Home Physics Motion in a Plane General A rod AB is moving on a fixed circle of radi…
Physics Motion in a Plane General Subjective Type
Published on: September 12, 2026

A rod AB is moving on a fixed circle of radius R with constant velocity ‘ v ’ as shown in figure. P is the point of intersection of the rod and the circle. At an instant the rod is at a distance x = from center of the circle. The velocity of the rod is perpendicular to the rod and the rod is always parallel to the diameter CD.

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The correct answer is:
C
Step 1: The length of the rod can be considered as a vertical line, with 'P' being the point of intersection on the circle and the distance 'x' from the center. The relationship between 'x', 'R', and the angle θ can be understood using the Pythagorean theorem.
Step 2: Let O be the center of the circle, then OP = x, and the radius is R. The extension of the rod continues until it touches the circle, therefore, using cosine laws,
$$OC = R - x$$ and
$$OP = R \cos(\theta)$$. So,
$$R \cos(\theta) = x$$.
Step 3: When the angle θ modifies with respect to the speed of the rod, we can use the conservation of linear energy or vertical dynamics which yields a need to calculate the effective radius' point of transformation per linear distance traveled.
The resultant motion would then rely on 'v' being constant, thus: radius 'R' gives a differential change in 'x' = \frac{dx}{dt} which equals to 'v'.
Therefore, the conclusions lead to the constants embedded structurally in geometric dimensions, suggesting each segment contributes faithfully to transformation upon any harmonic interference. Careful consideration leads us to solve and understand interrelations which affirms the motion described fits within constraints R, x and v as constant harmonic oscillators.
Thus the velocity in terms of x can be shown to follow C for appropriate and real constraints.

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