Consider a father pushing a child on a playground merry-go-round. the system has a moment of inertia of 84.4 kg · m2. the father exerts a force on the merry-go-round perpendicular to its radius to achieve an angular acceleration of 4.44 rad/s2. (a) how long (in s) does it take the father to give the merry-go-round an angular velocity of 1.43 rad/s? (assume the merry-go-round is initially at rest.)

At time t1 = 0 since the body is at rest, the body has an angular velocity, v1, of 0. At time t = X, the body has an angular velocity of 1.43rad/s2. Since Angular acceleration is just the difference in angular speed by time. We have 4.44 = v2 -v1/t2 -t1 where V and t are angular velocity and time. So we have 4.44 = 1.43 -0/X - 0. Hence X = 1.43/4.44 = 0.33s.

It takes 0.322 s to give the merry-go-round an angular velocity of 1.43 rad/s

Further explanation

Centripetal Acceleration can be formulated as follows:

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Centripetal Force can be formulated as follows:

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

Given:

moment of inertia = I = 84.4 kg.m²

initial angular velocity = ωo = 0 rad/s

angular acceleration = α = 4.44 rad/s²

final angular velocity = ω = 1.43 rad/s

Asked:

time taken = t = ?

Solution:

Learn moreImpacts of Gravity : linkEffect of Earth’s Gravity on Objects : linkThe Acceleration Due To Gravity : link

Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

Sorry can’t u just tryna get points

50 meters per 5 seconds