Question

The flywheel of a steam engine runs with a constant angular velocity of 150 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in 1.6 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 75.0 rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 37 cm from the axis of rotation? (d) What is the magnitude of the net linear acceleration of the particle in (c)?

Answer 1

a)
`\ddot n = -1.563\,(rev)/(min^(2))`, b)
`\Delta n = 7197.697\,rev`, c)
`a_(t) = 1.009* 10^(-3)\,(m)/(s^(2))`, d)
`a = 22.823\,(m)/(s^(2))`

Step-by-step explanation:

a) Constant angular acceleration is:

`\ddot n = (\dot n - \dot n_(o))/(\Delta t)`

`\ddot n = (0\,(rev)/(min) - 150\,(rev)/(min))/((1.6\,h)\cdot \left(60\,(min)/(h) \right))`

`\ddot n = -1.563\,(rev)/(min^(2))`

b) The amount of revolutions required to stop the flywheel is:

`\Delta n = (\dot n^(2)-\dot n_(o)^(2))/(2\cdot \ddot n)`

`\Delta n = (\left(0\,(rev)/(min) \right)^(2)-\left(150\,(rev)/(min) \right)^(2))/(2\cdot \left(-1.563\,(rev)/(min^(2)) \right))`

`\Delta n = 7197.697\,rev`

c) The tangential acceleration of the particle is:

`a_(t) = \left(1.563\,(rev)/(min^(2)) \right)\cdot \left((1)/(3600)\,(min^(2))/(s^(2))\right)\cdot \left(2\pi\,(rad)/(rev)\right)\cdot (0.37\,m)`

`a_(t) = 1.009* 10^(-3)\,(m)/(s^(2))`

d) The radial acceleration of the particle is:

`a_(r) = \left[\left(75\,(rev)/(min) \right)\cdot \left((1)/(60)\,(min)/(s) \right)\cdot \left(2\pi\,(rad)/(rev) \right)\right]^(2)\cdot (0.37\,m)`

`a_(r) = 22.823\,(m)/(s)`

The net linear acceleration is:

`a = \sqrt{a_(r)^(2)+a_(t)^(2)}`

`a = \sqrt{\left(22.823\,(m)/(s^(2)) \right)^(2)+\left(1.009* 10^(-3)\,(m)/(s^(2)) \right)^(2)}`

`a = 22.823\,(m)/(s^(2))`