Nonuniform Circular Motion-Find time given r, a-sub-t, V-sub-0

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The discussion revolves around calculating the time it takes for a car to make a turn while undergoing nonuniform circular motion. Given the radius of 35 meters, an initial speed of 8.0 m/s, and a tangential acceleration of 0.4 m/s², participants explore different equations to solve for time. One contributor suggests using the equation for angular displacement under constant angular acceleration, while another proposes a translational kinematics approach as a simpler alternative. The calculations lead to a time estimate of approximately 4.6 seconds. Ultimately, the consensus is that using translational kinematics can clarify the solution process.
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Nonuniform Circular Motion--Find time given r, a-sub-t, V-sub-0

Homework Statement



A car turns through 60 degrees while traveling on the circumference of a 35m radius circle. The care is traveling at 8.0 m/s as it enters the turn and maintains a constant tangential acceleration of .4m/s2 throughout the turn. How long did it take the car to make the turn?


(A) 2.3s (B) 4.2s (C) 4.9s (D) 8.4s

DATA SUMMARY:
at: .4m/s2
Vo: 8.0m/s
R:35m

Homework Equations



at=d|v|/dt=.4m/s2

|V|= \omega *r

\theta= \omega*r

3. The attempt at solution

at=d|v|/dt=.4m/s2

|V|=.4t +8.0 (from integrating at)

\omega= (.4t+8.0)/35 (from |V|=\omega*r)

\theta= \omega*r

(\pi/3= ((.4t+8.0)/35)t= (.4/35) + (8.0t/35)

t=4.6

I get the feeling that there's one crucial equation I'm missing...
 
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\theta = \omega_{0} t + \frac{1}{2} \alpha t^{2}
should be the equation to be used instead.

\theta = \omega t is true only under a constant angular velocity - the general form is \theta = \int \omega dt
which yields the equation above for constant angular acceleration.

You could also use a translational kinematics approach instead, which I think should be less confusing than rotational dynamics.
 
Last edited:


Thank you!

I got it using translational kinematics.
 
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