Question

A flywheel within large water pump is a solid disk with mass 21 kg and a radius of 0.45 m. Starting from rest it begins to rotate with a constant acceleration of 2.4 rad/sec2 for 5.1 sec after which it rotates at a constant rate.

1) What is the moment of interia of the flywheel?

2) After 5.1 seconds, what is it's angular speed?

3) Though what angle has it rotated in those 5.1 seconds?

4) How many revolutions has it completed in thoe 5.1 seconds?

5) What is the magnitude of the linear velocity of a piece of metal on the very outer edge of the flywheel 4.1 seconds after it started to rotate?

6) What is the magnitude of the total linear acceleration of a piece of metal on the very outer edge of the flywheel 4.1 seconds after it started to rotate?

Answer 1

1) I = mr^2 /2 = 21 x 0.45^2 /2 = 2.13 kg m^2

2) wf = wi + alpha*t

wf = 0 + 2.4x5.1 =12.24 rad/s

3) wf^2 - wi^2 = 2 x alpha x theta

12.24^2 - 0 =2 x 2.4 x theta

theta =31.21 rad

4) revolutions = 31.21/2pi = 4.97 revolutions

5) wf = 0 + 2.4x4.1   = 9.84 rad/s

and linear velocity v = w *r

v = 9.84 x 0.45 = 4.43 m/s

6) tangential acc. a_t = alpha*r = 2.4 x 0.45 = 1.08 m/s^2

radial acc a_c = w^2 r = 9.84^2 x 0.45 = 43.57 m/s^2

total acc. = sqrt(a_C^2 + a_t^2) = 43.57 m/s^2