I have a rectangular shape with uniform mass, approx 6m x 1m. It rotates around an axis in the centre of its width but 1m from one end. The axis has some friction but this is not known.

What I need to work out is how much force is required to start rotating this object from different positions from the axis.

I fear I may be plunged out of my depth but if anyone could point me in the right direction it would be really appreciated!

  • 4
    $\begingroup$ You can not. If you have no idea of the stiction force of the center then one can not calculate this. You see if the opposing forces are 0 then no matter what the objects mass it will accelerate even with very small forces. The only thing innertia tells us is how fast it accelerates. $\endgroup$
    – joojaa
    Feb 17, 2017 at 11:06

2 Answers 2


If we neglect friction and the object is rotating about its centre of mass then it will experience an angular acceleration proportional to the torque applied and its moment of inertia.

Note that this is a direct analogue of the linear acceleration equation F = ma

But instead of force we use torque (force x radius) and instead of mass we use moment of inertia.

So as with linear acceleration, if we ingnore friction any force greater than zero will produce some acceleration.

The situation is slightly different if the axis of rotation is offset from the centre of mass. In this case the weight of the body will exert a torque on it ie it will behave as a pendulum.

So for your example with a 1m x 6m plank rotating around a point 1m from one end the centre of mass will lie 3m from each end so it can be simplified to a point mass connected to a pivot by a mass-less rigid bar 2m long.

There will be a constant force acting on the point mass of Mg acting vertically downwards. This will create a torque when it is displaced at some angle x from the horizontal so, again any torque greater than zero will produce some displacement but in order to get it horizontal you need to support the its full wieght so :

the minimum constant torque to rotate it in a full circle is given by :

T > M.g.r

Where M is the mass of the object, g is acceleration due to gravity and r is the distance from centre of mass to the pivot point.

Strictly speaking this is the torque required to hold it stationary in the horizontal position.

To work out the holding torque for other positions you need to use trigonometry to work out the resultant toque from the vertical force due gravity.


Since the friction is not known, the torque to overcome that friction can't be known either. You therefore can't do what you want since the problem is under-constrained.


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