The flywheel of a steam engine runs with a constant angular speed of 164 rev/min. When steam is shut off, the friction of the bearings and the air brings the wheel to rest in 1.6 h. I found the constant angular acceleration as 1.71rev/min^2.
They also ask you to find What is the magnitude of the tangential component of the linear acceleration of a particle that is located at a distance of 44 cm from the axis of rotation when the flywheel is turning at 82.0 rev/min?
I tried everything but cant gwt the right answer.
http://www.jiskha.com/search/index.cgi?query=+The+flywheel+of+a+steam+engine+runs+with+a+constant+angular+speed+of+164+rev%2Fmin.+When+steam+is+shut+off%2C+the+friction+of+the+bearings+and+the+air+brings+the+wheel+to+rest+in+1.6+h.+I+found+the+constant+angular+acceleration+as+1.71rev%2Fmin^2.+
To find the magnitude of the tangential component of the linear acceleration of a particle located at a distance from the axis of rotation, we can use the following equation:
acceleration = r * (angular acceleration)
Here, "r" represents the distance of the particle from the axis of rotation, and "angular acceleration" is the constant angular acceleration that you have determined to be 1.71 rev/min^2.
First, we need to convert the given angular speed from rev/min to rev/sec:
angular speed = 82.0 rev/min
= 82.0 rev/min * (1 min/60 sec)
= 1.3667 rev/sec
Next, we can substitute the values into the formula:
acceleration = (0.44 m) * (1.71 rev/min^2) [Note: convert the distance from cm to m]
= 0.44 m * (1/100 cm/m) * (1.71 rev/min^2) * (1 min/60 sec)^2 [Note: convert rev/sec to rev/min]
Simplifying the equation and calculating the result:
acceleration = 0.44 * 1.71 * (1/100) * (1/60)^2 m/s^2
= 5.692 x 10^-4 m/s^2
Therefore, the magnitude of the tangential component of the linear acceleration of the particle is approximately 5.692 x 10^-4 m/s^2.