Question

A particle is tied to a light inextensible string of length L and whirled in a vertical circle with other end of string as centre. At a certain instant when the string makes angle q = 53ۜ° with lower vertical its speed is found to be v = $\sqrt{5gL}$, then its acceleration at that instant of time is –

• A. 5 g
• B. 0.8 g
• C. 5.06 g
• D. g

Answer 1

Answer: (C)

5.06 g

Answer 2

a_t = g sin q = 0.8 g

a_r = $\frac{v^{2}}{L}$= 5g

\ a = $\sqrt{a_{r}^{2} + a_{t}^{2}}$= 5.06 g