Rotational Plus Translational Motion for Sphere of Yarn

[sweetpete28]

A massless string is wrapped around the equator of a solid sphere (mass M = 76.2 kg, radius R = 0.211 m). A girl holds the free end of the string, and the sphere is released from rest, Assume:
- the sphere is always parallel to the floor
- the string is always perpendicular to the radius of the sphere
- the string does not slip over the sphere

What is the length of the string that has been unwound when the sphere reaches an angular speed ω = 28.6 rad/s?

Can someone please help? Here is what I did but answer is wrong:

v = ωr = 6.0346 m/s

vf^2 = v0^2 + 2as
s = 1.86 m

1.86 = 1/2at^2
t= .615s

.615s * 28.6 = 17.59

17.59 * .211 = 3.7...but this wrong...please help?

 

[tiny-tim]

hi sweetpete28!

(try using the X² and X₂ buttons just above the Reply box )

v = ωr = 6.0346 m/s

vf^2 = v0^2 + 2as
s = 1.86 m

you're assuming that a = g

what about the tension? ​

 

[sweetpete28]

Ohhh

You're right! Tension would = mg since it does not fall...right? So there is 0 acceleration.

I know θ(t) multiplied by radius r will give me length string has been unwound...but how do I get what t equals? I'm still stuck...

 

[sweetpete28]

If mg is tension = 748 N and work done on sphere is 555J once it reaches 28.6 rad/s angular velocity...can I divide 555J by 748N to get length unwound since F times d = Work??

Which would give answer of 555/748 = .742 m?

 

[tiny-tim]

hi sweetpete28!

Ohhh

You're right! Tension would = mg since it does not fall...right? So there is 0 acceleration.

nooo, of course it accelerates, but at less than g

either call the tension T, and consider forces and torques,

or (probably easier) use conservation of energy ​

 

[sweetpete28]

If mg is tension = 748 N and work done on sphere is 555J once it reaches 28.6 rad/s angular velocity...can I divide 555J by 748N to get length unwound since F times d = Work?? (I used KE = 1/2MIω^2 to get work done on sphere = 555J).

Which would give answer of 555/748 = .742 m?