I already asked this so answer both for more points! A carnival ride initially rotates counterclockwise at 2.0 rad/s but comes to rest with a constant acceleration over an angular displacement of 6.0 rad. What is the angular acceleration?​

a) v = 13.8 m / s , b) a = 95.49 m / s² , c) a force that goes to the center of the carnival ride  and d)   μ = 0.10

Explanation:

For this exercise we will use the angular kinematics relationships and the equation that relate this to the linear kinematics

a) reduce the magnitudes to the SI system

    w = 1.1 rev / s (2pi rad / 1rev) = 6.91 rad / s

The equation that relates linear and angular velocity is

    v = w r

    v = 6.91  2

    v = 13.8 m / s

b) centripetal acceleration is given by

    a = v² / r = w² r

    a = 6.91² 2

    a = 95.49 m / s²

c) this acceleration is produced by a force that goes to the center of the carnival ride

d) Here we use Newton's second law

    fr -W = 0

    fr = W

    μ N = mg

Radial shaft

     N = m a

     N = m w² r

     μ m w²  r = m g

     μ = g / w² r

     μ = 9.8 / 6.91² 2

     μ = 0.10

Explanation:

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s