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At the instant the flywheel is turning at 75 rev/min, what is the tangential component of the linear accelrat?
The flywheel of a steam engine runs with a constant angular speed of 125 rev/min. When steam is shut off, the friction of the bearings stops the wheel in 1.7 h.
(a) What is the constant angular acceleration, in revolutions per minute-squared, of the wheel during the slowdown?
-1.22 rev/min2 <<<<-------ANSWER
(b) How many rotations does the wheel make before stopping?
rotations
6377.5 rotations <<<<-------ANSWER
(c) At the instant the flywheel is turning at 75 rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 50 cm from the axis of rotation?
______ mm/s2
(d) What is the magnitude of the net linear acceleration of the particle in (c)?
______m/s2
I JUST NEED HELP ON PARTS (d) AND (c). I ALREADY GOT AND POSTED THE CORRECT ANSWERS FOR PARTS (a) AND (b). THANKS.
I got part (d) its 31 m/s^2
To convert any angular value (velocity, acceleration, distance) to a tangential value, you must first have the angular value defined in terms of radians. Then, you simply multiply by the radius of interest. Since the angular acceleration is constant, the tangential acceleration is constant also, so the speed of 75rpm doesn't matter here. First, convert to radians/s^2
-1.22 rev/min^2 * 2PI radians / rev * (1 min)^2 / (60 sec)^2 = -.00213 rad/s^2
to convert this to tangential acceleration at 500 mm, multiply by 500mm = -1.065 mm/s^2
to find the magnitude of linear acceleration at that point, you must combine the centripetal acceleration with the tangential acceleration and get the magnitude. Use Pythagorean's theorem, where atotal = sqrt(atangential^2 + acentripetal^2)
Centripetal acceleration is v^2 / R, where v is the tangential velocity at that point and R is the radius to that point. This is where the 75rpm come in.
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