Question

Suppose you and a friend, each of mass 55 kg , go to the park and get on a 4.5-m-diameter merry-go-round. You stand on the outside edge of the merry-go-round, while your friend pushes so that it rotates once every 7.0 seconds. Part APart complete What is the magnitude of the (apparent) outward force that you feel?

a. 12 N
b. 50 N
c. 100 N
d. 200 N

Answer 1

Answer:

c. 100 N

Explanation:

$m$ = mass of each = 55 kg

$T$ = Time period of rotation = 7.0 sec

Angular frequency is given as

$w = \frac{2\pi }{T} \\w = \frac{2(3.14) }{7}\\w = 0.897 rads^{-1}$

$d$ = Diameter of the merry-go-round = 4.5 m

Radius of the merry-go-round is given as

$r = \frac{d}{2} = \frac{4.5}{2} = 2.25 m$

Magnitude of outward force is same as the centripetal force and is given as

$F = m r w^{2} = (55) (2.25) (0.897)^{2} \\F = 100 N$

Answer 2

Answer:

99.57 N

Explanation:

mass, m = 55 kg

diameter = 4.5 m

radius, r = 4.5 / 2 = 2.25 m

time period, T = 7 second

Angular velocity, ω = 2π/T = 2π/7 = 0.897 rad/s

Apparent froce is the centripetal force.

F = m x r x ω²

F = 55 x 2.25 x 0.897 x 0.897

F = 99.57 N

Thu, the apparent force is 99.57 N .