Help finding torque and force upon a uniform disk?

Question

A uniform disk with a mass of 120 kg and a radius of 1.4 m rotates initially with an angular speed of 1000 rev/min.

(a) A constant tangential force is applied at a radial distance of 0.6 m. What work must this force do to stop the wheel?

_____kJ

(b) If the wheel is brought to rest in 2.5 min, what is the magnitude of the torque produced by the force?

_____N·m

What is the magnitude of the force?

_____N

(c) How many revolutions does the wheel make in these 2.5 min?

_____rev

Answer 1

In rotational mechanics, "torques" do everything that forces used to do, "moments of inertia" do everything that masses used to do, and "angular velocities" do everything that velocities used to do.

That's a bit confusing, I know. But if you understand normal translational mechanics, with forces, masses, and velocities, then you understand the rotational mechanics, with just a substitution of ideas.

So:

A /torque/ is a force times its perpendicular distance from the rotation axis. (In this case, that distance is 0.6 m.) It is written with a T.

A /moment of inertia/ is written with a capital I = k m s², where m is the mass of an object, s is the "size" of the object (in this case the radius), and k is a special constant which depends on the geometry of the object (as well as how, on that geometry, you measure the "size" of the object).

You pick up a textbook to discover that for a solid disk, if you measure s as the radius of the disk, then k = 1/2. (You can look this up on Wikipedia, too, under "List of moments of inertia.")

So for your disk, I = (0.5) * (120 kg) * (1.4 m)² = 117.6 kg m².

A /angular velocity/ is probably the hardest of these terms. It's the speed that the outer edge is going, divided by the radius of the disk. So, in 1 minute, the outer edge makes 1000 revolutions. It goes a distance of 2 * pi * r each revolution, which is a distance of 2000 * pi * r. So it goes at a speed of 2000 * pi * r / minute, or 2000*pi*r/(60 s).

Dividing by r, that's an angular speed of ω = (2 * pi * 1000 / 60) per second

ω = 104.72 / s.

That's all of the hard work. Now comes the easy part: applying your traditional physics equations.

The kinetic energy is E = (1/2) m v². That becomes E = (1/2) I ω². Plug in the numbers, and you'll get the answer for (a).

You know that F = m (dv/dt). But that means that T = I (dω/dt).

In this case, dt (the change in time) is equal to 2.5 minutes, or 150 seconds, while dω is just equal to ω, because the change in ω is from the original value to 0. So, you know I, dω, and dt -- you then can just solve for T.

The magnitude of the force? Well, T = F * (0.6 m), as you'll remember from the above definition of torque. So divide by 0.6 meters.

(c) is a little tricky, because the disk is slowing down at a steady rate. The average rate that it spins at will therefore be (1000 rpm + 0 rpm) / 2 -- averaging the initial and final rpm's. (You can only do that at a constant rate.)

But that means that we can treat it as if it were a 500 rpm disk rotating for 2.5 minutes. That's just 1,250 revolutions.