Calculating Time for Ball to Start Rolling on Horizontal Surface

AI Thread Summary
The discussion focuses on determining the time it takes for a ball, initially sliding at 6.5 m/s on a horizontal surface with a kinetic friction coefficient of 0.3, to start rolling without slipping. Participants emphasize the need to establish equations for both translational and rotational motion, noting that the translational speed decreases while the rotational speed increases. The correct condition for rolling without slipping is when the translational speed equals the angular velocity times the radius of the ball. There are suggestions to isolate time by using the no-slip condition after deriving the equations for velocity and angular velocity. The conversation highlights the importance of accurately applying the concepts of kinetic friction and rotational inertia in solving the problem.
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
\mu = 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 \mu
ma = mg\mu
a = g\mu

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" \omega r (what is correct name in English?)

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

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

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 \mu
ma = mg\mu
a = g\mu
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" \omega r (what is correct name in English?)
That's the condition for "rolling without slipping".

v(t) = v0 - at = \omega r
\omega = \frac{v0 - g*mu*t}{r}
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 v = \omega r.
 
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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