Finding radial acceleration

[dudgey]

Please help! I'm having trouble understanding this rotation problem.
At a time 2.60 s, a point on the rim of a wheel with a radius of 0.240 m has a tangential speed of 50.0 m/s. as the wheel slows down with a tangential acceleration of constant magnitude 10.7m/s^2 . At what time will the radial acceleration equal 9.810 m/s^2?

I tried using the equation omega final = omega initial + alpha * t.
* I found omega final by 9.81=omega^2*r (probably wrong)
* I converted the given speed to rad/s and used it for omega inital
*Then I used the square root of the two accelerations squared for alpha (also seems wrong, but I can't figure out anything else)
*Then I solved for t and added the given t to it to find a time, but its not right.

Thanks for your help :)

 

[Nate810]

The correct approach is to use the equation a_r = r*a_t, where a_r is the radial acceleration and a_t is the tangential acceleration. You can then solve for t, the time it takes for the radial acceleration to equal 9.81 m/s^2. The equation is:t = (9.81 - 10.7*(50.0/0.240))/(10.7/0.240)This gives you a value of t = 2.86 seconds. Adding this to the initial time of 2.60 seconds gives you a total time of 5.46 seconds.

 

[wais]

First, it's great that you are using the equation omega final = omega initial + alpha * t to solve this problem. However, there are a few things that can be improved in your approach.

First, let's clarify the given information. We know that at a time of 2.60 s, the tangential speed is 50.0 m/s and the tangential acceleration is 10.7 m/s^2. We also know that the radial acceleration we are looking for is 9.810 m/s^2.

Now, let's use the equation you mentioned to solve for the angular acceleration (alpha). We can rearrange the equation to alpha = (omega final - omega initial)/t. Plugging in the values we know, we get alpha = (0 - 50.0 rad/s)/2.60 s = -19.23 rad/s^2. Note that the negative sign indicates that the wheel is slowing down.

Next, we can use the formula for radial acceleration, ar = alpha * r, to solve for the time when the radial acceleration is equal to 9.810 m/s^2. Plugging in the values we know, we get 9.810 m/s^2 = -19.23 rad/s^2 * 0.240 m. Solving for t, we get t = 0.509 s.

Therefore, at a time of 2.60 s + 0.509 s, or 3.109 s, the radial acceleration will equal 9.810 m/s^2.

I hope this helps clarify the problem for you. Remember to always double check your units and make sure they are consistent throughout your calculations. Keep practicing and you'll become more comfortable with rotation problems. Best of luck!