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In the special case where a bowling ball has initial translational velocity but no initial angular velocity, the bowling ball will experience a contact force due to Coulomb friction $\mu mg$.

In the special case where the bowling ball has initial translational velocity $v$ and initial angular velocity $\omega$ such that $v$ = $r$ x $\omega$, where $r$ is the radius of the ball, the contact force is $0$ and the ball is rolling without slipping.

In the general case where the bowling ball has initial translational velocity $v$ and initial angular velocity $\omega$; such that $\omega \ne 0$ and such that $v \ne r$ x $\omega$, is the contact force still $\mu mg$? Or will it depends on the relative speed between the ball and the surface; eg $v$ - ($r$ x $\omega)$ ?

Reference: https://bowlingknowledge.info/images/stories/what_makes_a_bowling_ball_hook.pdf

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  • $\begingroup$ Friction opposes relative motion it is of two types static and kinetic here since there is relative slipping when the case when v=/ rw the friction is normal force times coefficient of kinetic friction $\endgroup$ – Prateek Mourya Nov 21 at 5:07
  • $\begingroup$ in theory is still the same, in practice I don't know $\endgroup$ – Wolphram jonny Nov 21 at 6:05
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In the special case where the bowling ball has initial translational velocity $v$ and initial angular velocity ω such that $v = r ω$, where $r$ is the radius of the ball, the contact force is 0 and the ball is rolling without slipping.

No. It's not really a 'special case'. The friction force is always:

$$F_f=F_N \mu_k$$

where:

$$F_N=mg$$

Its direction is always opposite to the relative direction of motion of the two surfaces.

No matter what the initial angular velocity $\omega$ is, the restoring torque $F_f r=\mu_k mg r$ will ensure over time that $v=\omega r$. The time needed will depend on inertial moment $I$and friction coefficient $\mu$:

$$\mu_k mgr=\alpha I$$

where

$$\alpha= \frac{\text{d}\omega}{\text{d}t}$$

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enter image description here

If you write the equations of motion for this free body diagram you obtain:

$$m\,\ddot{x}=F-F_c$$ $$I\,\ddot{\varphi}=F_c\,r$$

you have two situation

I) rolling condition

this mean that $$x=\,r\,\varphi~,\Rightarrow~\ddot{x}=r\,\ddot{\varphi}$$

thus you obtain for the contact force $F_c$

$$F_c=\frac{F\,I}{I+m\,r^2}$$

II) sliding condition

in this case the contact force is a function of : $$F_c=F_c(s~,\mu~,N)$$

where

  • $s=\dot{x}-r\,\dot{\varphi}$ the sliding velocity
  • $\mu$ the friction coefficient
  • $N=m\,g$ the normal force

enter image description here

with $$s\mapsto \frac{\dot{x}-\omega\,r}{\omega\,r}$$ $$F_{c,\text{max}}=\mu_{\text{max}}\,N$$ $$F_{c,g}=\mu_{g}\,N$$

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