what would happen when an inflated balloon is place in an ice bath and the atmosphere pressure has not changed?
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Refer to the diagram shown below.
v = 35 km/h, the speed limit of the train
r = 150 m, the radius of the curve
ω = angular velocity
m = the mass of the strap
θ° = 15, the angle the strap makes with the vertical
T = tension in the strap
Note that
v = 35 km/h = 35*0.2778 m/s = 9.7223 m/s
The tangential velocity is v = rω, therefore the angular vcelocity is
ω = (9.7223 m/s)/(150 m) = 0.0648 rad/s
The centripetal force tending to make the train derail causes the strap to make an angle of 15 with the vertical.
Let θ = the maximum allowable angle at 35 km/h.
For horizontal equilibrium,
Tsin(θ) = mrω²
For vertical equilibrium,
Tcos(θ) = mg
Therefore
tan(θ) = (rω²)/g
= [(150 m)*(0.0648 rad/s)]/(9.8 m/s²)
= 0.0643
θ = tan⁻¹ 0.0643 = 3.7°
Because 15 > 3.7, we conclude that the train exceeded the 35 km/h speed limit when rounding the curve.
Answer: The train exceeded the 35 km/h speed limit.
v = 35 km/h, the speed limit of the train
r = 150 m, the radius of the curve
ω = angular velocity
m = the mass of the strap
θ° = 15, the angle the strap makes with the vertical
T = tension in the strap
Note that
v = 35 km/h = 35*0.2778 m/s = 9.7223 m/s
The tangential velocity is v = rω, therefore the angular vcelocity is
ω = (9.7223 m/s)/(150 m) = 0.0648 rad/s
The centripetal force tending to make the train derail causes the strap to make an angle of 15 with the vertical.
Let θ = the maximum allowable angle at 35 km/h.
For horizontal equilibrium,
Tsin(θ) = mrω²
For vertical equilibrium,
Tcos(θ) = mg
Therefore
tan(θ) = (rω²)/g
= [(150 m)*(0.0648 rad/s)]/(9.8 m/s²)
= 0.0643
θ = tan⁻¹ 0.0643 = 3.7°
Because 15 > 3.7, we conclude that the train exceeded the 35 km/h speed limit when rounding the curve.
Answer: The train exceeded the 35 km/h speed limit.