Question

A neighbor's child wants to go to a neighborhood carnival to experience the wild rides. The neighbor is worried about safety as one of the rides looks dangerous. and asks for your advice. the ride in question has a 10-lb chair which hangs freely from a 30-ft long chain attached to a pivot on the top of a tall tower. when a child enters the ride, the chain is hanging straight down. the child is then attached to the chair with a seat belt and shoulder harness.

When the ride starts up the chain rotates about the tower and reaches a maximum speed and remains at thta speed. It rotates the tower once every 3 seconds.

When you ask the operator he says that the ride is safe and sits in a stationary chair which creaks but holds the 200 pounds.

Has the operator shown that the ride is safe for a 50 lb child?

Answer 1

Answer:

No, the ride is not safe

Explanation:

From the diagram attached, it is seen that

$\sum f_{y} = 0$

$\tau cos \theta - W = 0$................(1)

$\sum f_{x} = 0$

$-ma + \tau sin \theta = 0$.......(2)

$a = w^{2} r$

From the diagram, $r = Lsin \theta$

$a = w^{2} L sin \theta$

$w = \frac{2\pi }{T}$

$a = \frac{4\pi ^{2}L sin \theta }{T^{2} }$............(3)

Put (3) into (2)

$\tau sin \theta = \frac{4\pi ^{2}m L sin \theta }{T^{2} }$

$\tau = \frac{4\pi ^{2}m L }{T^{2} }$

w = weight of chair + weight of child = 50 + 10 = 60 lb

g = 32 ft/s²

m = w/g = 60/32 = 1.875

L = 30 ft

T = 3 secs

$\tau = \frac{4\pi ^{2}*1.875*30 }{0.3^{2} }$

$\tau = 246.74 lbs$

since 246.74 lbs > 200 lbs, it is not safe because the stationary chair will creak

Answer 2

Answer:

According to the result obtained, the trip is not safe for a 50 lb child.

Explanation:

The sum of forces in y must be equal to zero:

∑Fy = 0

0 = W + Tcosθ

The sum of forces in x:

∑Fx = m*a = -m*W²*r = -T1sinθ

r = Lsinθ

m*W²*Lsinθ = T1sinθ

$m\frac{4\pi ^{2}L }{T^{2} } =T\\\frac{60}{32} \frac{4\pi ^{2} 30}{3^{2} } =T\\T=250lb$

Like 250 lb > 200 lb, the answer is no