Questions & explanations
1. A stone tied to a string of length 1 m is whirled in a vertical circle. At the bottom, its speed is 7 m/s. At what distance from the centre of the circle does the stone hit the ground after the string goes slack? (Take g = 10 m/s², ground at the level of the centre.)
- 1.0 m
- 0.5 m
- 1.5 m
- 2.0 m
Answer: 1.0 m
The string goes slack when T=0. Using radial equation T - mg cosθ = mv²/r, at slack v² = -gr cosθ. Energy conservation: v² = v_bottom² - 2gr(1 - cosθ). Equating gives cosθ = (2gr - v_bottom²)/(3gr) = (20 - 49)/(30) = -29/30. Then v² = -gr cosθ = 10*(29/30) = 29/3 m²/s². The launch angle from horizontal is θ (since tangent perpendicular to radius), so v_x = v sinθ, v_y = -v cosθ. With sinθ = √(1 - cos²θ) = √(1 - 841/900) = √(59/900) = √59/30. The time to hit ground from height y = -r cosθ = 29/30 m: using y = v_y t + ½gt², with v_y = -v cosθ = -√(29/3)*(-29/30) = (29/30)√(29/3). Solve quadratic: t = √(2y/g) = √(58/300) = √(29/150). Horizontal distance = v_x t = (√(29/3)*√59/30)*√(29/150) = (29√59)/(30√450) = (29√59)/(30*15√2) = (29√59)/(450√2) ≈ 0.99 m ≈ 1.0 m.
2. A stone tied to a string is whirled in a vertical circle of radius 1 m. Its speed at the bottom is 6 m/s. Which graph best describes the variation of tension T with angle θ measured from the bottom?
- T decreases linearly from bottom to top, becoming zero at the top.
- T decreases, becomes zero at some angle above horizontal, then remains zero.
- T decreases, becomes zero at the horizontal, then increases again.
- T remains constant throughout the motion.
Answer: T decreases, becomes zero at some angle above horizontal, then remains zero.
Using energy conservation, speed at angle θ: v² = v_b² - 2gr(1-cosθ). Tension T = mv²/r + mg cosθ. With v_b=6 m/s, r=1 m, g=10 m/s², T becomes zero when cosθ = - (v_b²/(gr) - 2)/3 = - (36/10 - 2)/3 = - (3.6-2)/3 = -1.6/3 ≈ -0.533, giving θ ≈ 122° (above horizontal). For larger θ, T would be negative, so string slackens and T=0. Thus T decreases, becomes zero at an angle above horizontal, then remains zero.
3. Two masses 0.2 kg and 0.3 kg are connected by a light string passing through a smooth ring on a horizontal table. The string is taut and both masses move in concentric circles about the ring. The total string length is 1 m. If the tension in the string is 10 N, what is the angular speed (in rad/s) of the masses?
- 10.0
- 7.07
- 9.13
- 11.5
Answer: 9.13
Using the condition that the net force on the ring is zero, the tensions on both sides are equal. For each mass, tension provides centripetal force: T = m₁ω²r₁ = m₂ω²r₂. Thus m₁r₁ = m₂r₂. Also r₁ + r₂ = L = 1 m. Solving gives r₁ = m₂L/(m₁+m₂) = 0.3/0.5 = 0.6 m. Then ω = √(T/(m₁r₁)) = √(10/(0.2×0.6)) = √(10/0.12) ≈ 9.13 rad/s.
4. A car takes a turn of radius 100 m on a road with friction coefficient 0.5. The road is banked at 30° (tan30° ≈ 0.577). Which configuration gives the highest maximum safe speed?
- Flat road: 22.4 m/s
- Banked road with friction: 38.9 m/s
- Banked road without friction: 24.0 m/s
- Flat road: 31.6 m/s
Answer: Banked road with friction: 38.9 m/s
For a banked road with friction, the maximum safe speed is v_max = √(r g (tanθ + μ)/(1 - μ tanθ)). Substituting r=100 m, g=10 m/s², tanθ=0.577, μ=0.5 gives v_max = √(1000 × 1.077 / 0.7115) ≈ √1514 ≈ 38.9 m/s, which is higher than the flat road (22.4 m/s) and the frictionless banked road (24.0 m/s).
5. A conical pendulum and a simple pendulum have the same string length L. Which graph correctly shows the variation of the time period T of the conical pendulum with the angle θ that the string makes with the vertical?
- T increases with θ, starting from T = 2π√(L/g) at θ = 0.
- T is constant and equal to 2π√(L/g) for all θ.
- T decreases with θ, starting from T = 2π√(L/g) at θ = 0.
- T decreases with θ, starting from T = 2π√(L/g) at θ = 0, but becomes zero at θ = 90°.
Answer: T decreases with θ, starting from T = 2π√(L/g) at θ = 0.
For a conical pendulum, the centripetal force is provided by the horizontal component of tension. The time period is T = 2π√(L cosθ / g). At θ = 0, cosθ = 1, so T = 2π√(L/g). As θ increases, cosθ decreases, so T decreases. Hence the correct graph shows T decreasing from the simple pendulum value.
6. The time period T of a conical pendulum of length L with angle θ is given by:
- T = 2π √(L / (g sinθ))
- T = 2π √(L / (g cosθ))
- T = 2π √(L sinθ / g)
- T = 2π √(L cosθ / g)
Answer: T = 2π √(L cosθ / g)
For a conical pendulum, the bob moves in a horizontal circle of radius r = L sinθ. The centripetal force is provided by the horizontal component of tension: T sinθ = mω²r. The vertical equilibrium gives T cosθ = mg. Eliminating T gives ω² = g/(L cosθ). Since T = 2π/ω, we get T = 2π √(L cosθ/g).
7. A car takes a banked curve of radius 100 m with banking angle 20° and coefficient of static friction 0.5. What is the maximum safe speed in m/s? (tan 20° = 0.364, g = 10 m/s²)
- 30.0
- 24.7
- 38.9
- 32.5
Answer: 32.5
Using the formula for maximum speed on a banked curve with friction: v_max = √(r g (tanθ + μ)/(1 - μ tanθ)). Substituting r=100 m, g=10 m/s², tan20°=0.364, μ=0.5 gives numerator = 0.864, denominator = 0.818, so v_max = √(1000 × 0.864 / 0.818) ≈ √1056 ≈ 32.5 m/s.
8. A stone tied to a string is whirled in a vertical circle of radius 1 m. A small constant tangential drag force of 0.5 N acts opposite to the velocity. If the stone has mass 0.1 kg and the minimum speed at the bottom to just complete the loop in the ideal case is √(5gr) ≈ 7.07 m/s, what is the minimum speed at the bottom required when drag is present?
- 9.07 m/s
- 7.07 m/s
- 6.07 m/s
- 8.07 m/s
Answer: 9.07 m/s
Using work-energy theorem: work done by gravity = -2mgr = -2 J, work done by drag = -F_drag × πr = -0.5π ≈ -1.57 J. At top, minimum KE = 0.5mgr = 0.5 J. So initial KE = 2 + 1.57 + 0.5 = 4.07 J. v = √(2×4.07/0.1) = √81.4 ≈ 9.02 m/s. Closest option is 9.07 m/s.
9. A stone tied to a string moves in a vertical circle of radius 2 m. If its speed at the bottom is 10 m/s, what is its speed at the top? (g = 10 m/s²)
- √80 m/s
- √20 m/s
- √60 m/s
- √140 m/s
Answer: √60 m/s
Using conservation of mechanical energy: total energy at bottom = total energy at top. (1/2)mv_bottom² = (1/2)mv_top² + mg(2r). Cancel m, multiply by 2: v_bottom² = v_top² + 4gr. So v_top² = v_bottom² − 4gr = 100 − 4×10×2 = 60, hence v_top = √60 m/s.
10. A stone tied to a string of length 1 m is whirled in a vertical circle. At the bottom, its speed is 6 m/s. At what angle from the bottom does the string go slack? (Take g = 10 m/s²)
- cos⁻¹(2/3)
- cos⁻¹(-2/3)
- cos⁻¹(0.533)
- cos⁻¹(-0.533)
Answer: cos⁻¹(-0.533)
Using energy conservation and radial force equation: T(θ) = m v_bottom²/r - mg(2 - 3 cosθ). Set T=0, substitute v_bottom=6, r=1, g=10: 36 - 10(2 - 3 cosθ)=0 → 36 -20 +30 cosθ=0 → 16 +30 cosθ=0 → cosθ = -16/30 = -0.533. Thus θ = cos⁻¹(-0.533).
11. A car takes a banked turn of radius 100 m at angle 15°. If μ_s = 0.2, what is the maximum safe speed? (g = 10 m/s², tan15° = 0.27)
- √[1000(0.27-0.2)/(1-0.2×0.27)]
- √[1000(0.27-0.2)/(1+0.2×0.27)]
- √[1000(0.27+0.2)/(1+0.2×0.27)]
- √[1000(0.27+0.2)/(1-0.2×0.27)]
Answer: √[1000(0.27+0.2)/(1-0.2×0.27)]
For maximum safe speed, friction acts down the slope. Using N cosθ = mg + f sinθ and N sinθ + f cosθ = mv²/r with f = μ_s N, we get v_max = √[rg(tanθ + μ_s)/(1 - μ_s tanθ)]. Substituting r=100 m, g=10 m/s², tanθ=0.27, μ_s=0.2 gives option d.
12. A conical pendulum has string length 1 m and the bob moves in a horizontal circle of radius 0.6 m. A particle slides in a smooth hemispherical bowl of radius 1 m, moving in a horizontal circle at a depth 0.2 m below the rim. What is the ratio of their time periods?
- √(0.6) : √(0.8)
- √(0.8) : √(0.2)
- 1 : 1
- √(0.2) : √(0.8)
Answer: 1 : 1
For conical pendulum, vertical distance from suspension to plane is L cosθ = √(L² - r²) = √(1 - 0.36) = 0.8 m, so T₁ = 2π√(0.8/g). For bowl, vertical distance from centre to plane is R - h = 1 - 0.2 = 0.8 m, so T₂ = 2π√(0.8/g). Ratio = 1:1.