3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
A particle moves on a circle of radius R with constant speed v. Consider the time interval during which its position vector, drawn from the centre, turns through 120∘. Let aˉ be the magnitude of the average acceleration (change of velocity divided by elapsed time) over this interval. The ratio aˉ/(v2/R) is ______. Exactly one option is correct.
A1
B23
C2π33
D2π3
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Answer: C
The velocity turns through the same angle θ=120∘ as the radius, its magnitude staying v. So ∣Δv∣=2vsin(θ/2)=2vsin60∘=3v.
Elapsed time t=Rθ/v=3v2πR.
aˉ=2πR/(3v)3v=2π33Rv2≈0.827Rv2.
The average is smaller than the instantaneous value v2/R because the acceleration vectors over the arc point in different directions and partly cancel. Answer (C).
Difficulty: Hard
Question 2Level 2 · ExamSingle correct
Two small bobs, of masses m and 3m, hang from the same fixed point O by light inextensible strings of lengths 25cm and 52cm. Each bob moves uniformly in a horizontal circle (conical pendulum) about the vertical line through O, and both have the same angular speed 7rad/s. Take g=9.8m/s2 and neglect air resistance. At an instant when both strings lie in one vertical plane, with the bobs on opposite sides of the vertical through O, the distance between the bobs is ______. Exactly one option is correct.
A33cm
B50cm
C69cm
D63cm
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Answer: D
For a conical pendulum with string length L at angle θ: Tcosθ=Mg and Tsinθ=Mω2Lsinθ, so T=Mω2L and Lcosθ=g/ω2.
The depth of each bob below O is h=Lcosθ=g/ω2=9.8/49=0.20 m. It does not depend on mass or string length, so both bobs move at the same level.
Radii: r1=0.252−0.202=0.15 m and r2=0.522−0.202=0.48 m.
On opposite sides at the same level the separation is r1+r2=0.63 m =63 cm. (On the same side it would be 33 cm.) Answer (D).
Difficulty: Hard
Question 3Level 3 · AdvancedSingle correct
A smooth hollow right circular cone is fixed with its axis vertical and its vertex pointing downward. Its semi-vertical angle (between the axis and the surface) is 30∘. A small particle slides on the inner surface in a horizontal circle with constant speed. The plane of the circle is 0.40m above the vertex. Take g=10m/s2. The time period of the motion is closest to ______. Exactly one option is correct.
A0.40s
B0.73s
C1.26s
D2.18s
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Answer: B
Only gravity and the normal force N act. The normal is perpendicular to the surface; the surface makes 30∘ with the vertical axis, so N makes 30∘ with the horizontal pointing inward and upward.
Vertical: Nsin30∘=mg. Horizontal, toward the axis: Ncos30∘=mv2/r.
Dividing: v2=grcot30∘. With r=htan30∘, v2=gh. The speed depends only on the height: v=10×0.4=2 m/s.
r=0.4tan30∘=0.2309 m, so T=2πr/v=0.726 s ≈0.73 s.
The horizontal component of N supplies the centripetal force; there is no separate 'centripetal force'. Answer (B).
Difficulty: Very hard
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