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.