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Suppose you have an object rolling down the incline at 30 degrees.

enter image description here

Given the point of contact is instantaneously at rest, I decided to analyse torques at that point. Therefore, the only force creating a torque is the force of gravity, here being $$ F = m g \sin(\theta) \Rightarrow \text{Torque} = I(\alpha) = m g \sin(\theta) R. $$ Using both the right-hand rule and the fact that angular velocity is clockwise, torque should be negative. But the expression obtained suggests a positive torque. What am I missing here? Also, if the torque is negative, then the tangential acceleration should be negative too, but it makes sense to say that tangential acceleration is positive. Can someone please clarify these two points?

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You have not considered the direction of torque in you equation. Since the Torque is caused by Frictional Force $F_r$ which is in the direction $-i$. the torque is $\vec{\tau}= \vec{r}\times\vec{F_r}$ and $\vec{r} $ is in $j$ direction So the cross product yields

$\vec{\tau}= -|r||F_r|\hat{k}$

which ofcourse is in the negative direction of positive z-axis

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  • $\begingroup$ Of course. And that makes perfect sense if we are considering the friction force causing torque. But what if we are looking at the point of contact, at which friction exerts no torque, and only gravity causes a torque? $\endgroup$ – Ptheguy Oct 1 '15 at 16:37
  • $\begingroup$ But friction does exert a force, otherwise the ball would slide down the slope without rolling. Suppose the slope has friction until halfway down, then it has none. The ball would gain angular momentum (spin) as before until half way. While sliding the rest of the way, no torque, so the spin would remain constant even as the ball gained in speed. In fact it would gain speed (accelerate) even faster since all the loss of potential energy would go into linear acceleration and none into angular acceleration, during the sliding part of the trip. $\endgroup$ – Aabaakawad Oct 2 '15 at 17:44
  • $\begingroup$ In the rolling regime, the friction adds a torque such that there is no relative motion between ball surface and slope surface at point of contact. In the sliding regime, there will develop relative motion. $\endgroup$ – Aabaakawad Oct 2 '15 at 17:50
  • $\begingroup$ Sir, how does friction cause torque if I am analyzing the rotation with respect to the point of contact? It goes through the axis of rotation so it doesn't exert torque. $\endgroup$ – Ptheguy Oct 3 '15 at 3:13
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the method i use for determining the direction is quite simple. basically when you are to find the direction of a cross product say a x b just put your hand on a such that it is facing the b vector. thus for finding the direction of torque put your hand on the radius ( in the direction perpendicular to the force) such that it faces the force and curl your fingers the direction of curl will give you the direction of rotation.

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@vishwaas is correct, but here is an explanation without vector math. By convention, $x$-$y$-$z$ ordinates obey their own RHR such that rotating an imaginary right hand from positive $x$-axis to positive $y$-axis around the $z$-axis (counter-clockwise) puts positive $z$ in the towards-the-observer direction (out of the page). The rotation of, and the torque on, the rolling object is clockwise, putting the direction of the torque vector (by RHR) away from the observer (into the page). So the torque vector is negative in the $z$ direction.

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  • $\begingroup$ I know the torque is negative. But Torque=I(α)=mgsin(θ)R isn't negative. So what did I miss? $\endgroup$ – Ptheguy Oct 1 '15 at 19:19
  • $\begingroup$ @PedZad, Actually, to be precise, torque is a 3-D vector with x-value 0, y-value 0, and z-value -mg𝗌ɪɴ(θ)R Your formula gives it's magnitude (correctly). I should have said the torque is negative in the z direction I'll fix that in my answer. $\endgroup$ – Aabaakawad Oct 2 '15 at 1:26

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