Two horizontal curves on a bobsled run are banked at the same angle, but one has twice the radius of the other. The safe speed (no friction needed to stay on the run) for the smaller radius curve is v. What is the safe speed on the larger radius curve?
A) v/2
B) v /√ 2
C) 2v
D) v√2

I shall answer it here so this can help other students out
First, FN = mg/cos
On a banked curve, FNx is the only acceleration force
FNx = mv^2/r
FNx = FNsin
mg/cos * sin = mv^2/r
cancel out the mass (m) and sin1/cos is tan
gtan = v^2/r
rgtan = v^2
√(rgtan) = v (this is the velocity for the smaller radius.)
call larger radius velocity X

for larger radius, it becomes 2r, so √(2rgtan) = X

understand you can do √xy = √x √y

√(rgtan) * √2 = X (√(rgtan) = v)
v * √2 = X (where X is velocity with larger radius)

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prince6193 prince6193 · Answer 1 · 2021-12-09T04:29:05+01:00

Answer:

Answer:

safe speed for the larger radius track u= √2 v

Explanation:

The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.

Also given that r_1= smaller radius

r_2= larger radius curve

r_2= 2r_1..............i

let u be the speed of larger radius curve

now, \sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}∑F=

r

1

mv

2

=

r

2

mu

2

................ii

form i and ii we can write

v^2= \frac{1}{2} u^2v

2

=

2

1

u

2

⇒u= √2 v

therefore, safe speed for the larger radius track u= √2 v