Flywheel decelerating from 800 rev/min to rest in 6 seconds

[2502floyd]

Can anyone please please please help?

A flywheel initially rotating at a speed of 800 rev/min, is brought to rest with uniform angular deceleration in 6 secs.

a. How many revolutions does the flywheel make before coming to rest?

b. Determine the magnitude and direction of the resultant linear acceleration of a point A on the flywheel 0.2s before coming to rest. Draw a vector diagram showing the magnitude and direction of the resultant linear acceleration and its radial and tangentail components. A is positioned at a fixed radius of 160mm from the axis of rotation.

c. At what time will both the radial and tangential components of acceleration be equal in magnitude.

 

[NateTG]

Do you know the rotational kinematic equations?

When you ask for help here, you're much more likely to get responses if you indicate that you've tried something. This is a homework help forum rather than a do your homework for you forum.

 

[2502floyd]

I have attempted part a
and I make it 40 revolutions.

But part b I don't know where to start.

 

[NateTG]

I have attempted part a
and I make it 40 revolutions.

But part b I don't know where to start.

Well, if you know enough calculus you can write equations that describe the position of a paticle at the edge of the wheel, and take derivatives.

Alternatively, if you determine the angular speed and acceleration of the wheel at the moment that the problem is asking for, you should be able to determine the centripetal (radial) and tangential acceleration of a particle at the edge of the wheel.

Try answering the following questions (roughly in order):

What is the angular acceleration at the requested time? (This should be easy.)
What is the tangential acceleration of a point at the edge of the wheel based on the radius, and the angular acceleration?
What is the angular velocity at the requested time?
What is the centripetal acceleration of a point at the edge of the wheel at the requested time?

 

[faust9]

Have you ever seen this:

$$\vec {\rm a}=(\frac{d^2r}{dt^2}-r\omega^2)\hat{{\rm e}}_r+(r\alpha+2\frac{dr}{dt}\omega)\hat{{\rm e}}_\theta$$

r is a constant thus it's derivative is zero so you're left with:

$$\vec {\rm a}=-r\omega^2\hat{{\rm e}}_r+r\alpha\hat{{\rm e}}_\theta$$

Once you find the above, you'll have direction and magnitude is a simple calculation.

If you were able to calculate the 40 revs then finding the point at 0.2s should be just as easy. Find $\alpha$ (you should have this already) and use that to find $\omega_f$ by integrating:

$$\alpha \int_0^{(6s-0.2s)}\ dt=\int_{\omega_o}^{\omega_f}\ d\omega$$

$\omega$ was given, you have alpha, you have r, and the above yields $\omega_f$ which is used in the acceleration equation.

As for c: what is the orientation of velocity and acceleration to any function?