Physics 13.1 Moment of Inertia Application (2 of 11) Acceleration=? of the 'True' Yo-Yo

In this video I will find the acceleration, a=?, of a "true" yo-yo of mass m.

Next video in this series can be seen at:

[Numaram-rqir]

Sir I pray for your wellbeing after every video I watch of yours. You make school so much easier for me. Thank you.

[aytugsezer]

Hi.Dear respectable Michel van Biezen.
I am from Turkey and I am studiying Physic at Hacettepe University.You are bettter than my dear Prof.Dr. .... You are amazing teacher.I wish ı could meet with u.Thank u for your all lessons.

[inesbilkic]

Why do we use the same setup for a yo-yo as for a hanging mass? If I set up a situation in which there is a force of tension pulling up a hanging mass, the set up is ma = mg - T. Now, if I have a yo-yo that is the same mass and if we make a situation in which there is an equal force of tension that is pulling up on it, the acceleration can also be found using the setup ma = mg - T. By this logic, we will solve for both the yo-yo and the hanging mass having an equal acceleration, but this simply cannot be possible because of the conservation of energy principle. As the yo-yo rotates, it converts some of its potential energy to rotational kinetic energy and some to translational kinetic energy. However, the hanging mass converts all of its potential energy to translational kinetic energy, so it MUST move faster. I know I am going wrong somewhere here, please help me figure out where!

[Monafide]

Very thankful for your help, excellent explanation!

[ptyptypty]

to quote Bill Nye.... "SCIENCE RULES !" if I were a politician I would add to the End of this video, "I approve this Message" :) well done Michel...

[animesh]

what if the mass of the smaller disc is considered?
do we add the MOI of inner and outer disc

[Yourejusatube]

Thanks for the video! You're a great teacher!

[mr.perfecttube]

How to deal with problems in which the yo-yo is released in an upward accelerating elevator?

[guilanfarman]

What a great video! Thanks a lot Professor. Why do we use Rout to calculate the moment of inertia(even though there is an empty disc inside the yo-yo)?

[tiredOfSleep-ur]

Sir, thanks a lot for your videos. It usually help me see things more clearly most of the time, but today I really have a hard time understanding why the angular acceleration is referred to the a/R (inner) instead of a/R (outer). I figure that since the you calculate inertia with R (outer), R (outer) should be used for the angular acceleration as well. Also, T can be expressed with the acceleration in which case I am certain the acceleration refers the entire yoyo, not just the inner loop. What do you think, sir?

[edgarburguete]

Thanks a lot. I have one question, why do the tangencial acceleration and the downward acceleration are treated as the same?

[Andrew-dhws]

why do you only calculate the moment of inertia for one disk even tho on a yo yo there are two disks? should you do 2 * ( 1/2 * m * r^2)?

[suzy]

Sir how to calculate acceleration if there are 2 strings one string attach to outer and another one attach to inner? It easy to find torque but when sigma torque = I*alpha and alpha =a/R .which R do we choose?

[mr.jamesdavidrobert]

If a yo-yo was fired vertically with sufficient escape velocity and recoiled, then refired, could a man put himself into space through this method of propulsion?

[IOIIOHilmiHikmatulloh]

Dear Mr. Michel van biezen, i imagine the rope attached to my finger, so what happen if i let the yoyo drop and at the same time i pulled my finger with some acceleration, is angular acceleration (alpha) of the yoyo is getting bigger or smaller ? And why ? Please answer sir, im really curious, thank you

[geenathdamsara]

Why Did You Took Rout For I(Moment Of Interia)

[oldwolfp]

Sir, if we take Rmin as R/3 instead of R and take Rmax as R instead of 3R, the result changes. I couldn't understand that point, ratio is same but result changed cuz of square of Rmax.

[hrhbucket]

Thanks, this a a great video. I followed the logic of how to calculate the 'downward' acceleration 'a' but is it possible to derive a similar formula for the angular acceleration alpha. I tried and came up with the formula alpha = 2g / ( (Rout*Rout/Rin) + 2*Rin). Trouble is that I tried plugging some figures into this and it does not work. Could you help ?

[김의현-od]

I thought the equation for linear acceleration = sqrt[a(tan)^2 +a(rad)^2] ?
What happened to the radial acceleration?

[geenathdamsara]

I Did'nt Understand What You Did With R(out )And R(in)