Calculating Time for Ball to Start Rolling on Horizontal Surface

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SUMMARY

The discussion focuses on calculating the time it takes for a ball to transition from sliding to rolling on a horizontal surface, given an initial speed of 6.5 m/s and a kinetic friction coefficient of 0.3. The key equations involve translational acceleration due to friction and the relationship between linear velocity and angular velocity. The correct approach requires setting the translational speed equal to the rolling condition, represented by the equation v = ωr, where ω is the angular velocity. The solution involves deriving expressions for both translational and rotational speeds as functions of time and applying the no-slip condition to isolate the time variable.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with concepts of kinetic friction and its coefficient
  • Knowledge of rotational dynamics, specifically the moment of inertia
  • Ability to manipulate equations involving linear and angular velocities
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  • Study the relationship between linear acceleration and angular acceleration in rolling motion
  • Learn about the no-slip condition and its application in rotational dynamics
  • Explore the derivation of equations for translational and rotational motion
  • Investigate the effects of varying friction coefficients on rolling motion
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domagoj412
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Homework Statement


Ball starts to slide with initial speed v0 = 6.5 m/s on horizontal surface. After what time will ball start to roll? Kinetic friction between the ball and the surface is 0.3.
v0 = 6.5 m/s
[tex]\mu[/tex] = 0.3
I = 1/2 mr^2

Homework Equations


I have solved this problem but I'm not sure if is correct, so please check...


The Attempt at a Solution



Fk = N [tex]\mu[/tex]
ma = mg[tex]\mu[/tex]
a = g[tex]\mu[/tex]

This is the acceleration which is opposite to direction of the ball so it is slowing it down.
So velocity when it starts to roll must be equal to the "rolling velocity" [tex]\omega[/tex] r (what is correct name in English?)

v(t) = v0 - at = [tex]\omega[/tex] r
[tex]\omega[/tex] = [tex]\frac{v0 - g*mu*t}{r}[/tex]

Ek = Fk*s + (1/2) I[tex]\omega[/tex]^2
When I loose the masses and substitute [tex]\omega[/tex]:

1/2 v0^2 = g*mu*(v/t) + (1/5) * (v0 - g*mu*t)

And that is one equation with one unknown (t).
But equation is pretty big and I suppose I got something wrong...
 
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domagoj412 said:
I = 1/2 mr^2
Careful: What's the rotational inertia of a ball?

The Attempt at a Solution



Fk = N [tex]\mu[/tex]
ma = mg[tex]\mu[/tex]
a = g[tex]\mu[/tex]
This is good for the translational acceleration. What about rotation?

This is the acceleration which is opposite to direction of the ball so it is slowing it down.
So velocity when it starts to roll must be equal to the "rolling velocity" [tex]\omega[/tex] r (what is correct name in English?)
That's the condition for "rolling without slipping".

v(t) = v0 - at = [tex]\omega[/tex] r
[tex]\omega[/tex] = [tex]\frac{v0 - g*mu*t}{r}[/tex]
Not exactly. The translational speed decreases while the rotational speed increases. Write expressions for both speeds as a function of time and solve for when they meet the condition [itex]v = \omega r[/itex].
 
There's several things wrong here. You solved for the correct equation for velocity as a function of time, but you don't set that equal to omega*r. You need to find the equation for omega as a function of time.

Once you have velocity and omega as functions of time, you can use the no slip condition to isolate t. (the rolling velocity is called the angular velocity in english, represented by omega)

Edit: Totally stole my thunder, doc.
 

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