Question

what is the radius of the bobsled turn in m, assuming it is ideally banked and there is no friction between the ice and the bobsled?

Answer 1

Answer on Question #65593, Physics / Mechanics | Relativity

What is the radius of the bobsled turn in m, assuming it is ideally banked and there is no friction between the ice and the bobsled?

Solution:

Using a free body diagram with corresponding force equations.

$\mathrm{mg} = \mathrm{F}_{\mathrm{N}}$ cos $\theta$

$F_{N} = mg / \cos \theta$

ma = F_N sin θ

ma = mg x sin θ / cos θ

ma = mg x tg θ

Centripetal acceleration can be found

$a = v^2 / R$

$\mathrm{mv}^2 / \mathrm{R} = \mathrm{mg} \times \mathrm{tg} \theta$

$R = m v^{2} / mg \times t g$

$R = v^{2} / g \times t g \theta$

Answer: $R = v^2 / g \times tg \theta$

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