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# Rotational version of Newton's second law

David explains moment of inertia and the rotational version of Newton's second law and shows how to solve an example problem. Created by David SantoPietro.

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• should we not find the torque of object #3?
• Since there is not a force applied to object #3, it does not affect the total torque. T=F*d, F is 0. But it does have mass, so it is factored into the rotational inertia.
• I got (alpha) = 2.167m/s^2 for the last problem. What did I miss and how would I make the answer into rad/s^2?
• upper line: you end with 450-120 (N.m)=330 N.m
Newton: I think about it from the basic formula F = mass times acceleration (F=ma). From this you take just the units: Newton = kg . m/s^2, so now you know how to write Newton in basic units.
bottom line - unit of the Moment of inertia: I=m.r^2, so unit is kg.m^2

So up you have: N.m = kg . m/s^2 (N) times meter = kg. m^2/s^2
down you have: kg.m^2

Kilograms will cancel out, also meters square (m^2) will cancel out. What will last is 1/s^2.

But I would like know too, how we got radians. I guess it is because we callculated how big angle have masses come by the speed that they have earned from the applied forces.
• When he derives the rotational acceleration formula, isn't he assuming that the angle is 90 degrees and that sin theta is 1? What if it isn't 90 degrees? Would the formula be torque/mr^2sin(theta)?
• Yes, he derived the formula only for the case he was explaining, where the angle was 90.
And you're right. For all cases, you can use ` α = τ/(mr^2 * sinϴ) ` .
• I'm probably going to embarrass myself with this question but still, here it comes.

If the rotational inertia is proportional to the square of r (aka the distance vector) how come that it's so much easier to open a door pushing from the knob? If i make the door swing exerting a force on a point closer to the "lateral" edge (thus increasing r) it requires much less effort than, say, pushing/pulling really close to the axis around which the door is rotating.
Is common sense failing me or am i missing something important here?
• Shorter version:
Torque is not equal to F but dependent on it.
Constant T -> greater the radius->lesser the ANGULAR acceleration
Basically, rate of change of the rate of change of the angle is lesser.
You can try pushing a door with the same T, the closer you are, the greater the angular displacement, extend that idea to angular acceleration.
Good question XD
correct me if wrong
• How to know if it is rad or deg ?
• If you calculate something in SI units, then the angle will always come out in radians.

Also, degrees will use the ° symbol (and sometimes "deg"), while radians will either use nothing at all, or the unit "rad".
• Why is it that when you are further from the axis, angular acceleration increases? Looking at the formula for angular acceleration, if r increases, wouldn't rotational inertia increase even more than torque because rotational inertia is mr^2 while torque is just Fr? I'm asking because for the practice called "Angular Acceleration and angular second law" it repeatedly says that increasing the radius increases angular acceleration because it increases torque.
• The moment of inertia is the property of the object(like mass),which doesn't change when the place where the force is applied changes.
• How come he didn't take gravity into account.

Doesn't the first mass (3) have a force of 30 Newtons? Because 3kg * 10 m/s/s = 30 Newtons.

Therefore you could calculate torque. 3 meters * 30 newtons = 90 NM

And then you could do the rest for all three of the masses, and also get different torques for the other two while taking gravity into account.
• He is not taking gravity into account. He is just saying there are only two forces acting on two of the balls. This is in an alternate dimension and will probably never happen. He is trying to simply things. In real life, there would be so much to take into account it would be nearly impossible to get the correct answer!
• What effect does the centripetal acceleration have on the angular acceleration? As the mass begins pulling on the string and the string on the mass does that deduct from the total angular acceleration? Is the centripetal force accounted for in this equation?
• Short answer is they are related through angular velocity, but torque causing angular acceleration and rotation causing centripetal acceleration are independent of each other. The amount of centripetal acceleration does not impact the amount of angular acceleration.
The angular acceleration is α(t) = dω/dt.
The centripetal acceleration is a(t) = v²/r = ω²r.
Rearranging terms
ω = √(a/r)
α(t) = dω/dt = d/dt[√(a/r)] = 1/[2r*√(a/r)] * da/dt
Not a particularly useful result but that is how they are related.
• Can't you just get rid of r's?
I mean a = T/I = Fr/mr^2 = F/mr.

If not, why?