Is the tension of the string in a yoyo the same when descending and ascending?

[weesiang_loke]

Homework Statement

For a yoyo made of two uniform discs of radius R and each has mass M/2. A short shaft connecting the discs has a very small radius of r. A string is wrapped around the shaft and the yoyo is released with initial speed of zero. Assume that the string is vertical all times and energy in the system is conserved

My question is whether the tension of the string when the yoyo is descending and the tension of string when it is ascending is the same.

Homework Equations

-

The Attempt at a Solution

I think that the tension is the same for both cases as the net force acting downwards should be the same if we want the yoyo to accelerate when it moves downwards and decelerate when it moves up again. So in this case the energy is conserved. Is this the correct explanation?

 

[chaoseverlasting]

The tension will not be the same. This is because when the yoyo is going down, say it rotates in the clockwise direction. What this does is change the direction of the frictional force acting on the string as it's wrapped around the smaller metal rod. When the yoyo is coming up, it's spinning the in the anticlockwise direction and the tension is in the opposite direction. Thus, the tension in both the cases will be different.

 

[weesiang_loke]

so what i get for descending motion:
assuming the tension in the string is T and the mass of the whole yoyo is M. so the net force acting in downwards direction:
we have M*g - T = M*a -----> force equation
T*r = I*\alpha where I is the moment of inertia and \alpha is the angular acceleration.

and \alpha = \frac{<i>a</i>_tangetial}{<i>r</i>}
I = 2* (0.5*(M/2)*R²
so by substition of T, we will get
a_tangential = \frac{2<i>g</i>}{2 + \frac{<i>R</i>²}{<i>r</i>²}}

and so the tension in string during the descending of the yoyo is \frac{<i>M*g*R</i>²}{2<i>r</i>² + R²}.


but i don't know how to determine the tension in the ascending motion. any hint?

 

[tiny-tim]

hi weesiang_loke!

For a yoyo made of two uniform discs of radius R and each has mass M/2. A short shaft connecting the discs has a very small radius of r. A string is wrapped around the shaft and the yoyo is released with initial speed of zero. Assume that the string is vertical all times and energy in the system is conserved

My question is whether the tension of the string when the yoyo is descending and the tension of string when it is ascending is the same.

F = ma (using linear forces) takes no notice of rotation …

T will always equal mg - a, irrespective of the rotation

for an ordinary hand-held yoyo, i don't see how energy could possibly be conserved … the wrist muscles supply energy that keeps the yoyo going

however, if the top of the yoyo string is fixed, with the yoyo rolling-and-falling down the string only under gravity, bouncing off something, and then rolling up again, then the acceleration should be equal at each height on both trips, and so T will be equal also

(and of course work done, = ∫ T dh, will be the same)

(and the angular velocity and acceleration, ω and α, will be determined by the rolling constraints v = ωr and a = αr and by the conservation of energy equation )​

 

[weesiang_loke]

ok. got it. thanks.