Re: Why doesn't a rollerblader leaning into a tight turn fall over?

The net torque about any point in the rollerblader must be zero if the 
rollerblader is not to fall over.  If the torque is computed about his 
center of gravity, then the torque caused by the normal force of the 
ground up on his wheeels is balanced by the torque due to inward force of 
static friction of the ground on his wheels.  If these torques are 
computed about the contact point of the wheels on the ground, however, 
there seems to be a non-zero, net 
torque due to his weight since the normal and static friction forces act 
through the contact point and would not contribute to the torque about 
this point.  The moment of inertia of the wheels negligable.  How can 
there be a net torque about the contact point and none about the cg?


Hello Stephen,

I think that you were missing one of the forces in your analysis, as you 
will see.  It will be helpful if we first examine all of the forces 
associated with our skater, and then apply them to distances to calculate 
torques:

At the CG, gravity produces a force equal to his weight in the downward 
direction. The other force at the CG is effectively acting to ‘pull’ the 
skater in a horizontal direction, towards the outside of the curved turn 
that he is in.  Lets call the downward force “F1” and the horizontal 
force “F2”.

Now to find ‘torques’, or moments associated with the contact point of 
the skates we need to define moment arms or ‘vector’ distances through 
which the forces are acting:  Lets say that the CG of our skater when 
standing straight up is 4 feet above the ground  (wearing the skates, of 
course).  

If he leans over 20 degrees in the turn, his CG will move closer to the 
floor (equal to 4’ x cos 20o = 3.76’).  This value (3.76’) is the moment 
arm through which F2 acts.  At this 20o angle, his CG will also move to 
the center of the curve (away from his feet), a distance equal to 4’ x 
sin 20o =  1.37’.

Now we can define the torques at the wheels:

The “Tipping” torque equals:   F1 x 1.37’
The “Righting” torque equals  F2 x 3.76’

If the skater successfully makes it through the turn, the “Tipping” 
and “Righting” torques will exactly equal each other.

Remember that there is no torque produced at the skate/floor contact 
point**.  If we drew a force diagram, we would show the skate/floor 
contact as a "pivot point".  The only effective torques at this pivot 
point would be the Tipping and Righting torques above.

As a practical side note- A skater will automatically assure that 
his ‘Tipping’ and ‘Righting’ torques are equal when making a successful 
turn.  He will do this unconscientiously by leaning in to produce just 
enough Tipping force to match the Righting force.



** For the purpose of explaining your scenario.  It might be clearer to 
think of an ice skater on one leg for this example.  This would eliminate 
the “anti-tipping” effect of having two skates on the ground with 4 
wheels across.

I hope that this helps you.

Best Regards,