Pendulum bob with given tension

Assumption (figure not shown): the cord makes $\theta=40^\circ$ with the vertical, and the tension is $T=2.5W$. The bob moves on a circle of radius $L=2$ m, so use normal and tangential components.

          O
          |\
          | \  L = 2 m
          |40\
          |   \
                O  bob (W)

Normal direction (towards O)

$$T-W\cos\theta=\frac{W}{g}\frac{v^{2}}{L}$$ $$v^{2}=gL\,(2.5-\cos40^\circ)=9.81(2)(2.5-0.7660)=34.02\ \text{m}^2/\text{s}^2$$ $$v=5.833\ \text{m/s}$$

Accelerations

$$a_n=\frac{v^{2}}{L}=\frac{34.02}{2}=17.01\ \text{m/s}^2$$

Tangential direction (only the weight component acts):

$$W\sin\theta=\frac{W}{g}a_t\ \Rightarrow\ a_t=g\sin40^\circ=6.306\ \text{m/s}^2$$ $$a=\sqrt{a_n^{2}+a_t^{2}}=\sqrt{17.01^{2}+6.306^{2}}=18.14\ \text{m/s}^2$$

The acceleration makes $\tan^{-1}(6.306/17.01)=20.3^\circ$ with the cord.

Answer: $v=5.83$ m/s; $a=18.14$ m/s$^2$ ($a_n=17.01$, $a_t=6.31$ m/s$^2$).