Relating the Translational and Rotational Variables of a steam engine...?

Question

The flywheel of a steam engine runs with a constant rotational velocity of 150 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in 2.2 h.

(a) What is the constant rotational acceleration, in revolutions per minute-squared, of the wheel during the slowdown? (b) How many rotations does the wheel make during the slowdown?

(c) How many rotations does the wheel make before stopping?

(d) At the instant the flywheel is turning at 75 rev/min, what is the tangential component of the translational acceleration of a flywheel particle that is 50 cm from the axis of rotation? (e) What is the magnitude of the net translational acceleration of the particle in (d)?

Can you explain your answers, and show the equations you used to solve this?

Answer 1

(a) ω = ωo + αt
0 = 150 + α * 132
α = -150/132
≈ -1.136 rev/s²

(b) θ = ωot + ½αt²
= 150 *132 - ½*1.136*132²
≈ 9903 revolutions (nearest rev)

(c) See (b)

(d) 75 rpm = 75*2π rad/min
α = rω²
= 50 *(75*2π)²
≈ 11103300 cm/min²
= 111033 m/min²
= 111033/3600 m/s²
≈ 30.84 m/s²

(e) ???? I do not understand what it is you are seeking.

Answer 2

As anwered before:

The starting angular velocity is 300 rpm. The wheel stops in 2.5 h. The acceleration (if constant) is 300 rpm / 2.5 h. You need the answer in revs / min^2, so you must convert 2.5 h to minutes (multiply by 60), then divide and get your answer.

The average rotational speed is (1/2)*300 rpm. Therefore in 2.5 h (2.5*60 min) it will make 150*2.5*60 revolutions.

Tangential acceleration aT = r*a, where a is the angular acceleration that you calculated in part (a). r is given as 50cm. Radial acceleration aR = w^2*r where w = angular velocity (given as 75rpm) and r = 50cm. The net linear acceleration is the vector sum of radial and tangential acceleration √[aT^2 + aR^2]