If you had two bodies with the same weight but one having mass concentrated more in the center, while the other had most mass concentrated on the outside, but both had the same center of mass and gravity, and an equal force was applied on the same spot on both, would the torque induced be different?
[Physics] Does mass concentration affect the torque induced by a force
massnewtonian-mechanicstorque
Related Solutions
Yes, it will affect angular velocity since different mass distribution have different moment of inertia $I$ in general. The effect of torque $\tau$ on the angular velocity $\omega$ of the object is given by $$\tau=\frac{d}{dt}(I\omega)$$
The moment of inertia of a point mass is given by $I=mr^2$, so in your case, the radius differ by 10 time so moment of inertia differ by 100 times and so does the angular velocity. Note that when you mention torque, you dont necessary need to specify which point it acts on.
In these cases it always helps to draw a diagram:
The green vectors represent the force of gravity $w=mg$ (dashed) and its components along the inclined plane and perpendicular to it. The red forces are the normal force of the plane on the ball $n$, the force of friction $F$, and their vector sum (dashed).
Now the sphere rotates about the contact point - that is the point that doesn't move. In that frame of reference, noting that the red vectors all pass through the center of rotation we compute the torque as the force of gravity $w$ times the perpendicular distance to the pivot point $d= r\sin\theta$, i.e. $$\Gamma = w\cdot r \sin\theta$$ and we consider the moment of inertia of the ball about this pivot to be $$I = \frac25 mr^2 + mr^2=\frac75 mr^2$$ (by the parallel axes theorem).
As you pointed out, by considering the motion about the contact point, the value of $F$ doesn't seem to come into play. But remember that the center of mass of the sphere must accelerate as though all forces are acting on it; after canceling out the normal forces, that leaves us with $mg\sin\theta$ down the slope, and $F$ going the other way. The difference between these two forces gives rise to the acceleration of the sphere's c.o.m. so we can compute $F$ from
$$mg \sin\theta - F = m a$$
To compute $a$, we first need the angular acceleration $\dot\omega$which is found from
$$\dot \omega = \frac{\Gamma}{I} = \frac{mgr\sin\theta}{\frac75 m r^2} = \frac{5g\sin\theta}{7r}$$
The linear acceleration $a$ is of course the angular acceleration multiplied by the radius of the sphere, so
$$a = \frac57 g\sin\theta$$
From which it follows that
$$F = \frac{2}{7} m g \sin \theta$$
And if we know that, we can now compute the angular acceleration of the sphere about its center. The torque seen in the frame of reference of the sphere is
$$\Gamma' = Fr = \frac{2}{7} m g r \sin\theta$$
Now we use the moment of inertia of the sphere about its center in order to compute the angular acceleration, and find
$$\dot \omega = \frac{\Gamma'}{\frac25 mr^2} \\ = \frac{\frac{2}{7} m g r \sin\theta}{\frac{2}{5} m r^2}\\ =\frac{5 g \sin\theta}{7 r}$$
which is the same result as before.
So there is no contradiction. The forces of friction and gravity work together to cause the rotation - the difference in apparent torque comes about from the fact that you are working in different (and non-inertial) frames of reference, but if you do the calculation carefully you get the same answer.
Best Answer
The definition of torque is force times lever arm. As you state it, force and lever arm are identical, so is the torque.
There are differences. Because the moment of inertia differs, the applied torque will cause a different angular velocity (if the force is not in line with the centre of mass).