跳到主要内容

题型分析 — Motion in a Circle

Question Type 1: 弧度定义与角速度计算​

如何识别​

直接问 define the radian 或计算 ω\omega、TT、ff。

标准解题方法​

定义题

radian: angle subtended at centre of circle when arc length equals radius. 必须包含 arc length = radius 或 angle subtended when arc length = radius。

评分标准​

评分要点
  • B1: arc length = radius 两个要素齐全
  • C1: ω=2π/T\omega = 2\pi/T 公式正确
  • A1: 数值正确，单位 rad s−1^{-1}

完整原题​

Example 1 — 9702_s24_qp_42 (2 marks):

Define the radian.

Details

📝 MS 展开查看 MS:

  • B1: angle (subtended) at the centre of a circle
  • B1: (when) arc length = radius

Example 2 — 9702_w23_qp_42 (4 marks):

The minute hand of a clock revolves at constant angular speed around the face of the clock. The tip of the minute hand is at a distance of 2.8 cm from the centre of the clock face. A small piece of modelling clay is attached to the tip of the minute hand. (i) Calculate the angular speed ω\omega of the minute hand. (ii) Calculate the magnitude of the centripetal acceleration of the piece of modelling clay.

Details

📝 MS 展开查看 MS:

  • C1: ω=2π/T\omega = 2\pi/T
  • C1: T=60×60=3600T = 60 \times 60 = 3600 s
  • A1: ω=1.75×10−3\omega = 1.75 \times 10^{-3} rad s−1^{-1}
  • C1: a=rω2a = r\omega^2
  • A1: a=2.8×10−2×(1.75×10−3)2=8.6×10−8a = 2.8 \times 10^{-2} \times (1.75 \times 10^{-3})^2 = 8.6 \times 10^{-8} m s−2^{-2}

Example 3 — 9702_s20_qp_41 (2 marks):

The period of rotation of the stars is 44.2 years. Calculate the angular velocity ω\omega.

Details

📝 MS 展开查看 MS:

  • C1: ω=2π/T\omega = 2\pi/T
  • C1: T=44.2×365×24×3600T = 44.2 \times 365 \times 24 \times 3600
  • A1: ω=4.5×10−9\omega = 4.5 \times 10^{-9} rad s−1^{-1}

常见陷阱​

常见陷阱
  • ω\omega 单位必须是 rad s−1^{-1}，不是 Hz
  • 周期 TT 的单位必须是秒(s)
  • 年转秒：乘以 365×24×3600365 \times 24 \times 3600

Question Type 2: 向心加速度与向心力计算​

如何识别​

给出 mm, rr, vv 或 ω\omega，求力或加速度。

标准解题方法​

计算步骤
  1. 写出公式 F=mv2/rF = mv^2/r 或 F=mrω2F = mr\omega^2
  2. 代入数值（注意单位）
  3. 得出结果

评分标准​

评分要点
  • C1: 写出正确公式
  • C1: 正确代入
  • A1: 正确答案 + 单位

完整原题​

Example 1 — 9702_w21_qp_41 (4 marks):

(a) With reference to velocity and acceleration, describe uniform circular motion. (b) An object of mass 0.35 kg is attached to one end of a string. The object moves in a horizontal circle of radius 0.62 m at constant speed. The string makes an angle of 30° with the vertical. Calculate: (i) the centripetal acceleration (ii) the tension in the string

Details

📝 MS 展开查看 MS:

  • B1: speed is constant
  • B1: direction (of velocity) changes continuously
  • B1: acceleration (directed) towards centre of circle
  • C1: Tsin⁡θ=mv2/rT\sin\theta = mv^2/r or Tsin⁡30=maT\sin30 = ma
  • C1: Tcos⁡30=mgT\cos30 = mg, so T=0.35×9.81/cos⁡30=3.96T = 0.35 \times 9.81 / \cos30 = 3.96 N
  • C1: a=Tsin⁡30/m=3.96×0.5/0.35a = T\sin30 / m = 3.96 \times 0.5 / 0.35
  • A1: a=5.7a = 5.7 m s−2^{-2}

Example 2 — 9702_s23_qp_41 (4 marks):

A sphere of mass 0.15 kg is attached to a string and moves in a horizontal circle at constant speed. The centripetal acceleration is a=7.2a = 7.2 m s−2^{-2}. The radius is 0.45 m. Determine: (i) the speed of the sphere (ii) the period of the circular motion

Details

📝 MS 展开查看 MS:

  • C1: a=v2/ra = v^2/r
  • C1: v=ar=7.2×0.45=1.8v = \sqrt{ar} = \sqrt{7.2 \times 0.45} = 1.8 m s−1^{-1}
  • A1: v=1.8v = 1.8 m s−1^{-1}
  • C1: ω=v/r=1.8/0.45=4.0\omega = v/r = 1.8/0.45 = 4.0 rad s−1^{-1}, T=2π/ωT = 2\pi/\omega
  • A1: T=1.57T = 1.57 s

常见陷阱​

常见陷阱
  • 向心力是合力，不是独立力
  • 注意 v=rωv = r\omega 中 rr 必须是运动半径
  • 张力不等于向心力（除非水平且无重力）

Question Type 3: 竖直圆周运动​

如何识别​

物体在竖直圆轨道上运动，最低点/最高点的力分析。

标准解题方法​

方法

最低点：N−mg=mv2/rN - mg = mv^2/r，所以 N=mg+mv2/rN = mg + mv^2/r 最高点：N+mg=mv2/rN + mg = mv^2/r，所以 N=mv2/r−mgN = mv^2/r - mg 临界条件（刚好过最高点）：N=0,v=grN = 0, v = \sqrt{gr}

评分标准​

评分规则
  • B1: 确定力的方向
  • C1: 牛顿第二定律方程
  • C1: 代入
  • A1: 答案

完整原题​

Example 1 — 9702_w22_qp_42 (5 marks):

An object of mass 0.20 kg moves in a vertical circle of radius 0.80 m. At the top of the circle, the speed is 3.0 m s−1^{-1}. (i) Determine the centripetal acceleration of the object. (ii) Describe how the two forces acting on the object give rise to this centripetal acceleration.

Details

📝 MS 展开查看 MS:

  • C1: a=v2/r=3.02/0.80a = v^2/r = 3.0^2/0.80
  • A1: a=11.25a = 11.25 m s−2^{-2}
  • B1: weight (downwards) and contact/normal force (downwards)
  • B1: both forces act towards centre
  • B1: resultant of these two forces = centripetal force

常见陷阱​

注意
  • 最高点重力和法向力都指向圆心（同向）
  • 最低点法向力减去重力等于向心力
  • 若速度太小，物体在最高点会掉下来