Uniform Circular Motion — Question 9

Question 9

Two rotating masses impose unequal cord tensions

Two beads rotate together on a frictionless horizontal table at angular speed $\omega$. A 1 kg bead is 1 m from a fixed center, and a 2 kg bead lies on the same radial line at radius 2 m. Light taut cords join center to inner bead and inner to outer bead. Their tension limits are 50 N and 32 N, respectively.

1. Find both tensions as functions of $\omega$.

2. Find the largest allowed $\omega$ and identify which cord limits it.

3. At that limit, find both speeds and the inward net force on the inner bead.

Question 9 - Solution

Two rotating masses impose unequal cord tensions

(a) Begin with the outer bead.

For the outer bead, the outer cord alone supplies inward force: $$T_o=m_o\omega^2r_o=2\omega^2(2)=4\omega^2.$$ The inner bead has two opposed tensions: $$T_i-T_o=m_i\omega^2r_i=\omega^2\Longrightarrow\boxed{T_i=5\omega^2,\quad T_o=4\omega^2}\quad\text{(SI values)}.$$

(b) Apply both independent limits.

Apply both tension limits: $$5\omega^2\leq50\Longrightarrow\omega^2\leq10\ \mathrm{s^{-2}},\qquad 4\omega^2\leq32\Longrightarrow\omega^2\leq8\ \mathrm{s^{-2}}.$$ $$\boxed{\omega_{\max}=\sqrt8=2\sqrt2\ \mathrm{rad/s}}.$$ The outer cord limits the motion. At that speed, $T_o=32$ N and $T_i=40$ N, below its 50 N capacity.

(c) Use each bead’s own radius.

Each speed uses its own radius: $$\boxed{v_i=\omega r_i=2\sqrt2\ \mathrm{m/s}},\qquad \boxed{v_o=\omega r_o=4\sqrt2\ \mathrm{m/s}}.$$ $$\boxed{F_{{\mathrm{net}},i}=T_i-T_o=40-32=8\ \mathrm N\text{ inward}}.$$ This agrees with $m_i\omega^2r_i=1(8)(1)=8$ N. The outer cord pulls the inner bead outward.