A coin rests 15.0 cm from the center of a turntable. The coefficient of static friction between the coin and turntable surface is 0.390. The turntable starts from rest at t = 0 and rotates with a constant angular acceleration of 0.630 rad/s2. (a) After 3.00 s, what is the angular velocity of the turntable?
You are given angular velocity initial, angular acceleration, and time.
use wf=wi+at
wf=0+(0.630rad/s)*(3s)
To determine the angular velocity of the turntable after 3.00 seconds, we can use the formula:
ω = ω₀ + αt
where:
ω is the angular velocity after time t,
ω₀ is the initial angular velocity,
α is the angular acceleration, and
t is the time.
We are given:
ω₀ = 0 (since the turntable starts from rest),
α = 0.630 rad/s², and
t = 3.00 s.
Plugging in these values into the formula, we get:
ω = 0 + (0.630 rad/s²)(3.00 s)
= 1.89 rad/s
Therefore, after 3.00 seconds, the angular velocity of the turntable is 1.89 rad/s.
To find the angular velocity of the turntable after 3.00 s, we can use the kinematic equation for rotational motion:
ω = ω₀ + αt
where:
ω is the final angular velocity (unknown),
ω₀ is the initial angular velocity (which is 0, since the turntable starts from rest),
α is the angular acceleration (given as 0.630 rad/s²),
t is the time interval (3.00 s).
Plugging in the values, we get:
ω = 0 + (0.630 rad/s²)(3.00 s)
ω = 1.89 rad/s
Therefore, after 3.00 s, the angular velocity of the turntable is 1.89 rad/s.