Final answer:
The radius of the bobsled turn banked at 75.0° and taken at 30.0 m/s, assuming it is ideally banked, is approximately 24.6 meters. The centripetal acceleration of the bobsled in this turn is approximately 36.6 m/s². This acceleration is moderate but not excessively large.
Step-by-step explanation:
(a) What is the radius of a bobsled turn banked at 75.0° and taken at 30.0 m/s, assuming it is ideally banked?
Assuming the turn is ideally banked, we can use the formula:
tan(θ) = (v^2) / (g * r)
where θ is the angle of banking, v is the velocity, g is the acceleration due to gravity, and r is the radius of the turn.
Plugging in the given values, we have:
tan(75°) = (30.0 m/s)^2 / (9.8 m/s² * r)
Solving for r, we get:
r = (30.0 m/s)^2 / (9.8 m/s² * tan(75°)) ≈ 24.6 meters
Therefore, the radius of the bobsled turn is approximately 24.6 meters.
(b) Calculate the centripetal acceleration.
The centripetal acceleration can be calculated using the formula:
a_c = v^2 / r
Substituting in the given values, we have:
a_c = (30.0 m/s)^2 / 24.6 m ≈ 36.6 m/s²
Therefore, the centripetal acceleration is approximately 36.6 m/s².
(c) Does this acceleration seem large to you?
Based on the given values, the centripetal acceleration is 36.6 m/s², which is moderate but not excessively large.