Finding the speed of rolling ball

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

In rolling motion, friction doesn't work. So, applying work energy theorem,
$\frac 1 2 m v_0 ^2 =\frac 1 2 mv^2 + \frac 1 2 I_{center} \omega ^2$
$taking~ I_{ center} = \frac 2 5 m R^2 ~and ~ v =\omega R , v = \sqrt {( \frac 5 7 v_0 )}$

 

[haruspex]

Homework Statement

View attachment 209073

Homework Equations

The Attempt at a Solution

In rolling motion, friction doesn't work. So, applying work energy theorem,
$\frac 1 2 m v_0 ^2 =\frac 1 2 mv^2 + \frac 1 2 I_{center} \omega ^2$
$taking~ I_{ center} = \frac 2 5 m R^2 ~and ~ v =\omega R , v = \sqrt {( \frac 5 7 v_0 )}$

I assume you meant the v₀ to be outside the square root.
You are overlooking the energy lost to friction while sliding. We want the velocity just as it transitions to rolling. Up until then it has been sliding, as well as rotating.

 

[efdee]

Thanks to friction the ball can move.
Heat dissipation will be absent.
Remember v = omega R(ball) and eliminate omega.

 

[Pushoam]

I assume you meant the v0 to be outside the square root.

Yes.

You are overlooking the energy lost to friction while sliding.

Yes, I considered this motion as rolling motion from start unknowingly.
Friction doesn't work in rolling motion. But, it does work in sliding motion.

Another attempt:
$v = v_0 - f_r t /m = \omega R$
$\tau = f_r R = \frac 2 5 m R^2 \alpha$
$\omega = \alpha t = \frac {5 f_r } {2 m R} t \\ \frac { f_r t } m = \frac 2 5 v \\ v = \frac 5 7 v_0$

So, done. Thanks.

 

[haruspex]

Yes.

Yes, I considered this motion as rolling motion from start unknowingly.
Friction doesn't work in rolling motion. But, it does work in sliding motion.

Another attempt:
$v = v_0 - f_r t /m = \omega R$
$\tau = f_r R = \frac 2 5 m R^2 \alpha$
$\omega = \alpha t = \frac {5 f_r } {2 m R} t \\ \frac { f_r t } m = \frac 2 5 v \\ v = \frac 5 7 v_0$

So, done. Thanks.

There is an interesting quick method on this one.
Since the friction acts at ground level, angular momentum about an axis at ground level is conserved.

 

[Pushoam]

There is an interesting quick method on this one.
Since the friction acts at ground level, angular momentum about an axis at ground level is conserved.

Thanks for this insight. This is really interesting.