Question

The flywheel of a steam engine runs with a constant angular velocity of 190 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in 2.4 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 95.0 rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 38 cm from the axis of rotation? (d) What is the magnitude of the net linear acceleration of the particle in (c)?

Answer 1

(a)
`\alpha = - 1.32\ rev/m^(2)`

(b)
`\theta = 13674\ rev`

(c)
`\alpha_(tan) = 8.75* 10^(- 4)\ m/s^(2)`

(d)
`a = 22.458\ m/s^(2)`

Solution:

As per the question:

Angular velocity,
`\omega = 190\ rev/min`

Time taken by the wheel to stop, t = 2.4 h =
`2.4* 60 = 144\ min`

Distance from the axis, R = 38 cm = 0.38 m

Now,

(a) To calculate the constant angular velocity, suing Kinematic eqn for rotational motion:

`\omega' = \omega + \alpha t`

`omega'` = final angular velocity

`\omega` = initial angular velocity

`\alpha` = angular acceleration

Now,

`0 = 190 + \alpha * 144`

`\alpha = - 1.32\ rev/m^(2)`

Now,

(b) The no. of revolutions is given by:

`\omega'^(2) = \omega^(2) + 2\alpha \theta`

`0 = 190^(2) + 2* (- 1.32) \theta`

`\theta = 13674\ rev`

(c) Tangential component does not depend on instantaneous angular velocity but depends on radius and angular acceleration:

`\alpha_(tan) = 0.38* 1.32* (2\pi)/(3600) = 8.75* 10^(- 4)\ m/s^(2)`

(d) The radial acceleration is given by:

`\alpha_(R) = R\omega^(2) = 0.32(80* (2\pi)/(60))^(2) = 22.45\ rad/s`

Linear acceleration is given by:

`a = \sqrt{\alpha_(R)^(2) + \alpha_(tan)^(2)}`

`a = \sqrt{22.45^(2) + (8.75* 10^(- 4))^(2)} = 22.458\ m/s^(2)`