A 53 kg child sits in a swing supported by two chains, each 2.6 m long. If the tension in each chain at the lowest point is 441 N, find the child’s speed at the lowest point. (Neglect the mass of the seat.) Answer in units of m/s.
To solve this, use these two fundamental equations for pendulum motion.
\(\text {F(net)}= 2T - m*g = m*a \hspace{1em} |\text {Force net,} \underbrace {\text{T is tension,}}_{\text {2 for the two chains}} \text {m is mass,}\\ \hspace{13em}\text { g is acceleration from gravity }(9.81 m/s^2), \\ \text { }\hspace{13em}\text {a is acceleration in the normal direction. } \\ \)
\(\text { And }\\ a = \dfrac {v^2}{ r} \hspace{1em} |\text { a is acceleration in the normal direction, v is the velocity,} \\ \hspace{4.7em} \text {r is the radius of the swing's arc. } \\\)
\(2T - m*g = m* (v^2 / r) \hspace{1em} | \text {solve for v } \\ \text { }\\ 2(441)- 53*9.81 = 53*\dfrac{v^2}{2.6}\\ \text { }\\ v = \sqrt{\dfrac{362.07}{53*2.6}}=4.21 m/s \hspace{1em} \Leftarrow \text {Solution}\\ \)
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