What is the moment of inertia of the yo-yo?

[kirafreedom]

Homework Statement

a yo-yo is spun from rest by pulling on the string with a constant tension of 2.0N. The radius of the inner rod on which the string is strung around is 0.50cm. the tension applied for 5.0seconds after which the yo-yo is observed to spin with an angular velocity of 15 rad/sec

a)from the information given above, what is the moment of inertia of the yo-yo

b) what is the total angle the yo-yo has traveled through in these 5.0 seconds

Now you press your finger against the outer rim of the yo-yo ( which has a radius of 4.0 cm) to bring it to a stop. You apply a constant force of 2.0N directed perpendicular to the rim of the yo-yo. The tension from part a) is no longer being applied to the yo-yo. The coefficient of kinetic friction between you finger and the edge of the yo-yo is 0.80.

c) how ling does it take for the yo-yo to come to a stop

Homework Equations

I=1/2Mr^2
f=ma

The Attempt at a Solution

a)
f=ma
2=m x 9.8
2/9.8=m
m=0.204kg
I=1/2Mr^2
I=1/2 x 0.204 x 0.5^2
I=0.026
I=0.03

b) I think you do 15 rad/sec x 5. seconds x 360°
c) have no idea

 

[Doc Al]

a)
f=ma
2=m x 9.8
2/9.8=m
m=0.204kg
I=1/2Mr^2
I=1/2 x 0.204 x 0.5^2
I=0.026
I=0.03

Assume that the only force you need to worry about is the tension applied to the yo-yo.

Use: Torque = I * alpha

What's alpha? What's the torque?

 

[SteamKing]

How many radians are in 360 degrees?

 

[haruspex]

f=ma
2=m x 9.8
2/9.8=m

That would be true if the yo-yo were simply suspended, statically, from the string. But there's acceleration involved here.
As Doc Al says, you don't need to worry about gravity in this question. That's because you can take moments about the centre of the spindle; the gravitational force acts through that point, so has no moment about it.

I=1/2Mr^2

That's true for a uniform disc radius r. The given radius is only for the central spindle. You don't know how big the yo-yo is overall, or how the mass is distributed in it. The point of the question is to calculate I from the external behaviour described.

 

[kirafreedom]

sorry, for question c
c) how ling does it take for the yo-yo to come to a stop

its ment to be 'how long does it take for the yo-yo to come to a stop'

 

[kirafreedom]

i still don't get how to answer question c

 

[Doc Al]

i still don't get how to answer question c

First you must solve questions a and b correctly! (You don't really need b to answer c.)

For question c, you'll have the moment of inertia (from part a) and the forces slowing the yo-yo (applied by the finger) so you can calculate the new angular acceleration (alpha).

 

[kirafreedom]

Sois the formula alpha= moment of inertia times the force applies by the finger

 

[Doc Al]

Sois the formula alpha= moment of inertia times the force applies by the finger

No. The formula is just Newton's 2nd law applied to rotation, which I gave before:
Torque = I * alpha

 

[kirafreedom]

But the question says how long does it take for the yoyo to stop. The formula u provided is torque=I x alpha

 

[Doc Al]

But the question says how long does it take for the yoyo to stop. The formula u provided is torque=I x alpha

That's just the first step: Find alpha. Once you have alpha, use some kinematics to solve for the time.

 

[kirafreedom]

what is 'some kinematics'

 

[Doc Al]

what is 'some kinematics'

For example: Basic Equations of 1-D Kinematics

(You'll need to apply it to rotational quantities, of course.)

 

[kirafreedom]

so if i am correct this is how u do it

alpha=3

v=v0+at
0=15+3t
-5=t

 

[Doc Al]

so if i am correct this is how u do it

alpha=3

v=v0+at
0=15+3t
-5=t

If alpha = 3 rad/s^2, then that would be correct. (For one thing, it's slowing down so alpha would be negative.)

You first need to calculate alpha.