Question

2. An object of mass 10 kg is whirled round a horizontal circle of radius 4 m by a revolving string inclined to the vertical. If the uniform speed of the object is 5 m/s, calculate:

(i) the tension in the string

(ii) the angle of inclination of the string to the vertical.

Answer 1

$\text{forces acting on an object:}$ $\text {}$

$F_g-\text{gravity}$

$F_g=mg=9.8×10=98N$

$F_c-\text{force centripetal acceleration}$

$F_c=ma;a=\frac{v^2}{R}=\frac{25}{4}=6.25m/s^2$

$F_c=10×6.25=62.5N$

$T-\text{string tension}$

$\text{X-axis projection}$

$T_x+F_g=0$

$|Tx|=F_g=98N$

$\text{Y-axis projection}$

$T_y= F_c$

$T_y=62.5N$

$T=\sqrt{T^2_x+T^2_y}=\sqrt{62.5^2+98^2}\approx116.23N$

$T=\sqrt{T^2_x+T^2_y}=\sqrt{62.5^2+98^2}\approx116.23N$$T=\sqrt{T^2_x+T^2_y}=\sqrt{62.5^2+98^2}\approx116.23N$

$\alpha-\text{angle of inclination of the string to the vertical}$

$\tan{\alpha}=\frac{F_c}{T_x}=\frac{62.5}{98}\approx0.64$

$\alpha=32.6\degree$

Answer:$T=116.23N;\alpha=32.6\degree$