A wheel initially rotating at an angular speed of 1.2 rad/s turns through 34 revolutions during the time that it is subject to angular accelaration of .4 rad/s/s. How long did the accelaration last?
So I used theta=Wot+.5at^2. converted rev to rad. and set it up like a quadrtaic equation giving 1.2t+.2t^2-213.6 giving me t equals 13.4 seconds but that is not right??
34 rev = 213.6 rad is OK
(1/2)*0.4 t^2 + 1.2 t -213.6 = 0 is OK
The positive root is
t = [-1.2 + sqrt(1.44 + 170.88)]/0.4 = 29.8 s
You seem to have used the quadratic equation incorrectly.
To find the duration of the acceleration, we can use the kinematic equation for rotational motion:
θ = ω_i*t + 0.5*α*t^2
where:
θ is the angle in radians,
ω_i is the initial angular velocity in rad/s,
α is the angular acceleration in rad/s^2,
t is the time in seconds.
Given:
ω_i = 1.2 rad/s,
θ = 34 revolutions = 34 * 2π rad (since 1 revolution = 2π rad),
α = 0.4 rad/s^2.
Let's substitute these values into the equation:
34 * 2π = (1.2 * t) + (0.5 * 0.4 * t^2)
Simplifying this equation gives:
68π = 1.2t + 0.2t^2
We can rearrange this quadratic equation to standard form:
0.2t^2 + 1.2t - 68π = 0
Now, solving this quadratic equation for t, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
where a = 0.2, b = 1.2, and c = -68π.
Substituting these values into the quadratic formula:
t = (-1.2 ± √(1.2^2 - 4*0.2*(-68π))) / (2*0.2)
Calculating this equation will give the two potential values for t. Keep in mind that the negative solution can be ignored since time cannot be negative in this context.
Therefore, only the positive value of t will be the duration of the acceleration.