A body only rotates on an inclined plane if it has friction?

AI Thread Summary
A body on an inclined plane requires friction to rotate, as torque is generated by weight only when considering the point of contact as the axis of rotation. Without friction, no force produces torque when the mass center is used as the axis, meaning there is no rotational acceleration. The discussion highlights that linear acceleration can create angular acceleration about a point not aligned with the motion, complicating the relationship between linear and rotational dynamics. The shape of the body also influences the torque produced by normal forces. Ultimately, the presence of friction is crucial for ensuring that the body can rotate effectively on the inclined plane.
LCSphysicist
Messages
644
Reaction score
162
Homework Statement
.
Relevant Equations
.
I think that the answer probably will be yes, but i have a question:
Well, with respect to center of mass that's right, but if i choose the axis passing through the contact point, the weight force will produce torque, even without the friction.
 
Physics news on Phys.org
If you "move the axis of rotation to the contact point", it is constrained by the plane so that it cannot rotate. As such, can it really be declared an axis of rotation?
 
LCSphysicist said:
Homework Statement:: .
Relevant Equations:: .

I think that the answer probably will be yes, but i have a question:
Well, with respect to center of mass that's right, but if i choose the axis passing through the contact point, the weight force will produce torque, even without the friction.
Torque about what point, and what shape of body?

If you take the mass centre as the axis, the weight does not produce torque. For a cylinder, the normal force won't either. With no friction, there is no force producing torque.

If you take the point of contact as axis the weight does have torque, so there will be angular acceleration about that point. But linear acceleration of a body constitutes angular acceleration about a point not in its line of motion, so this still does not ensure rotational acceleration.

If the body is not a sphere or cylinder nor lying on a flat face then the normal force might have moment about the mass centre.
 
haruspex said:
Torque about what point, and what shape of body?

If you take the mass centre as the axis, the weight does not produce torque. For a cylinder, the normal force won't either. With no friction, there is no force producing torque.

If you take the point of contact as axis the weight does have torque, so there will be angular acceleration about that point. But linear acceleration of a body constitutes angular acceleration about a point not in its line of motion, so this still does not ensure rotational acceleration.

If the body is not a sphere or cylinder nor lying on a flat face then the normal force might have moment about the mass centre.
" point. But linear acceleration of a body constitutes angular acceleration about a point not in its line of motion, so this still does not ensure rotational acceleration "
I can't get this part, can you say more about this?
 
LCSphysicist said:
" point. But linear acceleration of a body constitutes angular acceleration about a point not in its line of motion, so this still does not ensure rotational acceleration "
I can't get this part, can you say more about this?
Consider a particle moving horizontally with velocity v, height h, mass m. Wrt a point on the ground it has angular velocity vh and angular momentum mvh. Similarly angular acceleration and linear acceleration.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top