Answer 1:
1. First off, I'm assuming that this conelike object will be spinning around the central axis, much like a top. This problem can get a bit tricky, so let's take a moment to define a few things: h  The height of the conelike object. a  The bigger radius at one edge of the object. b  The smaller radius at one edge of the object (i.e. the part where it was cut). m  The mass of the object. alpha  This will be the angular acceleration of the object, in radians per second squared (meaning the rate of change of the angular velocity)
The equation for torque is I*alpha, where I is something that is called the moment of inertia. For this object spinning in this way, I would be 3/10 (b^{2} + (a^{3} (a + b)) / (a^{2} + a b + b^{2})), which means that: torque = 3/10 (b^{2} + (a^{3} (a + b)) / (a^{2} + a b + b^{2})) * alpha. In order to find the amount of work done, you will have to know how much the object has spun around. Since it is spinning, it will rotate some angle (in radians), and let's call this angle theta. The equation for work is torque * theta, so once you know how far the object has spun in radians, you can multiply that by the torque calculated earlier to get the work. work = 3/10 (b^{2} + (a^{3} (a + b)) / (a^{2} + a b + b^{2})) * alpha * theta.

Answer 2:
You're going to need calculus for this one; that's how the formulas of the moments of inertia for various solids were derived in the first place. In order to confidently answer your question, I would need to unpack my undergraduate college mathematics books, but here is my guess, assuming my recollection of college mathematics is correct: The moment of inertia of a massive, rotating ring is equal to the mass of the ring (m) multiplied by the distance of the ring from the axis of rotation (r). So, Ir = mr. The moment of inertia of a massive, rotating disk (Id) is equal to the integral of a rotating ring from zero to the radius of the disk (R). However, the mass of the "ring" over which we are integrating is itself equal to the density of the material (p) times 2*pi times the radius of the ring. So, m = p*2*pi*r. Thus, the moment of inertia of the disk is equal to the integral of (p*2*pi*r)*r*dr = p*r^{2}*2*pi*dr. Therefore, Id = [p*r^{3}*pi*2/3 + k], evaluated at r = R minus r = 0, k being the constant of integration. Thus, Id = p*R^{3}*pi*2/3. However, what you want is the moment of inertia for a truncated cone (Ic). Our truncated cone has a length d, the distance from one flat face to the opposite flat face. Let's call the radius of the small flat face Rs and the radius of the large flat face Rl, so Rl > Rs. So, by stacking the disks of radius R from Rl to Rs, and ASSUMING MY DERIVATION IS CORRECT, Ic should be: Ic = integral[d*p*R^{3}*pi*2*dR/3], evaluated from Rs to Rl. Ic = [d*p*R^{4}*2*pi/12 + k], evaluated from Rs to Rl, Ic = (Rl^{4}  Rs^{4})*d*p*pi/6, where, Ic = moment of inertia of the cone Rl = the radius of the larger flat face of the cone Rs = the radius of the smaller flat face of the cone d = the length of the cone from one flat face to the other p = the density of the cone pi ~ 3.14159.... (I'm sure you know about pi!). Rl, Rs, and d you can get pretty easily by measuring. Getting p may pose a bit more of a challenge. To get p, you will need the mass of the cone (m) and the volume of the cone (V). To get m, just weigh the sucker. The volume of a nontruncated cone is pi*h*R 2 /3, where h is the height of the cone and R the radius of the base, but you've got a truncated cone. There are two ways you can get V, then; you can either do a bit more calculus and integrate the area of a circle from Rs to Rl along d again, or you can cheat and calculate the height the cone would have if it were NOT truncated (i.e. h), and then calculate the volume of a cone of height h and base Rl, and then subtract the volume of a cone of height (h  d) and base Rs. I'm sure you can figure that one out, too. If the cone is denser than water, then you can really cheat and just use Archimedes' law: a sinking body displaces its volume in water. Once you have V,p = m/V, and you can plug that back into the moment of inertia formula and get the moment of inertia. I am sure you have formulas for torque and work once you have Ic.

Answer 3:
It's not quite clear what you're asking. Are you wondering how much energy is required to get an object (cone) spinning at a certain number of rotations per minute? Or how rapidly the cone will accelerate when a given torque is applied? Torque is defined as force times distance: force applied a certain distance from the center of rotation. The formulas for handling torque, rotation, and rotary acceleration are very similar to the formulas for linear force, movement, and acceleration: F = m * a: force equals mass times acceleration T = I * A : torque equals moment of inertia (I) times rotary acceleration (A or alpha)and E = (1/2)*m*v^{2}: kinetic energy equals mass times velocity squared E = (1/2)*I*u^{2}: kinetic energy equals moment of inertia times rotational velocity (usually in radians per second) squaredThe moment of inertia for a cone is (3/10)*m*r^{2}, where r=radius of cone. To find the moment of inertia for a truncated cone like the one you describe (as if someone sliced the bottom off an ice cream cone), calculate it for a full cone, then subtract the moment of inertia for the "missing" cone.
Click Here to return to the search form.
