Solve Flywheel Question: Magnitude & Direction of Resultant Linear Accel.

[paul9619]

Flywheel Question??

Hi all I have done the following question:

A flywheel , intially rotating at a speed of 600 rev/min, is brought to rest with uniform angular deceleration in 10 seconds.

A) How many revolutions does the flywheel make before coming to rest?

I used Pheta = (Wo + W)t/2 to get 314.16 rads. The I used 314.16/2Pi to get 50 revolutions

B)Determine the magnitude and direction of the resultant linear acceleration of a point A on the flywheel 0.5seconds before coming to rest. Assume the point A is positioned at a fixed radius of 150mm from the axis of rotation.

I firstly worked out angular acceleration to be -6.2832 rads/sec. I then found W to be 3.14 rads/sec. Using the formula for radial acceleration (ar) = w^2r I got 1.4804.
And for tangential acceleration i got -0.94248. Therefore the resultant linear acceleration using pythagorus is 1.755. The resultant angle is tan-1(-0.94/1.48) = -32.48 degrees.

Now what's confusing me is all the minus signs. Is this a legit answer? The minus angle is confusing me?

The final question is:

C) Draw a vector diagram showing the maginitude and direction of the resultant linear acceleration and its radial and tangential components??

Shall I draw the vector diagram using the answers above. So it will end up being in the 3rd quadrant?

Cheers

Do I just use the minus values worked out above.

 

[NotMrX]

Hi all I have done the following question:

B)Determine the magnitude and direction of the resultant linear acceleration of a point A on the flywheel 0.5seconds before coming to rest. Assume the point A is positioned at a fixed radius of 150mm from the axis of rotation.

I firstly worked out angular acceleration to be -6.2832 rads/sec. I then found W to be 3.14 rads/sec. Using the formula for radial acceleration (ar) = w^2r I got 1.4804.
And for tangential acceleration i got -0.94248. Therefore the resultant linear acceleration using pythagorus is 1.755. The resultant angle is tan-1(-0.94/1.48) = -32.48 degrees.

Now what's confusing me is all the minus signs. Is this a legit answer? The minus angle is confusing me?

It might help to draw a picture of the wheel at the time of 0.5 seconds with point A at the correct position. Then draw the accelleration components. The centripetal or radial accelaration points to the center and tangential accleration is pointing the opposite direction of movement.

 

[kyle8921]

I would like to first commend you on successfully solving the flywheel question and obtaining the correct answers. Your calculations and use of relevant equations are correct.

To answer your question, the minus signs in your answers are a result of the direction of the acceleration being opposite to the direction of rotation. In this case, the flywheel is decelerating, so the acceleration is in the opposite direction of its initial rotation. This is why your tangential acceleration is negative, indicating a decreasing speed.

For the vector diagram, you can use the magnitude and direction of the resultant linear acceleration and its radial and tangential components as calculated above. The resultant linear acceleration vector would be pointing in the negative x direction, while the radial and tangential components would be pointing in the negative and positive y directions, respectively. This would result in a vector diagram in the third quadrant, as you mentioned.

Overall, your answers and approach to the problem are correct. Keep up the good work!