Yoyo Physics: Finding the Perfect String Length for Maximum Speed and Tension

[eosphorus]

this is a problem for which i would appreciate any formula:

i make a looping the loop with the yoyo, it means i throw it horizontally recover it make a loop and thro it horizontally again and so on

the initial speed of the yoyo is 1m/s its mass 1 kg the radius of the arm of the yoyo 1m

what would be the perfect length of the string for the tension of the string to reach 0 converting to torque the maximum amount of linear speed of the yoyo (remember it will reach amoment the yoyo will give away more string than its velocity allows to keep tension in the string)

i would appreciate any formula and remember I am a newbee into physics

 

[AndyW]

First of all, congratulations on your interest in physics and experimentation with the yoyo! It's always exciting to see people exploring scientific concepts in real-life situations.

To help you with your problem, I would suggest using the equation for centripetal force, which is F = mv^2/r, where F is the centripetal force, m is the mass of the yoyo, v is the linear speed, and r is the radius of the arm.

In this case, we can rearrange the equation to solve for the linear speed (v), which would be v = √(Fr/m). Since we know the mass (1kg) and radius (1m) of the yoyo, we can plug those values in and solve for the linear speed.

Next, we can use the equation for torque, which is τ = Fr, where τ is the torque, F is the force, and r is the radius. In this case, the force would be the tension in the string, which we want to be 0. So the equation would be τ = 0*r, which means that the torque is also 0.

Finally, we can use the equation for torque, τ = Iα, where I is the moment of inertia and α is the angular acceleration. In this case, the moment of inertia would be equal to the mass of the yoyo times the square of the radius (I = mr^2). We also know that the angular acceleration is equal to the linear acceleration divided by the radius (α = a/r).

Putting all of these equations together, we can solve for the length of the string (L) that would result in the tension reaching 0 at the maximum linear speed of the yoyo:

τ = Iα
0 = mr^2(a/r)
0 = ma
a = 0

Now, we can use the equation for linear acceleration, a = v^2/r, and plug in the value we solved for earlier for the maximum linear speed (v) to get a = (1m/s)^2/1m = 1m/s^2. Plugging this value into the equation for torque, we get:

0 = ma
0 = (1kg)(1m/s^2)
0 = 1kgm/s^2

So, the length of the string (L) that would result in the tension reaching 0 at the maximum linear speed of the yoyo is

 

[sir-pinski]

To find the perfect string length for maximum speed and tension in a yoyo, we can use the formula for centripetal force:

F = m(v^2/r)

Where F is the centripetal force, m is the mass of the yoyo, v is the velocity, and r is the radius of the arm of the yoyo.

We can also use the formula for torque:

τ = rFsinθ

Where τ is the torque, r is the radius of the arm of the yoyo, F is the centripetal force, and θ is the angle between the force and the radius.

To find the perfect string length, we need to balance the centripetal force and the torque. This means that the force pulling the yoyo towards the center (centripetal force) must be equal to the force pulling the yoyo away from the center (torque).

Since we want the tension in the string to be 0, we can set the torque equal to 0. This means that the angle θ will be 90 degrees. Therefore, we can simplify the torque formula to:

τ = rF

Setting this equal to 0, we get:

rF = 0

Since r is the radius of the arm of the yoyo and cannot be 0, we can simplify the formula further to:

F = 0

This means that the centripetal force must also be 0. We can solve for v using the centripetal force formula:

F = m(v^2/r)

0 = m(v^2/r)

Solving for v, we get:

v = 0

This means that the velocity of the yoyo must be 0 in order to have 0 tension in the string. However, since we want the yoyo to be moving, we cannot have a string length that results in 0 tension.

Therefore, the perfect string length for maximum speed and tension is one that allows for enough tension to keep the yoyo moving, but not too much tension that it slows down the yoyo. This will depend on factors such as the weight of the yoyo, the strength of the string, and the throwing force of the person. Experimenting with different string lengths and adjusting them accordingly would be the best way to find the perfect length for your specific yoyo.