The flywheel of a steam engine runs with a constant angular velocity of 120 rev/

Question

The flywheel of a steam engine runs with a constant angular velocity of 120 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in 2.3 h. (a) What is the constant angular acceleration, in revolutions per minute-squared, of the wheel during the slowdown? (b) How many revolutions does the wheel make before stopping? (c) At the instant the flywheel is turning at 60.0 rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 29 cm from the axis of rotation? (d) What is the magnitude of the net linear acceleration of the particle in (c)? el inist is Units rev/min 2 (a) Number 5.6 Units rev (b) NumberT 8280 (c) Number2.73 (d) Number 7.12 Units | m/s 2 Units | m/s 2

Explanation / Answer

(A) wi = 120 rev/min

wf = 0

t = 2.3 x 60 min

wf = wi + alpha t

0 = 120 + (2.3 x 60 x alpha)

alpha = - 0.87 rev/min^2
(minus sign represents that angular acc is in opposite direction of angular velocity)

Ans: 0.86 rev/min^2

(B) wf^2 - wi^2 = 2 alpha rev

0^2 - (120^2) = 2(-0.87) rev

rev= 8280 rev

(C) a_t = alpha r

(alpha have to be in rad/s^2)

alpha = (0.87 x 2pi rad) / (60 s)^2 = 1.517 x 10^-3 rad/s^2

a_t = (1.517 x 10^-3)(0.29 )

a_t = 4.40 x 10^-4 m/s^2

(D) a_r = w^2 r

a_r = (60 x 2pi / 60)^2 (0.29) = 11.4 m/s^2  

a = sqrt(a_r^2 + a_t^2) = 11.4 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---