The large disk-shaped flywheel illustrated below has a radius R of 0.25 m. It is made to spin by the small wheel that contacts it at its rim. The small wheel applies a constant force of 1430 N. Friction in the bearing exerts a retarding torque 22 Nm on the large wheel. If the magnitude of the angular acceleration of the large wheel is 145 rad/s2, what is its mass?
So first calculate the total torque.
T1 is given = -22
T2 can be found using T=Fr
T=1430(.25)= 357.5
Total T= 335.5Nm
Then use T=Ia to get the moment of inertia
335.5=I (145)
I= 2.31
For a solid cylinder/disk, the moment of inertia is given by I=1/2mr^2
2.31=1/2m(.25)^2
m=73.9kg
To find the mass of the large disk-shaped flywheel, we can use the equation for torque:
Torque = Moment of inertia x Angular acceleration
The moment of inertia (I) of a disk is given by the equation:
I = (1/2) * m * r^2
where m is the mass of the disk and r is the radius.
In this case, the retarding torque is given as 22 Nm, and the angular acceleration is given as 145 rad/s^2. The radius is given as 0.25 m.
Substituting these values into the equation for torque, we get:
22 Nm = (1/2) * m * (0.25 m)^2 * 145 rad/s^2
We can rearrange the equation to solve for the mass (m):
m = (22 Nm) / [(1/2) * (0.25 m)^2 * 145 rad/s^2]
Calculating this expression, we find:
m = 22 Nm / (0.03125 m^2 * 145 rad/s^2)
m = 22 Nm / 0.02265625 m^2 rad/s^2
m ≈ 972.71 kg
Therefore, the mass of the large disk-shaped flywheel is approximately 972.71 kg.