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)]
-12.5 rad/s^2
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.
It seems like you've made a calculation error when converting units. Let's go through the calculation step by step to identify the mistake:
First, we need to convert the given angular deceleration rate from revolutions per minute (rev/min) to radians per second squared (rad/s^2). You started off correctly by multiplying the given rate by 2π rad/rev and then dividing by 60 min/sec.
alpha = -(250 rev/min) * (2π rad/rev) / (60 min/sec)
= -26.18 rad/s^2
Next, we need to calculate the tangential component of the linear acceleration. This can be found by multiplying the angular deceleration (alpha) by the distance from the axis of rotation (50 cm or 0.5 m).
acceleration = alpha * radius
= -26.18 rad/s^2 * 0.5 m
= -13.09 m/s^2 (rounded to two decimal places)
Therefore, the correct answer for the tangential component of the linear acceleration of the flywheel particle is -13.09 m/s^2.
It's worth noting that the sign of the acceleration is negative because the flywheel is decelerating, meaning its velocity is decreasing.