Asked by Jade
A carnival seat and a passenger has a combined weight of 193. As the seat is spun on a 18m rope tethered to a pole at an angle of 59.2 degrees, at what speed is the seat traveling?
Answers
Answered by
Damon
I do not know where your angle is
If from straight down then:
T = tension
m g = weight
A = angle from vertical
vertical equation
T cos A = m g = m(9.81)
so
T = m(9.81)/cos A
T sin A = m Ac = m v^2/R
where R = 18 sin A
so
m (9.81) sin A = m v^2 /(18 sin A)
v^2 = 9.81 * 18 * sin^2 A
note, mass cancels, does not matter, otherwise carnival ride would not work :)
If from straight down then:
T = tension
m g = weight
A = angle from vertical
vertical equation
T cos A = m g = m(9.81)
so
T = m(9.81)/cos A
T sin A = m Ac = m v^2/R
where R = 18 sin A
so
m (9.81) sin A = m v^2 /(18 sin A)
v^2 = 9.81 * 18 * sin^2 A
note, mass cancels, does not matter, otherwise carnival ride would not work :)
Answered by
Damon
T = m(9.81)/cos A
T sin A = m Ac = m v^2/R
where R = 18 sin A
so
m (9.81) sin A/cosA = m v^2 /(18 sin A)
v^2 = 9.81 * 18 * sin^2 A/cos A
check my arithmetic !!!
T sin A = m Ac = m v^2/R
where R = 18 sin A
so
m (9.81) sin A/cosA = m v^2 /(18 sin A)
v^2 = 9.81 * 18 * sin^2 A/cos A
check my arithmetic !!!
Answered by
Jade
The rope forms a triangle with the pole and the angle is between the rope and the point at which it connects to the pole.
Answered by
Damon
ok, that is may angle A so plug sin 59.2 and cos 59.2 in
but check my algebra first
but check my algebra first
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