Does the frictional force change if a bowling ball is slipping depending on the relative speed difference?

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 $$\omega)$$ ?

• 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 – Prateek Mourya Nov 21 at 5:07
• in theory is still the same, in practice I don't know – Wolphram jonny Nov 21 at 6:05

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

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

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

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