Question

An 80.0 -kg pilot in an aircraft moving at a constant speed of \(500 . \mathrm{m} / \mathrm{s}\) pulls out of a vertical dive along an arc of a circle of radius \(4000 . \mathrm{m} .\) a) Find the centripetal acceleration and the centripetal force acting on the pilot. b) What is the pilot's apparent weight at the bottom of the dive?

Short answer

Answer: The pilot's centripetal acceleration is 62.5 m/s², the centripetal force is 5000 N, and the apparent weight at the bottom of the dive is approximately 5784.8 N.

Step-by-step solution

Calculate the centripetal acceleration

We can find the centripetal acceleration using the formula \(a_c = \frac{v^2}{r}\), where \(v = 500\,\text{m/s}\) and \(r = 4000\,\text{m}\). $$ a_c = \frac{(500\,\text{m/s})^2}{4000\,\text{m}} = \frac{250000\,\text{m²/s²}}{4000\,\text{m}} = 62.5\,\text{m/s²}. $$ So, the centripetal acceleration is \(62.5\,\text{m/s²}\).

Calculate the centripetal force

Now, we can find the centripetal force using the formula \(F_c = m \times a_c\), where \(m = 80.0\,\text{kg}\) and \(a_c = 62.5\,\text{m/s²}\). $$ F_c = (80.0\,\text{kg}) × (62.5\,\text{m/s²}) = 5000\,\text{N}. $$ Therefore, the centripetal force acting on the pilot is \(5000\,\text{N}\).

Calculate the pilot's apparent weight

Finally, we find the pilot's apparent weight at the bottom of the dive using the formula \(W_{apparent} = m(g + a_c)\), where \(m = 80.0\,\text{kg}\), \(g = 9.81\,\text{m/s²}\), and \(a_c = 62.5\,\text{m/s²}\). $$ W_{apparent} = (80.0\,\text{kg})[ (9.81\,\text{m/s²}) + (62.5\,\text{m/s²})] = (80.0\,\text{kg})(72.31\,\text{m/s²}) = 5784.8\,\text{N}. $$ So, the pilot's apparent weight at the bottom of the dive is approximately \(5784.8\,\text{N}\).