- #1
How do you get that?Suyash Singh said:a=v^2/r
its the equation for centripetal accelerationharuspex said:How do you get that?
Centripetal force = mv^2/r for cylinderBvU said:More equations needed !
Tension = m(g-a)+mrw^2BvU said:And what role does that play here ?
What about the relation between ##\omega## and ##T## ?
What is ##I## ?
The only centripetal forces are internal to the cylinder. They have no consequence for the rate at which the cylinder descends.Suyash Singh said:Centripetal force = mv^2/r for cylinder
Please do not post answers at such an early stage. This is a homework forum. The idea is to provide hints, correct misunderstandings and point out errors.Dr Dr news said:I think T = W and a = (2/3) g
The tension would be equal to the weight only if the acceleration is zero. See OP's equation in #1, mg - ma = T.Dr Dr news said:I think T = W ...
Not quite. You are using T as a force, but on the right you have Iα, which is not a force. What have you forgotten?Suyash Singh said:T= 1/(2) (mr^2) a/r
Torque=I alphaharuspex said:Not quite. You are using T as a force, but on the right you have Iα, which is not a force. What have you forgotten?
Right, so what is the torque in this case?Suyash Singh said:Torque=I alpha
torque is the tangential forceharuspex said:Right, so what is the torque in this case?
No, a torque is not a force. How do you find the torque of a force about an axis?Suyash Singh said:torque is the tangential force
torque=R F sin(theta)haruspex said:No, a torque is not a force. How do you find the torque of a force about an axis?
Right, so correct your equation in post #15 and go from there.Suyash Singh said:torque=R F sin(theta)
where theta is the between R and F
T= mr alphaharuspex said:Right, so correct your equation in post #15 and go from there.
No, you hadSuyash Singh said:T= mr alpha
but the T is a force not a torque. Just replace the T with the torque that T exerts about the centre of the cylinder.Suyash Singh said:T= 1/(2) (mr^2) a/r
haruspex said:No, you had
but the T is a force not a torque. Just replace the T with the torque that T exerts about the centre of the cylinder.
What is the torque that the tension T exerts on the cylinder about the cylinder's axis?Suyash Singh said:what do you mean?
Torque=I alphaharuspex said:What is the torque that the tension T exerts on the cylinder about the cylinder's axis?
No, I mean using the equation you wrote in post #21 what is T's torque about the axis of the cylinder?Suyash Singh said:Torque=I alpha
I see no evidence that Suyash is doing that.Dr Dr news said:If you are taking moments about the string-cylinder contact point.
Torque=F R sin(theta)haruspex said:No, I mean using the equation you wrote in post #21 what is T's torque about the axis of the cylinder?
Yes, but that's a generic equation. You need to write it in terms of the variables in this question: T, r etc. What are F, R and θ here?Suyash Singh said:Torque=F R sin(theta)
F= tensionharuspex said:Yes, but that's a generic equation. You need to write it in terms of the variables in this question: T, r etc. What are F, R and θ here?
Suyash Singh said:F= tension
R= radius of cylinder
theta = 90 degrees
Torque=T R
haruspex said:Right, so correct your first equation in post #15 and go from there.
When a solid cylinder is dropped with a string attached to it, it experiences a combination of rotational and translational motion. The string exerts a tension force on the cylinder, causing it to rotate and accelerate downwards. This motion is governed by Newton's laws of motion and the principles of torque and angular momentum.
The speed at which the cylinder falls is affected by several factors, including the length and mass of the string, the mass and shape of the cylinder, and the force of gravity. Additionally, air resistance and friction can also play a role in the speed of the falling cylinder.
The length of the string affects the motion of the falling cylinder by changing the amount of tension and torque exerted on the cylinder. A longer string will result in a larger radius of rotation, causing the cylinder to fall slower. On the other hand, a shorter string will result in a smaller radius of rotation and a faster descent.
If the string is cut, the cylinder will continue to fall with only translational motion. This is because the tension force provided by the string is no longer present to cause rotation. However, if the string is cut at an angle, the cylinder may experience some rotational motion due to the force of gravity acting on the off-center center of mass.
Yes, the motion of a falling solid cylinder with string can be predicted using equations derived from Newton's laws of motion and the principles of torque and angular momentum. These equations can be used to calculate the acceleration, velocity, and position of the cylinder at any given time during its fall.