Question

A mischievous young man riding a ferris wheel decides to release a beanbag as he passes the top point in the ride; at this point in the ride, the young man is traveling due West. The ferris wheel has a radius of 3.29   m and goes around once every 1.17  minutes \ . The bottom of the wheel is 1.11  m off the ground, so that the release point of the bag will be 7.69  m  off the ground.

A)What is the velocity of the bean bag just before it is released by the mischievous young man? Give magnitude and direction. Give the direction as an angle in degrees measured counterclockwise from due East, when looking North.

B)What is the acceleration of the bean bag just before it is released by the mischievous young man? Give magnitude and direction. Give the direction as an angle in degrees measured counterclockwise from due East, when looking North.

C)For how much time is the bean bag in the air?

D)The young man finds that the bag does not land directly under the point of release. How far does the bag travel horizontally as it is falling?

It would help if you explain too thanks

Answer 1

A)

T = Time period of rotation of the wheel = 1.17 min = 1.17 (60) sec = 70.2 sec

Angular speed of the wheel is given as

w = 2/T = 2 (3.14)/70.2 = 0.0895 rad/s

r = radius of the wheel = 3.29 m

Speed of the wheel at the top is given as

v = r w

v = (3.29) (0.0895)

v = 0.29 m/s

the bean bag has same velocity at the top as the wheel

v_b = magnitude of velocity of bean bag = 0.29 m/s

Direction : towards west, so 180 deg counterclockwise to east direction.

B)

acceleration is same as the centripetal acceleration and acts towards the center

a_b = acceleration of bean bag = v_b²/r = (0.29)²/3.29

a_b = 0.026 m/s²

Direction :270 degree

C)

Consider the motion of the bag in vertical direction :

v_oy = initial velocity of the bag in vertical direction = 0 m/s (Since it moved horizontally just after release)

a_y = acceleration = - 9.8 m/s²

t = time of travel

y = vertical displacement = - 7.69 m

Using the kinematics equation

y = v_oy t + (0.5) a_y t²

- 7.69 = (0) t + (0.5) (- 9.8) t²

t = 1.25 sec

D)

x = horizontal distance moved by bag

v_ox = initial horizontal velocity of the bag = v_b = 0.29 m/s

t = time of travel = 1.25 sec

Since there is no acceleration along the horizontal direction, we have

x = v_ox t

x = (0.29) (1.25)

x = 0.3625 m