posted by winterWX .
The flywheel of a steam engine runs with a constant angular speed of 250 rev/min. When steam is shut off, the friction of the bearings stops the wheel in 2.1 h.
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?
The flywheel will decelerate at a uniform rate, since the frictional torque and moment of inertia can be sonsidered constant. That rate is
alpha = -(250 rev/min)*(2 pi rad/rev)*(60 min/sec)/[(2.1 hr)*(3600 sec/hr)]
Multiply that "alpha" by the radius to get the tangential component of acceleration for a point on the rim of the wheel. It will not matter what the rpm is at the time. RPM will affect the centripetal acceleration, however.
I tried that and got -6233 mm/s^2 or -6.23 m/s^2 but that is the wrong answer. What am I doing wrong? Thanks.
torque of 466 N·m acts on a flywheel. At the instant that the flywheel's angular speed is 76 rad/s, at what rate is work being done by the torque