# Circular motion

Could someone explain why this is wrong
(edited 1 year ago)
Original post by Student 999
Could someone explain why this is wrong

If I understand this correctly then
T = mg - mv^2/r
so as the speed increases, the tension decreases?

The centriptal force must be given by T-mg to ensure circular motion so
mv^2/r = T - mg
So
T = mg + mv^2/r
(edited 1 year ago)
Original post by mqb2766
If I understand this correctly then
T = mg - mv^2/r
so as the velocity increases, the tension decreases?

The centriptal force must be given by T-mg to ensure circular motion so
mv^2/r = T - mg
So
T = mg + mv^2/r

Is the idea where the centripetal force is the resultant force caused by how weight and tension interacts?
Original post by Student 999
Is the idea where the centripetal force is the resultant force caused by how weight and tension interacts?

For circular motion, a (centripetal) force of mv^2/r must act on the body towards the center of the circle (normal to the motion). So that must be given by T-mg, when the object is at the bottom of the circle.
(edited 1 year ago)
Original post by mqb2766
For circular motion, a (centripetal) force of mv^2/r must act on the body towards the center of the circle (normal to the motion). So that must be given by T-mg.

From how I understand it T-mg must = centripetal force to display circular motion since the overall resultant must be centripetal acting towards the centre. Is that correct?
Original post by Student 999
From how I understand it T-mg must = centripetal force to display circular motion since the overall resultant must be centripetal acting towards the centre. Is that correct?

Yes, when the object is at the bottom of the circle. It will exhibit circular motion if a force of mv^2/r always acts perpendicular to the motion, towards the center. At the top of the circle,
T+mg = mv^2/r
and when the "string" is horizontal
T = mv^2/r
Original post by mqb2766
Yes, when the object is at the bottom of the circle. It will exhibit circular motion if a force of mv^2/r always acts perpendicular to the motion, towards the center. At the top of the circle,
T+mg = mv^2/r
and when the "string" is horizontal
T = mv^2/r

In my first sketch am I confusing myself with a spring system? Whereas the second sketch is the correct idea
Original post by Student 999
In my first sketch am I confusing myself with a spring system? Whereas the second sketch is the correct idea

It would help to see the original question / scenario youre considering. Im presuming that its an inextensible, light string with an object of mass m at the end and its moving in a vertical circle?

For it to be moving in a vertical circle, there must be an inwards (towards the center / radial) centripetal force mv^2/r. The varying tension in the string will provide this and the three scenarios mentioned before give it at 4 points
vertical down T = mg + mv^2/r
vertical up T = -mg + mv^2/r
horizontal T = mv^2/r
Obviously, you can resolve the weight to get the more general expression
T = mg cos(theta) + mv^2/r
where theta is the angle with the downwards vertical.

Not fully watched it, but looks about right with the normal playing the same role as tension
Much appreciated

Original post by mqb2766
It would help to see the original question / scenario youre considering. Im presuming that its an inextensible, light string with an object of mass m at the end and its moving in a vertical circle?

For it to be moving in a vertical circle, there must be an inwards (towards the center / radial) centripetal force mv^2/r. The varying tension in the string will provide this and the three scenarios mentioned before give it at 4 points
vertical down T = mg + mv^2/r
vertical up T = -mg + mv^2/r
horizontal T = mv^2/r
Obviously, you can resolve the weight to get the more general expression
T = mg cos(theta) + mv^2/r
where theta is the angle with the downwards vertical.

Not fully watched it, but looks about right with the normal playing the same role as tension