An iron disk has a radius of 0.625 m and a mass of 217 kg. The disk is mounted on its axis so that it is free to spin. How much force is required if the force is applied at a distance 19.5 cm from the axis of rotation?
I = (1/2)m r^2
F (.195) = I alpha
since you did not say what the angular acceleration alpha is, I can not do anything more for you
α = 1.00 rad/s2
well, then you have your answer
F = (1/2) 217 (.625)^2 / .195
To find the force required to spin the iron disk, we can use the formula for torque:
Torque = Force x Distance.
Given:
Radius of the iron disk (r) = 0.625 m.
Mass of the disk (m) = 217 kg.
Distance from the axis of rotation (d) = 19.5 cm = 0.195 m.
First, let's find the moment of inertia (I) of the disk. The moment of inertia is a measure of an object's resistance to rotational motion and depends on its shape and mass distribution. For a solid disk rotating about its axis, the moment of inertia is given by:
I = (1/2) x m x r^2.
Plugging in the values:
I = (1/2) x (217 kg) x (0.625 m)^2.
Calculating:
I = (1/2) x 217 kg x 0.390625 m^2.
I = 42.853515625 kg·m^2.
Now, we can calculate the force required using the torque formula. Rearranging the formula, we have:
Force = Torque / Distance.
The torque is given by the moment of inertia multiplied by the angular acceleration (alpha). Since the disk is mounted on its axis and can freely spin, let's assume there is no angular acceleration (alpha = 0). Therefore, the torque becomes:
Torque = I x alpha = I x 0 = 0.
Plugging in the values into the force formula:
Force = 0 / 0.195 m.
Since the torque is zero, we find that no force is required to keep the disk spinning when the distance from the axis is 19.5 cm. In this case, the disk will continue to spin without any additional force applied.