Answer:
To solve this problem, we will use the kinematic equations of rotational motion:
1. ωf = ωi + αt (final angular velocity = initial angular velocity + angular acceleration x time)
2. θ = ωit + 1/2 αt^2 (angle rotated = initial angular velocity x time + 1/2 x angular acceleration x time^2)
3. vf = ri (final tangential velocity = radius x final angular velocity)
4. ar = rα (centripetal acceleration = radius x angular acceleration)
where ω is angular velocity, α is angular acceleration, t is time, θ is angle, v is tangential velocity, r is radius, and a is acceleration.
a) Using equation 1, we can find the final angular velocity:
ωf = ωi + αt
ωf = 0.250 rev/s + 0.900 rev/s^2 x 0.200 s
ωf = 0.430 rev/s
b) Using equation 2, we can find the angle rotated:
θ = ωit + 1/2 αt^2
θ = 0.250 rev/s x 0.200 s + 1/2 x 0.900 rev/s^2 x (0.200 s)^2
θ = 0.0350 rev
c) Using equation 3, we can find the tangential velocity at t = 0.200 s:
vf = ri
vf = (0.750 m/2) x 0.430 rev/s x 2π rad/rev
vf = 1.62 m/s
d) Using equation 4, we can find the centripetal acceleration at t = 0.200 s:
ar = rα
ar = 0.750 m/2 x 0.900 rev/s^2 x 2π rad/rev
ar = 2.12 m/s^2
The magnitude of the resultant acceleration of a point on the rim at t = 0.200 s is the vector sum of the tangential and centripetal accelerations:
a = √(at^2 + ar^2)
a = √(vf^2/r^2 + ar^2)
a = √((1.62 m/s)^2/(0.750 m/2)^2 + (2.12 m/s^2)^2)
a = 4.58 m/s^2
Therefore, the angular velocity of the turntable after 0.200 s is 0.430 rev/s, it has spun through an angle of 0.0350 rev, the tangential speed of a point on the rim of the turntable at t = 0.200 is 1.62 m/s, and the magnitude of the resultant acceleration of a point on the rim at t = 0.200 s is 4.58 m/s^2.
Step-by-step explanation: