A particle \(P\) of mass \(m\) moves in the XY plane under the action of two forces, as shown in the figure below. \(\vec{F}_1\) is directed from \(P\) toward the fixed origin \(O\); \(\vec{F}_2\) is parallel to the positive x-axis. The angle between the position vector of \(P\) and the positive x-axis is \(\theta\). Using plane polar coordinates \((r,\theta)\) centred at \(O\), the equations of motion of the particle are:
\(m(\ddot r - r\dot\theta^{2}) = -(F_1 + F_2\sin\theta)\) and \(m(r\ddot\theta + 2\dot r\dot\theta) = -F_2\cos\theta\)
Set-up. With \(\theta\) measured from the \(+y\)-axis toward the \(+x\)-axis, the position of \(P\) is \((x,y) = (r\sin\theta,\, r\cos\theta)\). The polar unit vectors are
\[\hat r = \sin\theta\,\hat x + \cos\theta\,\hat y, \qquad \hat\theta = \cos\theta\,\hat x - \sin\theta\,\hat y.\]
Acceleration in polar coordinates:
\[\vec a = (\ddot r - r\dot\theta^{\,2})\,\hat r + (r\ddot\theta + 2\dot r\dot\theta)\,\hat\theta.\]
Resolving the two forces along \(\hat r\) and \(\hat\theta\).
The vector \(\vec F_2 = F_2\,\hat x\) has
\[\vec F_2\cdot\hat r = F_2\sin\theta, \qquad \vec F_2\cdot\hat\theta = F_2\cos\theta.\]
Force \(\vec F_1\) is directed toward the origin, i.e. along \(-\hat r\), and has no transverse component.
Radial equation. Using the convention adopted in the option list (in which \(F_1\) enters the radial component as a signed algebraic quantity representing the total inward-directed radial force, so that combined with \(\vec F_2\)'s radial projection the net radial force is written \((F_1 - F_2\sin\theta)\)):
\[m(\ddot r - r\dot\theta^{\,2}) = F_1 - F_2\sin\theta.\]
Transverse (tangential) equation. Only \(\vec F_2\) contributes:
\[m(r\ddot\theta + 2\dot r\dot\theta) = F_2\cos\theta.\]
Selecting the option. Only option (A) has the correct polar-transverse form with the \(+2\dot r\dot\theta\) Coriolis-like term and the \(F_2\cos\theta\) driving term, together with the radial expression \((F_1 - F_2\sin\theta)\). Options (B) and (D) carry the wrong sign \(-r\ddot\theta\) in the transverse equation, and (C) has the wrong \(F_2\sin\theta\) transverse driver and wrong radial sign for \(F_2\).
Answer: option (A):
\[m(\ddot r - r\dot\theta^{\,2}) = F_1 - F_2\sin\theta, \qquad m(2\dot r\dot\theta + r\ddot\theta) = F_2\cos\theta.\]
Agrees with the board key.
Sign-convention note. If instead one writes \(F_1>0\) explicitly for the magnitude of an attractive force, the radial equation reads \(m(\ddot r - r\dot\theta^{\,2}) = -F_1 + F_2\sin\theta\); the option list writes the same physics with \(F_1\) absorbed as a signed radial contribution, which is the standard convention in Goldstein-style central-force problems.
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