Physics Solution: Motion of Connected Discs
Analysis in Center of Mass (CM) Frame
When disc A is released, the Center of Mass (CM) moves with constant velocity, while the two masses rotate about the CM.
- Masses: $m_1$ (Disc A), $m_2$ (Disc B).
- Separation: $l$.
- Initial Velocity: $v_{A,initial}=0$, $v_{B,initial}=u$.
1. CM and Rotational Parameters:
$$V_{CM} = \frac{m_2 u}{m_1 + m_2}$$
The angular velocity of rotation about the CM:
$$\omega = \frac{v_{rel}}{l} = \frac{u}{l}$$
Radius of rotation for block B ($r_B$):
$$r_B = \frac{m_1 l}{m_1 + m_2}$$
2. Trajectory Equation for B:
The coordinates of B in the lab frame (setting $y$ along the string and $x$ along the velocity $u$):
$$x(t) = V_{CM} t + r_B \sin(\omega t)$$
$$v_x(t) = \frac{dx}{dt} = V_{CM} + r_B \omega \cos(\omega t)$$
Substituting variables:
$$v_x(t) = \frac{m_2 u}{m_1+m_2} + \left( \frac{m_1 l}{m_1+m_2} \right) \left( \frac{u}{l} \right) \cos(\omega t)$$
$$v_x(t) = \frac{u}{m_1+m_2} [m_2 + m_1 \cos(\omega t)]$$
3. Calculation of q (Loop Width):
The loop is formed when the particle moves backwards in the lab frame ($v_x < 0$). The width $q$ is the distance between the two turning points where $v_x = 0$.
$$m_2 + m_1 \cos(\omega t) = 0 \Rightarrow \cos(\omega t) = -\frac{m_2}{m_1}$$
Let $\phi = \cos^{-1}\left( \frac{m_2}{m_1} \right)$. Since $\cos(\omega t) < 0$, the solutions in the first cycle corresponding to the loop are:
$$\omega t_1 = \pi - \phi \quad \text{and} \quad \omega t_2 = \pi + \phi$$
The distance $q = |x(t_2) - x(t_1)|$. However, since the particle moves backward during the loop, the displacement is negative. We calculate magnitude:
$$x(t_2) - x(t_1) = V_{CM}(t_2 - t_1) + r_B (\sin(\omega t_2) - \sin(\omega t_1))$$
Term 1 (Linear drift):
$$t_2 – t_1 = \frac{1}{\omega} [(\pi+\phi) – (\pi-\phi)] = \frac{2\phi}{\omega} = \frac{2l \phi}{u}$$
$$V_{CM}(t_2 – t_1) = \left( \frac{m_2 u}{m_1+m_2} \right) \left( \frac{2l \phi}{u} \right) = \frac{2 m_2 l \phi}{m_1+m_2}$$
Term 2 (Oscillation):
Using $\sin(\pi+\phi) = -\sin\phi$ and $\sin(\pi-\phi) = \sin\phi$:
$$\sin(\omega t_2) – \sin(\omega t_1) = -\sin\phi – \sin\phi = -2\sin\phi$$
From $\cos\phi = m_2/m_1$, we find $\sin\phi = \sqrt{1 – (m_2/m_1)^2} = \frac{\sqrt{m_1^2 – m_2^2}}{m_1}$.
$$r_B (\dots) = \left( \frac{m_1 l}{m_1+m_2} \right) \left( -2 \frac{\sqrt{m_1^2 – m_2^2}}{m_1} \right) = – \frac{2l \sqrt{m_1^2 – m_2^2}}{m_1+m_2}$$
Total q:
Since oscillation pulls back more than drift pushes forward during the loop:
$$q = | \text{Term 1} + \text{Term 2} | = -(\text{Term 2}) – \text{Term 1}$$
$$q = \frac{2l \sqrt{m_1^2 – m_2^2}}{m_1+m_2} – \frac{2 m_2 l \phi}{m_1+m_2}$$
Answer:
$$q = \frac{2l}{m_1+m_2} \left[ \sqrt{m_1^2 – m_2^2} – m_2 \cos^{-1}\left(\frac{m_2}{m_1}\right) \right]$$
Analysis in Center of Mass (CM) Frame
When disc A is released, the Center of Mass (CM) moves with constant velocity, while the two masses rotate about the CM.
- Masses: $m_1$ (Disc A), $m_2$ (Disc B).
- Separation: $l$.
- Initial Velocity: $v_{A,initial}=0$, $v_{B,initial}=u$.
1. CM and Rotational Parameters:
$$V_{CM} = \frac{m_2 u}{m_1 + m_2}$$ The angular velocity of rotation about the CM: $$\omega = \frac{v_{rel}}{l} = \frac{u}{l}$$ Radius of rotation for block B ($r_B$): $$r_B = \frac{m_1 l}{m_1 + m_2}$$
$$V_{CM} = \frac{m_2 u}{m_1 + m_2}$$ The angular velocity of rotation about the CM: $$\omega = \frac{v_{rel}}{l} = \frac{u}{l}$$ Radius of rotation for block B ($r_B$): $$r_B = \frac{m_1 l}{m_1 + m_2}$$
2. Trajectory Equation for B:
The coordinates of B in the lab frame (setting $y$ along the string and $x$ along the velocity $u$): $$x(t) = V_{CM} t + r_B \sin(\omega t)$$ $$v_x(t) = \frac{dx}{dt} = V_{CM} + r_B \omega \cos(\omega t)$$ Substituting variables: $$v_x(t) = \frac{m_2 u}{m_1+m_2} + \left( \frac{m_1 l}{m_1+m_2} \right) \left( \frac{u}{l} \right) \cos(\omega t)$$ $$v_x(t) = \frac{u}{m_1+m_2} [m_2 + m_1 \cos(\omega t)]$$
The coordinates of B in the lab frame (setting $y$ along the string and $x$ along the velocity $u$): $$x(t) = V_{CM} t + r_B \sin(\omega t)$$ $$v_x(t) = \frac{dx}{dt} = V_{CM} + r_B \omega \cos(\omega t)$$ Substituting variables: $$v_x(t) = \frac{m_2 u}{m_1+m_2} + \left( \frac{m_1 l}{m_1+m_2} \right) \left( \frac{u}{l} \right) \cos(\omega t)$$ $$v_x(t) = \frac{u}{m_1+m_2} [m_2 + m_1 \cos(\omega t)]$$
3. Calculation of q (Loop Width):
The loop is formed when the particle moves backwards in the lab frame ($v_x < 0$). The width $q$ is the distance between the two turning points where $v_x = 0$. $$m_2 + m_1 \cos(\omega t) = 0 \Rightarrow \cos(\omega t) = -\frac{m_2}{m_1}$$ Let $\phi = \cos^{-1}\left( \frac{m_2}{m_1} \right)$. Since $\cos(\omega t) < 0$, the solutions in the first cycle corresponding to the loop are: $$\omega t_1 = \pi - \phi \quad \text{and} \quad \omega t_2 = \pi + \phi$$ The distance $q = |x(t_2) - x(t_1)|$. However, since the particle moves backward during the loop, the displacement is negative. We calculate magnitude: $$x(t_2) - x(t_1) = V_{CM}(t_2 - t_1) + r_B (\sin(\omega t_2) - \sin(\omega t_1))$$ Term 1 (Linear drift): $$t_2 – t_1 = \frac{1}{\omega} [(\pi+\phi) – (\pi-\phi)] = \frac{2\phi}{\omega} = \frac{2l \phi}{u}$$ $$V_{CM}(t_2 – t_1) = \left( \frac{m_2 u}{m_1+m_2} \right) \left( \frac{2l \phi}{u} \right) = \frac{2 m_2 l \phi}{m_1+m_2}$$ Term 2 (Oscillation): Using $\sin(\pi+\phi) = -\sin\phi$ and $\sin(\pi-\phi) = \sin\phi$: $$\sin(\omega t_2) – \sin(\omega t_1) = -\sin\phi – \sin\phi = -2\sin\phi$$ From $\cos\phi = m_2/m_1$, we find $\sin\phi = \sqrt{1 – (m_2/m_1)^2} = \frac{\sqrt{m_1^2 – m_2^2}}{m_1}$. $$r_B (\dots) = \left( \frac{m_1 l}{m_1+m_2} \right) \left( -2 \frac{\sqrt{m_1^2 – m_2^2}}{m_1} \right) = – \frac{2l \sqrt{m_1^2 – m_2^2}}{m_1+m_2}$$ Total q: Since oscillation pulls back more than drift pushes forward during the loop: $$q = | \text{Term 1} + \text{Term 2} | = -(\text{Term 2}) – \text{Term 1}$$ $$q = \frac{2l \sqrt{m_1^2 – m_2^2}}{m_1+m_2} – \frac{2 m_2 l \phi}{m_1+m_2}$$
The loop is formed when the particle moves backwards in the lab frame ($v_x < 0$). The width $q$ is the distance between the two turning points where $v_x = 0$. $$m_2 + m_1 \cos(\omega t) = 0 \Rightarrow \cos(\omega t) = -\frac{m_2}{m_1}$$ Let $\phi = \cos^{-1}\left( \frac{m_2}{m_1} \right)$. Since $\cos(\omega t) < 0$, the solutions in the first cycle corresponding to the loop are: $$\omega t_1 = \pi - \phi \quad \text{and} \quad \omega t_2 = \pi + \phi$$ The distance $q = |x(t_2) - x(t_1)|$. However, since the particle moves backward during the loop, the displacement is negative. We calculate magnitude: $$x(t_2) - x(t_1) = V_{CM}(t_2 - t_1) + r_B (\sin(\omega t_2) - \sin(\omega t_1))$$ Term 1 (Linear drift): $$t_2 – t_1 = \frac{1}{\omega} [(\pi+\phi) – (\pi-\phi)] = \frac{2\phi}{\omega} = \frac{2l \phi}{u}$$ $$V_{CM}(t_2 – t_1) = \left( \frac{m_2 u}{m_1+m_2} \right) \left( \frac{2l \phi}{u} \right) = \frac{2 m_2 l \phi}{m_1+m_2}$$ Term 2 (Oscillation): Using $\sin(\pi+\phi) = -\sin\phi$ and $\sin(\pi-\phi) = \sin\phi$: $$\sin(\omega t_2) – \sin(\omega t_1) = -\sin\phi – \sin\phi = -2\sin\phi$$ From $\cos\phi = m_2/m_1$, we find $\sin\phi = \sqrt{1 – (m_2/m_1)^2} = \frac{\sqrt{m_1^2 – m_2^2}}{m_1}$. $$r_B (\dots) = \left( \frac{m_1 l}{m_1+m_2} \right) \left( -2 \frac{\sqrt{m_1^2 – m_2^2}}{m_1} \right) = – \frac{2l \sqrt{m_1^2 – m_2^2}}{m_1+m_2}$$ Total q: Since oscillation pulls back more than drift pushes forward during the loop: $$q = | \text{Term 1} + \text{Term 2} | = -(\text{Term 2}) – \text{Term 1}$$ $$q = \frac{2l \sqrt{m_1^2 – m_2^2}}{m_1+m_2} – \frac{2 m_2 l \phi}{m_1+m_2}$$