A 1.14kg ball is connected by means of two massless strings to a vertical, rotating rod. The strings are tied to the rod and form two sides

The angle the strings make with the horizontal must be 30°, so the radius of rotation must be

r = 1.59*cos(30°) metres

 

Resolve forces in vertical and horizontal directions.  These must each be balanced.

 

Vertical:  T1sin(30°) = T2sin(30°) + mg  ...(1)  

where T1 is tension in upper string (58.2N); T2 that in lower string; m is mass of ball (1.14kg) and g is gravitational acceleration (9.8m/s²)

 

Horizontal:  T1cos(30°)+T2cos(30°) = mv²/r ...(2) 

where v is speed of ball.

 

From (1)  T2 = T1 - mg/sin(30°)   ...(3)

                   =  58.2 - 1.14*9.8/sin(30°)

 

$${\mathtt{T2}} = {\mathtt{58.2}}{\mathtt{\,-\,}}{\frac{{\mathtt{1.14}}{\mathtt{\,\times\,}}{\mathtt{9.8}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{30}}^\circ\right)}}} \Rightarrow {\mathtt{T2}} = {\mathtt{35.856}}$$

T2 ≈ 35.9 Newtons

 

Use (3) in (2)

T1cos(30°) + T1cos(30°) - mgcos(30°)/sin(30°) = mv²/r 

v² = (r/m)*(2T1cos(30°) - mgcos(30°)/sin(30°))

    = (1.59*cos(30°)/1.14)*(2*58.2*cos(30°) - 1.14*9.8*cos(30°)/sin(30°))

 

$${\mathtt{v}} = {\sqrt{\left({\frac{{\mathtt{1.59}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{30}}^\circ\right)}}{{\mathtt{1.14}}}}\right){\mathtt{\,\times\,}}\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{58.2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{30}}^\circ\right)}{\mathtt{\,-\,}}{\frac{{\mathtt{1.14}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{30}}^\circ\right)}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{30}}^\circ\right)}}}\right)}} \Rightarrow {\mathtt{v}} = {\mathtt{9.919\: \!048\: \!659\: \!810\: \!567}}$$

v ≈ 9.92 metres/sec

 

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