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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.

physics
Guest Mar 1, 2017

Best Answer 

 #2
avatar+25975 
+15

Although your final value of 4.21m/s is correct GingerAle, you might want to check the values involved under the square root sign!

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Alan  Mar 1, 2017
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3+0 Answers

 #1
avatar+644 
+15

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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Theory & Formulas: Complements of Christiaan Huygens, Galileo Galilei, and Sir Isaac Newton, et al. 

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GingerAle  Mar 1, 2017
 #2
avatar+25975 
+15
Best Answer

Although your final value of 4.21m/s is correct GingerAle, you might want to check the values involved under the square root sign!

.

Alan  Mar 1, 2017
 #3
avatar+644 
+10

Thank you Sir Alan. Lancelot would throw banana peels at me for that.  smiley

 

\(v = \sqrt{\dfrac{362.07}{53}*2.6} = 4.21 m/s \hspace{1em} \Leftarrow \text {Solution -- }\textcolor {red} {(Corrected \;work \;product \;for \;solution.)}\\ \)

GingerAle  Mar 1, 2017
edited by GingerAle  Mar 1, 2017
edited by GingerAle  Mar 1, 2017

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