I was doing a few very old circular motion questions and came across a question where I believe the answer is incorrect. The book's answer is $ 18.88 \; N$, whereas I get $19.68\;N$. Can someone verify my solution is correct (or incorrect!).
One end of a light inelastic string of length $2\;m$ is attached to a fix point $2.5 \;m$ vertically above the centre of a fixed sphere of radius $1.5\;m$. A small ball, of mass $2\;kg$, at the other end of the string is made to rotate in a horizontal circle on the smooth surface of the sphere with a speed of $2\;m/s$. Show the string must be taut and calculate the tension in it.
Below is my diagram showing the current situation.
The string must be taught since it makes a right angle triangle $1.5 : 2: 2.5$ or $3 : 4 : 5$. Using this I can find $\sin \theta = \frac{3}{5}$ and $\cos \theta = \frac{4}{5}$. I can also find the value of $r$, which I will need when calculating the the circular motion force. To find $r$, using the smaller right angled triangle.
$$ \sin \theta = \frac{r}{2} $$ $$ \frac{3}{5} = \frac{r}{2} $$ $$ r = \frac{6}{5} $$
$\vec{j} $ components: $$ T \cos \theta + N \sin\theta = 2g $$ $$ T \cdot \frac{4}{5} + N \cdot \frac{3}{5} = 2g $$ $$ 4T + 3N = 10g \label{eq:EQU1} \tag{1} $$
$\vec{i} $ components $F = ma$: $$ T \sin \theta - N \cos \theta = \frac{mv^2}{r} $$ $$ T \cdot \frac{3}{5} - N \cdot \frac{4}{5} = \frac{2 \cdot 2^2}{\frac{6}{5}} $$ $$ T \cdot \frac{3}{5} - N \cdot \frac{4}{5} = \frac{20}{3} $$ $$ 3T - 4N = \frac{100}{3} $$ $$ 9T - 12N = 100 \label{eq:EQU2} \tag{2} $$
Eliminate $N$ by $ (1) \times 4 + (2) $
$$ 25T = 40g + 100 $$ $$ T = 19.68 $$
The book's answer is $ 18.88\;N$. If someone can point out my error it would be greatly appreciated.