The flywheel of a steam engine runs with a constant angular veolicty of 160 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in 1.1h.

Question

a. What is the constant angular acceleration in rev/min^2 of the wheel during slowdown?

b. How many revs does the wheel make before stopping?

c. As the instant the wheel is turning at 80.0 rev/ min, what is the tangential component of the linear acceleration of a wheel particle that is 48 cm from the axis of rotation?

d. What is the magnitude of the net linear acceleration of the particle in c?

Answer 1

a. ω = ω₀+ εt; ε = - ω₀/t = - 160rev/min/66min = - 80/33 rev/min² = - 2.424 rev/min².

b. n = (ω + ω₀)/2·t = (0 + 160rev/min)/2·66min = 5280 rev

c. a_t = εr = - 80/33·2π/3600s²·48cm = - 0.2031 cm/s²

d. a = √a_t² + a_n²= √(- 0.2031)² + (80²·4π²/3600·48)² = 3368825 cm/s²