题目

Problem

Figure 2 shows a fairground ride that consists of a chair of mass $m$ attached to one end of a rigid arm of length $\dfrac{5a}4$ . The other end of the arm is freely hinged to the rim of a thin horizontal circular disc of radius $a$ . The disc rotates with constant angular speed $\omega$ about a vertical axis through the centre of the disc. As the ride rotates the arm remains in a vertical plane through the centre of the disc. The arm makes a constant angle $\theta$ with the vertical, where $\tan\theta=\dfrac34$

The chair is modelled as a particle and the arm is modelled as a light rod.

(a) Find the tension in the arm in terms of $m$ and $g$

(3)

(b) Find $\omega$ in terms of $a$ and $g$

(6)

题目中文翻译

图 2 所示的游乐设施由一个质量为 $m$ 的座椅与一根长为 $\dfrac{5a}4$ 的刚性臂组成，座椅连接于刚性臂的一端。臂的另一端自由铰接于半径为 $a$ 的水平薄圆盘的边缘。圆盘绕过其圆心的竖直轴以恒定角速度 $\omega$ 转动。设施转动时，臂始终位于过圆盘圆心的竖直平面内。臂与竖直方向的夹角恒为 $\theta$ ，其中 $\tan\theta=\dfrac34$ 。

将座椅视为质点，臂视为轻杆。

(a) 用 $m$ 和 $g$ 表示臂中的张力。

(3)

(b) 用 $a$ 和 $g$ 表示 $\omega$ 。

(6)

解答

(a)

解法一

思路

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座椅没有竖直加速度。轻杆上的拉力沿杆，因此竖直分量平衡重力。

答题过程

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Since $\tan\theta=3/4$ ,

\sin\theta=\frac35,\qquad \cos\theta=\frac45.

Let the tension in the arm be $T$ . Resolving vertically,

T\cos\theta=mg.

Thus

\frac45T=mg,\qquad \boxed{T=\frac54mg}.

(b)

解法一

思路

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圆周运动的半径要从旋转轴量到座椅，包括圆盘半径及杆的水平投影。再以张力水平分量提供向心力。

答题过程

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The chair lies outside the rim, so the radius of its horizontal circular path is

r=a+\frac{5a}{4}\sin\theta=a+\frac{5a}{4}\frac35=\frac{7a}{4}.

Resolving horizontally towards the vertical axis,

T\sin\theta=mr\omega^2.

Substituting part (a),

\frac54mg\frac35=m\frac{7a}{4}\omega^2.

Cancelling $m/4$ gives $3g=7a\omega^2$ . Therefore

\boxed{\omega=\sqrt{\frac{3g}{7a}}}.

M3 2021 June Q2

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程