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3 years ago

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Mary and her younger brother Alex decide to ride the 17 -foot-diameter carousel at the State Fair. Mary sits on one of the horse

s in the outer section at a distance of 8 feet from the center. Alex decides to play it safe and chooses to sit in the inner section at a distance of 5 feet from the center.
(a) What is Mary's angular speed ωM compared to that of Alex's angular speed ωA ? Give your answer as a multiple of ωA.ωM=ωA
(b) What is Mary's tangential speed vM compared to that of Alex's tangential speed vA? Give your answer as a multiple of vA.vM=vA

1 answer:

Elodia [21]3 years ago

7 0

Answer:

(a) ωM=ωA

(b) vA = (0.625)vM

Explanation:

The tangential velocity of the body is calculated by the following formula:

v = ω*R Formula (1)

where:

v is the tangential speed of the body in feet over second (ft/s)

ω is the angular speed in radians over second (rad/s)

R is the radius of the the body in the circular path in meters (m)

(a) What is Mary's angular speed ωM compared to that of Alex's angular speed ωA

The angular speed of Mary and Alex is determined by the angular speed of the carousel ( ωC),therefore it is the same for both:

ωM=ωC , ωA=ωC

ωM=ωA

(b) What is Mary's tangential speed vM compared to that of Alex's tangential speed vA?

Data:

RM = 8 ft : Mary radio

RA = 5 ft : Alex radio

We apply the formula (1)

v = ω*R

vM = ωC *8 : Mary's tangential speed

vA = ωC *5 : Alex's tangential speed

We establish the relationship between vA and vM like this:

vA/vM = (ωC*5) / (ωC*8)

We cancel ωC

vA = (5/8)vM

vA = (0.625)vM

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$u_0\cos\theta=10 \cos 53 ^{\circ}$

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Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).

$\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2$

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So, the total time in the air is 1.63 seconds.

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