $T_B$ is the tension in the part of the rod between B and the empty end and $T_A$ is the tension in the part of the rod between A and B. The radius of the circular path of ball B is $r_B = L=0.390$ m.
$F_{cB}=T_B - T_A$, and $F_{cB}=m r_B\omega^2$
$T_B=T_A + m r_B\omega^2$
$T_B = 14.8+0.410\times0.390\times(6.795)^2$
$T_B = 14.8+0.410\times0.390\times46.17$
$T_B = 14.8+0.410\times18.01$
$T_B = 14.8 + 7.38$
$T_B\approx22.2$ N

Answer:

(a) $14.8$ N
(b) $22.2$ N