题目

Problem

A car is travelling round a circular track. The track is banked at an angle $\alpha$ to the horizontal, as shown in Figure 4.

The car and driver are modelled as a particle.

The car moves round the track with constant speed in a horizontal circle of radius $r$ .

When the car is moving with speed $\dfrac12\sqrt{gr}$ round the circle, there is no sideways friction between the tyres of the car and the track.

(a) Show that $\tan\alpha=\dfrac14$

(5)

The sideways friction between the tyres of the car and the track has coefficient of friction $\mu$ , where $\mu<4$

The maximum speed at which the car can move round the circle without slipping sideways is $V$ .

(b) Find $V$ in terms of $\mu$ , $r$ and $g$ .

(7)

题目中文翻译

一辆汽车沿圆形跑道行驶。如图 4 所示，跑道的横向倾角为 $\alpha$ 。

将汽车与驾驶员视为一个质点。

汽车以恒定速率沿半径为 $r$ 的水平圆周行驶。

汽车以速率 $\dfrac12\sqrt{gr}$ 沿圆周行驶时，轮胎与跑道之间没有侧向摩擦力。

(a) 证明 $\tan\alpha=\dfrac14$ 。

(5)

轮胎与跑道之间的侧向摩擦系数为 $\mu$ ，其中 $\mu<4$ 。

汽车沿圆周行驶而不发生侧滑的最大速率为 $V$ 。

(b) 用 $\mu$ 、 $r$ 和 $g$ 表示 $V$ 。

(7)

解答

(a)

解法一

思路

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没有摩擦力时，支持力的竖直分量平衡重力，水平分量提供向心力。向心加速度水平指向圆心，不沿斜面法线。

答题过程

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Let the combined mass of the car and driver be $m$ , and let the normal reaction be $R$ . There is no sideways friction at speed $v_0=\sqrt{gr}/2$ .

The vertical acceleration is zero, so

R\cos\alpha=mg.\tag{1}

Resolving horizontally towards the centre of the circle,

R\sin\alpha=\frac{mv_0^2}{r} =\frac{m(gr/4)}{r}=\frac{mg}{4}.\tag{2}

Dividing (2) by (1),

\boxed{\tan\alpha=\frac14.}

解法二

思路

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沿横向坡面向下分解，可以消去支持力。不过汽车有水平向心加速度，它在坡面方向也有分量，不能把这一方向当成平衡。

答题过程

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At speed $v_0=\sqrt{gr}/2$ , the acceleration is horizontal towards the centre, with magnitude

a_c=\frac{v_0^2}{r}=\frac g4.

Resolve down the bank towards the inner edge of the track. The normal reaction has no component in this direction; the component of the weight is $mg\sin\alpha$ , and that of the acceleration is $a_c\cos\alpha$ . Thus

mg\sin\alpha=m\frac g4\cos\alpha.

Dividing by $mg\cos\alpha$ gives

\boxed{\tan\alpha=\frac14.}

(b)

解法一

思路

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最大速率时，汽车有向坡外、坡上侧滑的趋势，所以极限摩擦力沿坡向下。必须重新设支持力，不能沿用(a)中无摩擦时的数值。

答题过程

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At the maximum speed $V$ , the car is on the point of slipping up the bank. The limiting friction therefore acts down the bank, towards the inner edge. Let the normal reaction now be $S$ , and let the friction magnitude be $F$ . Then

F=\mu S.

Resolving vertically,

S\cos\alpha-F\sin\alpha=mg,

S(\cos\alpha-\mu\sin\alpha)=mg.\tag{3}

Resolving horizontally towards the circle’s centre,

S\sin\alpha+F\cos\alpha=\frac{mV^2}{r},

and hence

S(\sin\alpha+\mu\cos\alpha)=\frac{mV^2}{r}.\tag{4}

Dividing (4) by (3) eliminates $S$ :

\frac{V^2}{rg} =\frac{\sin\alpha+\mu\cos\alpha}{\cos\alpha-\mu\sin\alpha} =\frac{\tan\alpha+\mu}{1-\mu\tan\alpha}.

Using $\tan\alpha=1/4$ from part (a),

V^2=rg\frac{1/4+\mu}{1-\mu/4} =rg\frac{1+4\mu}{4-\mu}.

The condition $\mu<4$ makes the denominator positive and is consistent with a positive normal reaction in (3). Taking the positive square root,

\boxed{V=\sqrt{\frac{rg(1+4\mu)}{4-\mu}}.}

M3 2023 June Q4

题目

解答

(a)

解法一

思路

答题过程

解法二

思路

答题过程

(b)

解法一

思路

答题过程