Falling solid cylinder with string

[Suyash Singh]

Homework Statement

Homework Equations

m:mass of solid cylinder
T: tension in string
w:angular velocity

The Attempt at a Solution

m(g-a)=T
mg-ma=T

a=v^2/r=w^2r

now what?

 

[BvU]

More equations needed !

 

[haruspex]

a=v^2/r

How do you get that?

 

[Suyash Singh]

How do you get that?

its the equation for centripetal acceleration

 

[Suyash Singh]

More equations needed !

Centripetal force = mv^2/r for cylinder

 

[BvU]

And what role does that play here ?
What about the relation between $\omega$ and $T$ ?
What is $I$ ?

 

[BvU]

Make a sketch to see where forces act (free body diagram)

 

[Suyash Singh]

And what role does that play here ?
What about the relation between $\omega$ and $T$ ?
What is $I$ ?

Tension = m(g-a)+mrw^2

 

[haruspex]

Centripetal force = mv^2/r for cylinder

The only centripetal forces are internal to the cylinder. They have no consequence for the rate at which the cylinder descends.
You need to think about angular acceleration and its relationships to torque and to linear acceleration.

 

[Dr Dr news]

I think T = W and a = (2/3) g

 

[haruspex]

I think T = W and a = (2/3) g

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.
Did you read the guidelines?

 

[Dr Dr news]

My bad. You might start by considering the motion to be about the contact between the cylinder and the string like considering the motion of a tire down the road by analyzing the motion of the tire about the contact with the road.

 

[kuruman]

I think T = W ...

The tension would be equal to the weight only if the acceleration is zero. See OP's equation in #1, mg - ma = T.

 

[Suyash Singh]

Tension=I alpha
where I is moment of inertia
and alpha is angular acceleration
Tension=1/(2) (MR^2) alpha
but what is alpha?

 

[Suyash Singh]

oh ok so i found this formula
a=r alpha

So T= 1/(2) (mr^2) a/r
=1/2 mra
Also
T=m(g-a)

comparing we have
1/2mra=m(g-a)
1/2ra=g-a
(r/(2)+1)a=g
r+2/2=g/a
r+2=2g/a
a=2g/r+2

now what?

 

[haruspex]

T= 1/(2) (mr^2) a/r

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]

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?

Torque=I alpha

 

[haruspex]

Torque=I alpha

Right, so what is the torque in this case?

 

[Suyash Singh]

Right, so what is the torque in this case?

torque is the tangential force

 

[haruspex]

torque is the tangential force

No, a torque is not a force. How do you find the torque of a force about an axis?

 

[Suyash Singh]

No, a torque is not a force. How do you find the torque of a force about an axis?

torque=R F sin(theta)
where theta is the between R and F

 

[haruspex]

torque=R F sin(theta)
where theta is the between R and F

Right, so correct your equation in post #15 and go from there.

 

[Suyash Singh]

Right, so correct your equation in post #15 and go from there.

T= mr alpha
T=m(g-a)
r alpha =m(g-a)
a=mg-ma
a(m+1)=mg
a=mg/m+1

 

[haruspex]

T= mr alpha

No, you had

T= 1/(2) (mr^2) a/r

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]

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 do you mean?

 

[haruspex]

what do you mean?

What is the torque that the tension T exerts on the cylinder about the cylinder's axis?

 

[Suyash Singh]

What is the torque that the tension T exerts on the cylinder about the cylinder's axis?

Torque=I alpha
=1/2 mr^2 a/r
=1/2 mar
=1/2 T r

 

[Dr Dr news]

If you are taking moments about the string-cylinder contact point, you need to use the parallel-axis theorem.

 

[haruspex]

Torque=I alpha

No, I mean using the equation you wrote in post #21 what is T's torque about the axis of the cylinder?

 

[haruspex]

If you are taking moments about the string-cylinder contact point.

I see no evidence that Suyash is doing that.

 

[Suyash Singh]

No, I mean using the equation you wrote in post #21 what is T's torque about the axis of the cylinder?

Torque=F R sin(theta)

 

[haruspex]

Torque=F R sin(theta)

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]

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?

F= tension
R= radius of cylinder
theta = 90 degrees

Torque=T R

 

[haruspex]

F= tension
R= radius of cylinder
theta = 90 degrees

Torque=T R

Right, so correct your first equation in post #15 and go from there.

 

[Suyash Singh]

Torque= T R
=m(g-a)R