Solving the Swing Tension: A 25kg Child's Story

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To determine the tension in the swing's chain for a 25 kg child swinging at 3.3 m/s, both gravitational force and centripetal force must be considered. At the bottom of the swing, the tension in the chain must counteract the gravitational force while providing the necessary centripetal force for circular motion. Newton's second law can be applied to relate these forces to the child's acceleration. The calculation involves summing the forces acting on the child and solving for tension. Understanding these principles is crucial for solving the problem accurately.
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Homework Statement


A 25 kg child is swinging on a 3.5 m swing. At the bottom of the swing
the speed of the child is moving at 3.3 m/s. What is the tension on the
swing’s chain to hold the child?


Homework Equations





The Attempt at a Solution


Can you help me i am confused about this.
 
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tnutty said:
A 25 kg child is swinging on a 3.5 m swing. At the bottom of the swing
the speed of the child is moving at 3.3 m/s. What is the tension on the
swing’s chain to hold the child?
…
Can you help me i am confused about this.

Hi tnutty! :smile:

The child is moving in a circle.

The forces on the child are the tension in the chain and the force of gravity.

Use good ol' Newton's second law to relate these forces to the child's acceleration. :smile:
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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