Solution to Question 46
Physics Principle: Work-Energy Theorem with Variable Forces.
System Analysis:
A bead of mass \(m\) and charge \(q\) moves along a rod. A fixed charge \(q\) is located at a perpendicular distance \(d\) from the center of the rod.
Motion stages:
1. From End A (start) to Center B.
2. From Center B to End C.
Variables:
\(v_0\): Unknown velocity of projection at start (End A).
\(v_1\): Velocity at the center (Point B).
\(v_2\): Velocity at the other end (End C).
\(U_A, U_B, U_C\): Electrostatic Potential Energy at points A, B, and C.
\(W_f\): Work done by friction.
Step 1: Energy Balance Equations
The work done by friction is symmetric. The normal force \(N\) depends on distance from the center, so \(N(x) = N(-x)\). Thus, the work done by friction from A to B is equal in magnitude to the work done from B to C. Let this magnitude be \(W_{fric}\).
Let \(\Delta U = U_A – U_B\) be the potential energy difference between the end and the center.
\(U_A = \frac{kq^2}{l}\) (Distance is \(l\))
\(U_B = \frac{kq^2}{d}\) (Distance is \(d\))
Since \(d < l\), \(U_B > U_A\), so potential energy increases going to the center.
Motion A \(\to\) B:
Change in KE + Change in PE = Work by Friction
$$ \frac{1}{2}m v_1^2 – \frac{1}{2}m v_0^2 + (U_B – U_A) = -W_{fric} $$
$$ \frac{1}{2}m v_0^2 – \frac{1}{2}m v_1^2 = (U_B – U_A) + W_{fric} \quad \dots(1) $$
Motion B \(\to\) C:
Change in KE + Change in PE = Work by Friction
$$ \frac{1}{2}m v_2^2 – \frac{1}{2}m v_1^2 + (U_C – U_B) = -W_{fric} $$
Note that \(U_C = U_A\).
$$ \frac{1}{2}m v_1^2 – \frac{1}{2}m v_2^2 = (U_B – U_A) + W_{fric} \quad \dots(2) $$
Step 2: Solve for \(v_0\)
Notice the Right Hand Side (RHS) of equations (1) and (2) is identical: \((U_B – U_A) + W_{fric}\).
Therefore, the Left Hand Sides must be equal:
$$ \frac{1}{2}m v_0^2 – \frac{1}{2}m v_1^2 = \frac{1}{2}m v_1^2 – \frac{1}{2}m v_2^2 $$
Wait, this would imply \(v_0^2 + v_2^2 = 2v_1^2\) ONLY if friction work cancels out or matches potential somehow. But actually, we need to eliminate \(W_{fric}\). Let’s rewrite strictly using the Work-Energy theorem to be safe.
Let’s sum the equations instead? No, we need to find \(v_0\).
From (2): \(W_{fric} = \frac{1}{2}m v_1^2 – \frac{1}{2}m v_2^2 – (U_B – U_A)\).
Substitute \(W_{fric}\) into (1):
$$ \frac{1}{2}m v_1^2 – \frac{1}{2}m v_0^2 + (U_B – U_A) = – [ \frac{1}{2}m v_1^2 – \frac{1}{2}m v_2^2 – (U_B – U_A) ] $$
$$ \frac{1}{2}m v_1^2 – \frac{1}{2}m v_0^2 + (U_B – U_A) = – \frac{1}{2}m v_1^2 + \frac{1}{2}m v_2^2 + (U_B – U_A) $$
Canceling \((U_B – U_A)\) from both sides:
$$ \frac{1}{2}m v_1^2 – \frac{1}{2}m v_0^2 = – \frac{1}{2}m v_1^2 + \frac{1}{2}m v_2^2 $$
$$ v_1^2 – v_0^2 = -v_1^2 + v_2^2 $$
$$ v_0^2 = 2v_1^2 – v_2^2 $$
Wait, this assumes the Work done by friction is the same as the change in mechanical energy without the friction term? Let’s re-evaluate.
Actually, look at eq (1) and (2) again.
(1) Loss in KE_A->B = Gain in PE + Work against Friction
(2) Loss in KE_B->C = Loss in PE + Work against Friction? No.
From B to C, Potential Energy Decreases (helps motion). Friction opposes motion.
Equation B->C: \(K_f – K_i = -\Delta U – W_f\)
$$ \frac{1}{2}m v_2^2 – \frac{1}{2}m v_1^2 = -(U_A – U_B) – W_{fric} = (U_B – U_A) – W_{fric} $$
Let’s restate clearly:
1) \( \frac{1}{2}m v_0^2 = \frac{1}{2}m v_1^2 + (U_B – U_A) + W_{fric} \)
2) \( \frac{1}{2}m v_1^2 = \frac{1}{2}m v_2^2 + (U_A – U_B) + W_{fric} \)
which is \( \frac{1}{2}m v_1^2 = \frac{1}{2}m v_2^2 – (U_B – U_A) + W_{fric} \)
From (2): \( W_{fric} = \frac{1}{2}m v_1^2 – \frac{1}{2}m v_2^2 + (U_B – U_A) \)
Substitute into (1):
$$ \frac{1}{2}m v_0^2 = \frac{1}{2}m v_1^2 + (U_B – U_A) + [ \frac{1}{2}m v_1^2 – \frac{1}{2}m v_2^2 + (U_B – U_A) ] $$
$$ \frac{1}{2}m v_0^2 = m v_1^2 – \frac{1}{2}m v_2^2 + 2(U_B – U_A) $$
Multiply by 2/m:
$$ v_0^2 = 2v_1^2 – v_2^2 + \frac{4}{m}(U_B – U_A) $$
Step 3: Substitute Potential Energy Terms
\( U_B = \frac{kq^2}{d} \), \( U_A = \frac{kq^2}{l} \) $$ U_B – U_A = kq^2 \left( \frac{1}{d} – \frac{1}{l} \right) $$ Substituting \( k = \frac{1}{4\pi\epsilon_0} \): $$ \frac{4}{m}(U_B – U_A) = \frac{4}{m} \frac{q^2}{4\pi\epsilon_0} \left( \frac{1}{d} – \frac{1}{l} \right) = \frac{q^2}{\pi\epsilon_0 m} \left( \frac{1}{d} – \frac{1}{l} \right) $$