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M3 2023 June Q3

A Level / Edexcel / M3

IAL 2023 June Paper · Question 3

题目

Problem

3.

A uniform solid right circular cone CC has base radius rr, height HH and vertex VV.

A uniform solid SS, shown in Figure 3, is formed by removing from CC a uniform solid right circular cone of height hh (h<H)(h<H) that has the same base and axis of symmetry as CC.

(a) Show that the distance of the centre of mass of SS from VV is

14(3H−h)\frac14(3H-h)
(5)

The solid SS is suspended by two vertical light strings. The first string is attached to SS at VV and the second string is attached to SS at a point on the circumference of the circular base of SS.

The solid SS hangs freely in equilibrium with its axis of symmetry horizontal.

The tension in the first string is T1T_1 and the tension in the second string is T2T_2

(b) Find T1T2\dfrac{T_1}{T_2}, giving your answer in terms of HH and hh, in its simplest form.

(3)
题目中文翻译

均匀实心正圆锥 CC 的底面半径为 rr，高为 HH，顶点为 VV。

从 CC 中挖去一个高为 hh (h<H)(h<H)、与 CC 有相同底面和对称轴的均匀实心正圆锥，形成图 3 所示的均匀立体 SS。

(a) 证明 SS 的质心到 VV 的距离为

14(3H−h)\frac14(3H-h)
(5)

用两根竖直轻绳悬挂立体 SS。第一根绳连接于 SS 的 VV 点，第二根连接于 SS 的圆形底面的圆周上的一点。

立体 SS 自由悬挂并处于平衡，其对称轴水平。

第一根绳中的张力为 T1T_1，第二根绳中的张力为 T2T_2。

(b) 用 HH 和 hh 表示 T1T2\dfrac{T_1}{T_2}，以最简形式作答。

(3)

解答

(a)

解法一

思路

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挖去的圆锥与原锥有同一个底面，并不是相似的小锥。两锥质量分别正比于高 H,hH,h；以原顶点 VV 为统一参考，去锥的质心距离为 H−h/4H-h/4。

答题过程

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Let the density be ρ\rho and set c=ρπr2/3c=\rho\pi r^2/3. The original cone has mass cHcH, the removed cone has mass chch, and the remaining solid has mass c(H−h)c(H-h).

The centre of mass of a uniform solid cone lies one quarter of its height from its base. Therefore the distances of the two centres of mass from VV are

3H4andH−h4\frac{3H}{4}\quad\text{and}\quad H-\frac h4

for the original and removed cones respectively.

Let xˉ\bar x be the distance of the centre of mass of SS from VV. Taking first moments, with the removed mass subtracted,

c(H−h)xˉ=cH3H4−ch(H−h4).c(H-h)\bar x =cH\frac{3H}{4}-ch\left(H-\frac h4\right).

Cancelling cc and multiplying by 44 gives

4(H−h)xˉ=3H2−4Hh+h2.4(H-h)\bar x=3H^2-4Hh+h^2.

Factorising the right-hand side,

3H2−4Hh+h2=(3H−h)(H−h).3H^2-4Hh+h^2=(3H-h)(H-h).

Since H>hH>h, H−h≠0H-h\neq0. Consequently

xˉ=(3H−h)(H−h)4(H−h)=14(3H−h).\boxed{\bar x=\frac{(3H-h)(H-h)}{4(H-h)} =\frac14(3H-h).}

(b)

解法一

思路

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轴水平、两绳竖直，因此力矩的力臂是沿轴的水平距离。以质心取矩可直接消去重力，第二根绳虽在圆周上，但其水平作用线位置仍在底面处。

答题过程

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Let GG be the centre of mass and write VG=xˉ=(3H−h)/4VG=\bar x=(3H-h)/4. The two vertical tensions have horizontal distances xˉ\bar x and H−xˉH-\bar x from GG.

Taking moments about GG, the weight has zero moment, so

T1xˉ=T2(H−xˉ).T_1\bar x=T_2(H-\bar x).

Therefore

T1T2= H−xˉxˉ= H−(3H−h)/4(3H−h)/4= H+h3H−h.\begin{align*} \frac{T_1}{T_2} =&\,\frac{H-\bar x}{\bar x}\\ =&\,\frac{H-(3H-h)/4}{(3H-h)/4}\\ =&\,\boxed{\frac{H+h}{3H-h}}. \end{align*}