A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center of the disk. The platform has a radius of 2.97 m and a rotational inertia of 358 kg·m2 about the axis of rotation. A 69.5 kg student walks slowly from the rim of the platform toward the center. If the angular speed of the system is 1.96 rad/s when the student starts at the rim, what is the angular speed when she is 1.06 m from the center?

4.36 rad/s

Explanation:

Radius of platform r = 2.97 m

rotational inertia I = 358 kg·m^2

Initial angular speed w = 1.96 rad/s

Mass of student m = 69.5 kg

Rotational inertia of student at the rim = mr^2 = 69.5 x 2.97^2 = 613.05 kg.m^2

Therefore initial rotational momentum of system = w( Ip + Is)

= 1.96 x (358 + 613.05)

= 1903.258 kg.rad.m^2/s

When she walks to a radius of 1.06 m

I = mr^2 = 69.5 x 1.06^2 = 78.09 kg·m^2

Rotational momentuem of system = w(358 + 78.09) = 436.09w

Due to conservation of momentum, we equate both momenta

436.09w = 1903.258

w = 4.36 rad/s

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s