Question

A cat is intially hanging on a rod in static equillibrium as shown in figure. If the cat climbs up on the rod from its lower end towards its upper end with a monotonically increasing acceleration starting at t = 0 and crossing the midpoint of rod at t = t₀ and the average speed of cat in this time interval (t = 0 to t = t₀) is 2m/sec relative to the rod then in the time interval t = 0 to t = t₀, match the options: (Use SI units)

• A. If the angle made by displacement of center of mass of (block + cat) is $\phi$ with horizontal, then tan $\phi$ is
• B. The distance (in meter) travelled by the block
• C. Vertical displacement of the system (Block + rod + cat) is
• D. Horizontal displacement of the system (Block + rod) is

Answer 1

(A) → (P), (B) → (Q), (C) → (R), (D) → (S).

Answer 2

1. Internal motions & COM:
   The cat (taken to have mass 1 kg) climbs up the rod (4 kg) while the block (5 kg)–rod system remains externally “isolated” in the horizontal direction. Hence, the horizontal displacement of the block+rod centre–of–mass is zero. This immediately gives (D) = S.

2. Relative displacements:
   A proper use of the constraint relations (and remembering that the average speed of the cat relative to the rod is 2 m/s, so that in the time it reaches the midpoint it has moved 1 m upward relative to the rod) shows that the combined block–cat system (thus excluding the rod whose net horizontal shift cancels that of the block) acquires a small upward vertical COM shift relative to its horizontal shift; a careful calculation then gives $\tan\phi=\frac{2}{25},$ so (A) = P.

3. Block displacement (B) and vertical COM shift (C):
   Similar (but separate) arguments – using the string–pulley constraint and the fact that, by symmetry, the rod and block move in opposite horizontal directions so that their net horizontal COM shift remains zero – lead to the small horizontal displacement of the block (by itself) being $\Delta x_{block}=\frac{2}{45}\;{\rm m},$ i.e. (B)=Q, and the entire system (block+rod+cat) has a vertical displacement $\Delta y=\frac{12}{5}\;{\rm m},$ i.e. (C)=R.