Question

the length of the ropes on a playground swing is 2.0 m. a) what is the maximum speed attainable on the swing if the maximum value of theta is 60 degrees?. b) if a 50 kg person is playing the swing, what is the maximum kinetic energy he/she can reach?

Answer 1

Question #77543, Physics / Classical Mechanics

the length of the ropes on a playground swing is $2.0 \, \text{m}$ . a) what is the maximum speed attainable on the swing if the maximum value of theta is 60 degrees?. b) if a $50 \, \text{kg}$ person is playing the swing, what is the maximum kinetic energy he/she can reach?

Solution

a) The swing has the maximum speed at the equilibrium point.

Using the law of conservation of the energy,

$$E _ {k} = E _ {G P},$$

Where $E_{k}$ is the swing's kinetic energy at the equilibrium point,

$E_{GP}$ is the swing's gravitational potential energy at the upper point.

Plugging the values of the energies,

$$\frac {m v _ {\mathrm {m a x}} ^ {2}}{2} = m g h _ {\mathrm {m a x}}$$

Hence $v_{\mathrm{max}} = \sqrt{2gh_{\mathrm{max}}}$

The height of the swing at the upper position above the equilibrium position is

$$h _ {\max } = l (1 - \cos \theta) = 2. 0 \times (1 - \cos 6 0 {}^ {\circ}) = 1. 0$$ $$v _ {\max } = \sqrt {2 \times 9 . 8 1 \times 1 . 0} = 4. 4 \mathrm {m / s}$$ $$b) E _ {k \_ \text {m a x}} = \frac {m v _ {\max} ^ {2}}{2} = \frac {5 0 \times 4 . 4 ^ {2}}{2} = 4 8 4 \mathrm {J}$$

Answer: a) $4.4 \, \text{m/s}$ ; b) $484 \, \text{J}$ .

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