Ball being hit starts to slide and roll

[ehild]

You might try using conservation of momentum about the point where slipping stops.
That would quickly give you the final angular velocity.
Knowing that it seems that the problem would be greatly simplified (energy loss, etc.)

Yes, it is a much simpler method to get the final speed. You can show it, as the problem is already solved (Post #40)

 

[Karol]

$$dW=(Fv+T\omega)dt,~~v=at=g\mu t,~~\omega=\alpha t=\frac{g\mu}{kr}t$$
$$Work=\int dW=...=\frac{mv_0^2k}{2(k+1)}$$
But that's the KE loss i found in post #42, so, no heat generated?

You might try using conservation of momentum about the point where slipping stops.
That would quickly give you the final angular velocity.

Which momentum, angular? it is mvr. now, what does it help me? which condition to apply at that point and what initial information exists? if i don't know the velocity, since i guess i have to find it with the conservation of momentum, then what else do i know?

 

[ehild]

$$dW=(Fv+T\omega)dt,~~v=at=g\mu t,~~\omega=\alpha t=\frac{g\mu}{kr}t$$
$$Work=\int dW=...=\frac{mv_0^2k}{2(k+1)}$$
But that's the KE loss i found in post #42, so, no heat generated?

Your formula for v is wrong. v= v₀-μgt. You miss a negative sign in the formula for W. The work is negative, and equal to the change of KE.
Energy is not lost. The mechanical energy lost is converted to other kind of energy, heat now.

 

[J Hann]

$$dW=(Fv+T\omega)dt,~~v=at=g\mu t,~~\omega=\alpha t=\frac{g\mu}{kr}t$$
$$Work=\int dW=...=\frac{mv_0^2k}{2(k+1)}$$
But that's the KE loss i found in post #42, so, no heat generated?

Which momentum, angular? it is mvr. now, what does it help me? which condition to apply at that point and what initial information exists? if i don't know the velocity, since i guess i have to find it with the conservation of momentum, then what else do i know?

M V R is the initial angular momentum.
Now what is the angular momentum about the point of contact when there is no more slipping?
(Hint: Use the parallel axis theorem)

 

[Karol]

Your formula for v is wrong. v= v0-μgt

$$dW=(Fv+T\omega)dt,~~v=v_0-at=v_0-g\mu t,~~\omega=\alpha t=\frac{g\mu}{kr}t$$
$$Work=\int dW=...=\frac{mv_0^2k(k+3)}{2(k+1)}>\frac{mv_0^2k}{2(k+1)}=\Delta EK$$
This time it makes sense to me since the friction work reduces total KE and also produces heat.

 

[ehild]

$$dW=(Fv+T\omega)dt,~~v=v_0-at=v_0-g\mu t,~~\omega=\alpha t=\frac{g\mu}{kr}t$$
$$Work=\int dW=...=\frac{mv_0^2k(k+3)}{2(k+1)}>\frac{mv_0^2k}{2(k+1)}=\Delta EK$$
This time it makes sense to me since the friction work reduces total KE and also produces heat.

Wrong. What did you take for F?
If the work is positive it increases the kinetic energy. But it is decreased as part of it was converted to heat.

 

[Karol]

$$dW=(Fv+T\omega)dt,~~v=v_0-at=v_0-g\mu t,~~\omega=\alpha t=\frac{g\mu}{kr}t$$
$$F=-mg\mu,~~t=\frac{v_0k}{(k+1)g\mu}$$
$$Work=\int dW=...=-\frac{mv_0^2k}{2(k+1)}=-\Delta EK$$
The work is entirely converted to decrease in KE? and which part is the heat?
I expected $Work>\vert \Delta KE \vert$

 

[ehild]

Work is not energy. The work done by all forces acting on an object is equal to the change of the KE of the object.
Some of the work is that of conservative forces. That work changes the potential energy. Some of the work is due to non-conservative forces.
If the potential energy does not change, but the KE decreases, the lost KE is transformed to internal energy, the KE of the random motion of particles.

 

[Karol]

I thank you very much ehild for your patience, Karol

 

[rcgldr]

Link to one of the prior threads for this problem:

https://www.physicsforums.com/threads/rolling-sliding-with-friction.159337