A wheel, starting from rest, has a constant angular acceleration of 1.3 rad/s2. In a 2.5-s interval, it turns through an angle of 98 rad. How long has the wheel been in motion at the start of this 2.5-s interval?
W = Wo + a t
T = To + Wo t + (1/2) a t^2
same old equations but now T is theta, the angle, W is angular velocity, and a is angular acceleration
so
Here To is zero, initial angle, Wo = 0, initial speed
T = .65 t^2 at start of interval
98 + T = .65 (t+2.5)^2
subtract the first from the second
98 = (.65)(5 t+6.25)
98 = 3.25 t + 4.06
t = 28.9 seconds
Check my arithmetic!!!
W = Wo + a T
To find out how long the wheel has been in motion at the start of the 2.5-second interval, we can use the basic kinematic equation for rotational motion:
θ = ω_0*t + 0.5*α*t^2
where:
θ is the angle turned by the wheel (98 rad),
ω_0 is the initial angular velocity (which is 0 since the wheel starts from rest),
α is the angular acceleration (1.3 rad/s^2),
t is the time interval (2.5 s).
Rearranging the equation, we get:
98 = 0 + 0.5*(1.3)*(2.5)^2
Simplifying, we have:
98 = 0.5*1.3*6.25
98 = 0.8125
Therefore, the time the wheel has been in motion at the start of the 2.5-second interval is 0.8125 seconds.
To determine how long the wheel has been in motion at the start of the 2.5-second interval, we need to find the initial time at which the wheel started moving.
We can use the following kinematic equation for rotational motion:
θ = ω₀t + (1/2)αt²
Where:
θ is the angle turned by the wheel (98 rad),
ω₀ is the initial angular velocity (0 rad/s since it starts from rest),
α is the angular acceleration (1.3 rad/s²),
t is the time interval (2.5 s).
Rearranging the equation to solve for t:
98 rad = (1/2)(1.3 rad/s²)t²
Multiplying both sides by 2 to eliminate the fraction:
196 rad = 1.3 rad/s²t²
Dividing both sides by 1.3 rad/s²:
t² = 196 rad / (1.3 rad/s²)
t² ≈ 150.77 s²
Taking the square root of both sides to solve for t:
t ≈ √(150.77 s²)
t ≈ 12.28 s
Therefore, the wheel has been in motion for approximately 12.28 seconds at the start of the 2.5-second interval.