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Physics, 09.12.2021 06:40 xwilliams83

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 sin*1/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)

Answers

ansver
Answer from: FombafTejanjr7503

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_1i

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

ansver
Answer from: Quest

i think its c if i'm wrong

ansver
Answer from: Quest

higher potential

explanation:

because it is in a high speed so it must move into a of high potential

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