Question

A yoyo of mass m=2 kg and moment of inertia ICM=0.09 kg m2 consists of two solid disks of radius R=0.3 m, connected by a central spindle of radius r=0.225 m and negligible mass. A light string is coiled around the central spindle. The yoyo is placed upright on a flat rough surface and the string is pulled with a horizontal force F=24 N, and the yoyo rolls without slipping.

(a) What is the x-component of the acceleration of the center of mass of the yoyo? (in m/s2)

a=


(b) What is the x-component of the friction force? (in N)

f=

Answer 1

:/ What did you do so far? Also, we're not really going to solve all your problems for you you have to at least make an effort

Answer 2

I solved the problem this way:

Using F = ma, and the given, I had F - friction force = ma, with friction force = 22 - 2a. With torque = I*alpha, I got 0.25*friction force - 0.1875*22 = 0.0625*alpha.

To put them all together, I got this equation, R(22-2a) - 0.1875*22 = (0.0625/0.25)*a
And then plugging the given, I got a = 1.83 m/s^2, but I don't think I'm doing this correctly.

Can someone please help me figure out where I went wrong right away. Thanks so much...

Answer 3

alpha isnt the same as a

Answer 4

y u don't say???

a= F*R(1-(r/R))/((I/R)+(m*R)) is correct or not!

Answer 5

No it is not correct.

Answer 6

then is this correct
a=[F*(R+r)]/[(l/R)+m*R)]
and friction f
f = force - m*a??

Answer 7

Yes a is correct. f is correct, you will get negative answer but input it positive. Tell me if it worked for you

Answer 8

Hi enjoy. I tried F=force - m*a = 22-(2*12.8333) =
-3.6666 in the negative x direction. I am down to my last chance. Like with all the problems. Thanks, I am trying to keep my head above water. I still have a long way to go. I also had trouble with Question no. 1.

Answer 9

You should get a negative answer, this means the friction is acting in the opposite way that the diagram shows therefore you must input it as positive!!!

Also how did you do on question 8!

Answer 10

It might be better to indicate the equation rather than answers, because the data changes. So, we have:
F-ff=ma (1)
I.alpha=F.R-ff.r (2)

and we have two equation ant variables.
but alpha is a/R or a/r

Answer 11

error, so we'll have:
F-ff=m.a -> ff=F-m.a (1)

I.alpha=F.r+ff.R (2)
alpha=a/R

then:
a=F.(R+r)/(I/R+m.R)

a=positive
ff=negative, and for a got green mark, but not for f, should I write ff with "-" or just the value?

I got a=16.3 f=-4.6

Answer 12

Hi Geoff, Thanks for the tip. I would have entered it as entered my answer as -3.666. I entered as positive 3.666 and it was correct. Good call.

Answer 13

my values were right, just had to write the value of f without the -.

Now, anyone has checked the falling ruler???

Answer 14

Whats your method for the falling ruler?