Physics Solution: Helical Friction on a Rod
(a) Working Principle
Kinetic friction always acts opposite to the direction of the resultant velocity of the surface relative to the contact.
When we simply push the rod, the velocity is purely axial, so the full force of friction opposes the push.
However, when we simultaneously rotate the rod, the resultant velocity becomes helical (spiral). The friction vector aligns with this helical path.
This means the friction force vector has two components: one axial and one tangential. The axial component of friction is significantly reduced because a portion of the total friction is now “used up” in the tangential direction. This makes it easier to insert or withdraw the rod.
(b) Calculating the Applied Torque
Let the axial velocity be $v$ and the angular velocity be $\omega$. The radius of the rod is $r$.
- Axial velocity component: $v_{axial} = v$
- Tangential velocity component: $v_{tan} = \omega r$
The friction force $f_{net}$ acts opposite to the resultant velocity vector. Let $\phi$ be the angle the resultant velocity makes with the axis of the rod.
$$ \tan \phi = \frac{v_{tan}}{v_{axial}} = \frac{\omega r}{v} $$
Since the rod moves with constant axial velocity under force $F$, the axial component of friction must balance $F$:
$$ f_{axial} = F $$
From vector resolution: $f_{axial} = f_{net} \cos \phi$
$$ \implies f_{net} = \frac{F}{\cos \phi} $$
The required Torque $\tau$ must balance the torque created by the tangential component of friction:
$$ \tau = f_{tan} \cdot r $$
$$ f_{tan} = f_{net} \sin \phi $$
Substituting $f_{net}$:
$$ \tau = \left( \frac{F}{\cos \phi} \sin \phi \right) r = F r \tan \phi $$
Finally, substitute $\tan \phi = \frac{\omega r}{v}$:
$$ \tau = F r \left( \frac{\omega r}{v} \right) $$
$$ \tau = \frac{F \omega r^2}{v} $$
(a) Working Principle
Kinetic friction always acts opposite to the direction of the resultant velocity of the surface relative to the contact. When we simply push the rod, the velocity is purely axial, so the full force of friction opposes the push. However, when we simultaneously rotate the rod, the resultant velocity becomes helical (spiral). The friction vector aligns with this helical path. This means the friction force vector has two components: one axial and one tangential. The axial component of friction is significantly reduced because a portion of the total friction is now “used up” in the tangential direction. This makes it easier to insert or withdraw the rod.
(b) Calculating the Applied Torque
Let the axial velocity be $v$ and the angular velocity be $\omega$. The radius of the rod is $r$.
- Axial velocity component: $v_{axial} = v$
- Tangential velocity component: $v_{tan} = \omega r$
The friction force $f_{net}$ acts opposite to the resultant velocity vector. Let $\phi$ be the angle the resultant velocity makes with the axis of the rod.
Since the rod moves with constant axial velocity under force $F$, the axial component of friction must balance $F$:
The required Torque $\tau$ must balance the torque created by the tangential component of friction:
Finally, substitute $\tan \phi = \frac{\omega r}{v}$: