Block A (0.40 kg)and block B (0.30 kg)are on a frictionless table (see figure).Spring 1 connects block A to a frictionless peg at 0 and spring 2 connects block A and block B.When the blocks are in uniform circular motion about 0,the springs have lengths of and ,as shown.The springs are ideal and massless,and the linear speed of block B is .If the spring constant of spring 1 is equal to ,the unstretched length of spring 1 is closest to
A) 0.51 m.
B) 0.52 m.
C) 0.53 m.
D) 0.54 m.
E) 0.55 m.

Question

Block A (0.40 kg)and block B (0.30 kg)are on a frictionless table (see figure).Spring 1 connects block A to a frictionless peg at 0 and spring 2 connects block A and block B.When the blocks are in uniform circular motion about 0,the springs have lengths of  and  ,as shown.The springs are ideal and massless,and the linear speed of block B is  .If the spring constant of spring 1 is equal to  ,the unstretched length of spring 1 is closest to 
A) 0.51 m. 
B) 0.52 m. 
C) 0.53 m. 
D) 0.54 m. 
E) 0.55 m.

Quizplus · Accepted Answer

When the blocks are in uniform circular motion, the net force on block A towards the center is provided by the tension in spring 1. The centripetal acceleration of block A is $\frac{v^2}{r}$, where $v$ is the speed of block B and $r$ is the radius of the circular motion. Thus, we have:
$$T = \frac{m_A v^2}{r}$$
where $m_A$ is the mass of block A. Since block A is also connected to block B by spring 2, the tension in spring 1 must also be equal to the sum of the forces on block A due to spring 2 and the gravitational force:
$$T = k_1(\Delta x_1) + k_2(\Delta x_2) + m_A g$$
where $k_1$ is the spring constant of spring 1, $\Delta x_1$ is the extension of spring 1 from its unstretched length, $k_2$ is the spring constant of spring 2, $\Delta x_2$ is the extension of spring 2 from its unstretched length, and $g$ is the acceleration due to gravity.

Since the springs are massless, the extensions of the springs must be equal to the lengths of the springs in the given configuration:
$$\Delta x_1 = \sqrt{r^2 - L_1^2} - L_{1,u}$$
$$\Delta x_2 = L_2 - L_{2,u}$$
where $L_1$ and $L_2$ are the lengths of springs 1 and 2 in the given configuration, and $L_{1,u}$ and $L_{2,u}$ are their respective unstretched lengths.

Substituting these expressions and simplifying, we get:
$$\sqrt{r^2 - L_1^2} = \frac{m_A}{m_A + m_B} \cdot \frac{v^2}{k_1} + \frac{k_2}{k_1} \cdot (L_2 - L_{2,u}) + \frac{m_A}{m_A + m_B} \cdot g$$

Solving for $r$, we get:
$$r = \sqrt{L_1^2 + \left(\frac{m_A}{m_A + m_B} \cdot \frac{v^2}{k_1} + \frac{k_2}{k_1} \cdot (L_2 - L_{2,u}) + \frac{m_A}{m_A + m_B} \cdot gight)^2}$$

Substituting the given values, we get:
\begin{align*}
r &= \sqrt{(0.118	ext{ m})^2 + \left(\frac{0.40	ext{ kg}}{0.40	ext{ kg} + 0.30	ext{ kg}} \cdot \frac{(11	ext{ m/s})^2}{44	ext{ N/m}} + \frac{0.20	ext{ N}}{44	ext{ N/m}} + \frac{0.40	ext{ kg}}{0.40	ext{ kg} + 0.30	ext{ kg}} \cdot 9.81	ext{ m/s}^2ight)^2} \
&\approx 0.535	ext{ m}
\end{align*}

Finally, the unstretched length of spring 1 is equal to $L_{1,u} = \sqrt{r^2 - L_1^2} \approx 0.53	ext{ m}$. Therefore, the correct answer is choice C.

Block a (0