Rotational Motion of the Yo-Yo

[Haveagoodday]

1. Homework Statement
A yo-yo roughly speaking consists of two round, uniform discs, sandwiched

around a third smaller disc. A string is wound around the middle disc, and so

the yo-yo may roll up and down as the string winds and unwinds. Consider

such a yo-yo, with the two bigger discs having radius R = 4.00 cm and mass

M = 30.0 g each; and the smaller disc in the middle having radius r = 0.700

cm and mass m = 5.00 g. The string is taken to be massless, and infinitely

thin.

a) What is the total moment of inertia of the yo-yo, around an axis going

through the centre of the discs? Indicate both the algebraic expression and

a number.

b) The end of the string is now fastened to something at a fixed position

(like a finger), and the yo-yo is let drop towards the floor. Identify the forces

acting on the yo-yo, and for each, indicate whether they provide torque, work,

impulse and/or acceleration to the yo-yo.

c) What is the acceleration of the yo-yo downwards; what is its angular

acceleration? How large is the string force?

d) How big a fraction of the total kinetic energy goes into the rotating motion?

The Attempt at a Solution

Can somebody check my solutions?
a)
ΣI= 1/2mr^2 + MR^2=4.812 kg*m^2
b)
There are forces like force of gravity and string force
gravity provides Work and acceleration
string provides torque and impulse

c) a=g= 9.80 m/s^2
Fg=2Mg=0.588 N
Fg=mg=0.049 N
τ=RFg=0.02352
τ=rFg=3.43*10^-4
α=Στ/ΣI=5.6 rad/s^2
Fs=mg/1-(mR^2/ΣI)= 0.388 N

d) haven't solved this one yet, does anybody have an idea for how to solve this one?

 

[SteamKing]

View attachment 91041

1. Homework Statement
A yo-yo roughly speaking consists of two round, uniform discs, sandwiched

around a third smaller disc. A string is wound around the middle disc, and so

the yo-yo may roll up and down as the string winds and unwinds. Consider

such a yo-yo, with the two bigger discs having radius R = 4.00 cm and mass

M = 30.0 g each; and the smaller disc in the middle having radius r = 0.700

cm and mass m = 5.00 g. The string is taken to be massless, and infinitely

thin.

a) What is the total moment of inertia of the yo-yo, around an axis going

through the centre of the discs? Indicate both the algebraic expression and

a number.

b) The end of the string is now fastened to something at a fixed position

(like a finger), and the yo-yo is let drop towards the floor. Identify the forces

acting on the yo-yo, and for each, indicate whether they provide torque, work,

impulse and/or acceleration to the yo-yo.

c) What is the acceleration of the yo-yo downwards; what is its angular

acceleration? How large is the string force?

d) How big a fraction of the total kinetic energy goes into the rotating motion?

The Attempt at a Solution

Can somebody check my solutions?
a)
ΣI= 1/2mr^2 + MR^2=4.812 kg*m^2

Are you sure about the units of I here? 1 kg-m² represents a relatively healthy inertia. I'm not sure a person could handle a yo-yo with this much inertia.

In the problem statement, the masses of the yo-yo parts are given in grams, not kilograms, and the radii of the various parts are given in centimeters, not meters.

b)
There are forces like force of gravity and string force
gravity provides Work and acceleration
string provides torque and impulse

c) a=g= 9.80 m/s^2
Fg=2Mg=0.588 N
Fg=mg=0.049 N
τ=RFg=0.02352
τ=rFg=3.43*10^-4

You're showing two different values of τ here. What are the units of τ? Why do you have two different values?

α=Στ/ΣI=5.6 rad/s^2
Fs=mg/1-(mR^2/ΣI)= 0.388 N

d) haven't solved this one yet, does anybody have an idea for how to solve this one?

 

[Haveagoodday]

Are you sure about the units of I here? 1 kg-m² represents a relatively healthy inertia. I'm not sure a person could handle a yo-yo with this much inertia.

In the problem statement, the masses of the yo-yo parts are given in grams, not kilograms, and the radii of the various parts are given in centimeters, not meters.
You're showing two different values of τ here. What are the units of τ? Why do you have two different values?

Sorry i ment to write 4.812*10^-5 kg*m^2 in a

 

[Haveagoodday]

Are you sure about the units of I here? 1 kg-m² represents a relatively healthy inertia. I'm not sure a person could handle a yo-yo with this much inertia.

In the problem statement, the masses of the yo-yo parts are given in grams, not kilograms, and the radii of the various parts are given in centimeters, not meters.
You're showing two different values of τ here. What are the units of τ? Why do you have two different values?

In c i calculated the torque for gravitational force for the outer parts and the inner part.

 

[haruspex]

gravity provides Work and acceleration

Does it also provide impulse or torque?
(The torque question is a bit unclear. The torque of a force depends on the axis you choose. Is it right to assume mass centre?)

c) a=g= 9.80 m/s^2
Fg=2Mg=0.588 N
Fg=mg=0.049 N

As with part a), it would be better to keep everything algebraic as long as possible, only plugging in numbers at the end. Makes it easier for others to follow your working. But you do need to use distinct variables, not keep reusing a, m, F etc. with different meanings.

τ=rFg=3.43*10^-4

I see you now consider Fg to be generating torque, but about which axis? Is that the axis you used for calculating I?

 

[hooolahooop]

I have a question about the situation in b), wouldn't the string force also provide acceleration? and the acceleration of the yoyo isn't the same as g (9.80) ,some of the energy is transferred to the spinning motion, right?

 

[haruspex]

I have a question about the situation in b), wouldn't the string force also provide acceleration? and the acceleration of the yoyo isn't the same as g (9.80) ,some of the energy is transferred to the spinning motion, right?

The linear acceleration is determined by the net force. I don't much like the question of whether the tension provides acceleration. It doesn't do any work, but does influence the acceleration. Yes, some of the energy goes into spinning the yoyo.

Let's try to get the rotational acceleration right. What axis do you want to use? You can use either mass centre of the yoyo or any point fixed in space. Having chosen it, what is the net torque about that axis and what is the yoyo's moment of inertia about it?

 

[hooolahooop]

The linear acceleration is determined by the net force. I don't much like the question of whether the tension provides acceleration. It doesn't do any work, but does influence the acceleration. Yes, some of the energy goes into spinning the yoyo.

Let's try to get the rotational acceleration right. What axis do you want to use? You can use either mass centre of the yoyo or any point fixed in space. Having chosen it, wad is the net torque about that axis and what is the yoyo's moment of inertia about it?

The question are asking for acceleration of the yoyo downwards i used different equations and calculated the total moment of inertia, where i got that

T(torque) = r * Fs(spring force)

T (torque) = I * alpha --> I * alpha = r*Fs ----> alpha = a/r

Ma = Mg - Fs

Fs = [I (total) *(a/r)]/r ----> Ma = Mg - (I*a/r^2)

solve for a -- > a = (M*g*r ^2) /[ r^2*M + I(total)] where r is the radius of the small disc because the string is around this disci used an axis through the middle( center of mass)

 

[statii]

What do you use as unit for radius and mass? (m and kg?) Also what did you get as total inertia?

 

[haruspex]

The question are asking for acceleration of the yoyo downwards i used different equations and calculated the total moment of inertia, where i got that

T(torque) = r * Fs(spring force)

T (torque) = I * alpha --> I * alpha = r*Fs ----> alpha = a/r

Ma = Mg - Fs

Fs = [I (total) *(a/r)]/r ----> Ma = Mg - (I*a/r^2)

solve for a -- > a = (M*g*r ^2) /[ r^2*M + I(total)] where r is the radius of the small disc because the string is around this disci used an axis through the middle( center of mass)

Yes, that all looks right.

 

[jensjensen]

For angular acceleration, i got a(alpha)=g/(R+I/(mR))