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.