Space Time & Motion A.2 Forces & Momentum (id: 711b63d55)

本题目来源于试卷: Space Time & Motion A.2 Forces & Momentum，类别为 IB物理学

[单选题]
A ball of mass 0.4kg is attached ttl-35jwnp;d -1kqaj fo a string of length 0.2m and(in ,ru(f n e5f;ul0fl8 jhg4r is moving in a vertical circular path wj fu( fg50 rie4;8l nrl,funh(ith tangential speed of $\sqrt3ms^{-1}$ at the instant shown in the diagram.

Given that$\sin (60^{\circ})$=$\frac{\sqrt3}{2}$ and $\cos (60^{\circ})$=$\frac{1}{2}$,what is the tension in the string at the shown position?

A. (6+$2\sqrt3$)N
B. 8N
C. 4N
D. (6-$2\sqrt3$)N


参考答案:  B


本题详细解析:

$\underline{\text{Explanation}}$:
The ball is rotating in a vertical circular path, then the forces acting on it is the tension of the string and the weight.

The centripetal force acting on the ball is the vector sum of the tension of the string and the component of the weight that is the same direction as the string

$$ $F_c=T-mgcos(60\mathrm{°})$

The centripetal force

$$ $F_c={\displaystyle \frac{m{v}^2}{r}}$

Then

$$ $T={\displaystyle \frac{m{v}^2}{r}}+mgcos(60\mathrm{°})$

$$ $T={\displaystyle \frac{(0.4){(\sqrt{3})}^2}{0.2}}+\left(0.4\right)\left(10\right)\left({\displaystyle \frac{1}{2}}\right)$

$$ = $8N$

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