Rotational Dynamics of rope

In summary: The T1 force gets canceled out by the friction force on the cylinder (which we know is 1/2 the weight of the cylinder).In summary, the conversation discussed the setup of a system involving a wooden cylinder, a rope, a crate, and a crank handle. The goal was to determine the tangential force required to raise the crate with a specified acceleration. The conversation also touched on a similar problem involving a cylinder, string, pulley, and block, and the desired acceleration of the block after the system is released from rest. Various equations and hints were provided to aid in solving the problems.
  • #1
Swatch
89
0
A rope is wrapped around a wooden sylinder with I=2.9 and radius = 0.25m
A 50 kg crate is suspended to the free end of the rope and is pulled upwards with an acceleration of 0.80 m/s*s
A crank handle is attached to the axle of the wooden cylinder and when turned rotates about the axle in a circle of radius 0.12 m
What tangential force F applied tangentially to the rotating crank is required to raise the crate with the acceleration mentioned. Ignore the mass of the rope and I of axle and crank.

I did:

Total torque of the cylinder = FRh - McgRcy = Icy*A
(where Rh=radius of crank handle circle Mc=mass of crate Rcy= radius of cylinder I=moment of inertia of cylinder A=angular acceleration)

A= ay/Rcy (where ay = translational acceleration of the crate)

When I solve for F I get F=1098 N when I should get 1200. Could someone give me hint to what I'm doing wrong, please.
 
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  • #2
Swatch said:
When I solve for F I get F=1098 N when I should get 1200. Could someone give me hint to what I'm doing wrong, please.
You have not considered the acceleration of the crate. The tension in the rope must be greater than the weight of the crate in order for the crate to accelerate.
 
  • #3
You may also use the same equation with moment of inertia of the system(in place of cyli.only) about the axis of rotation, which is I + Mc(Rcy)^2, as the crate is moving in a straight line distance Rcy from the axis of rotation.
 
  • #4
Yes I forgot the acceleration. Got the right answer. Thanks
 
  • #5
Actually I got another problem I'm stuck with:

A uniform solid cylinder with mass M and radius 2R rest on a horizontal tabletop. A string is attached by a yoke to a frictionless axle through the center of the cylinder so the cylinder can rotate about the axle. The string runs over a disk shaped pulley mass M and radius R. A block of mass M is suspended from the free end of the string. The string doesn't slip and the cylinder roll without slipping. Find the magnitude of the acceleration af the block after the system is released from rest.

What I have done so far is:

Total torque for the cylinder =f*2R = 0.5*MR^2a/R
so the friction force, f = 0.5Ma

Net translational force is : T1 - f =Ma
so T1 = 3/2*Ma (where T1 is the tension in the string between the cylinder and pulley)

For the pulley net torque = -T1 + T2 = 0.5*MR^2a/R
so the tension between the pulley and block is T2=Ma((R+3)/2)

For the block the net force is Mg-T2-T1

When I try to solve for a I get R and some other wrong stuff in the final answer. According to the right answer a =g/3

What am I doing wrong. A hint would be appreciated.
 
  • #6
Swatch said:
What I have done so far is:

Total torque for the cylinder =f*2R = 0.5*MR^2a/R
so the friction force, f = 0.5Ma
OK.

Net translational force is : T1 - f =Ma
so T1 = 3/2*Ma (where T1 is the tension in the string between the cylinder and pulley)
OK.
For the pulley net torque = -T1 + T2 = 0.5*MR^2a/R
so the tension between the pulley and block is T2=Ma((R+3)/2)
Oops. T1 and T2 are forces; you need the torque they produce.
 
  • #7
Thanks again. :biggrin:
 
  • #8
Swatch said:
For the block the net force is Mg-T2-T1

For anyone referencing this thread (like I am), the force on the block is just Mg-T2.
 

1. What is rotational dynamics of rope?

Rotational dynamics of rope is a branch of physics that deals with the motion of a rope that is being rotated around a fixed axis. It involves studying the forces and torques that act on the rope, as well as the rotational motion and equilibrium of the rope.

2. How does the shape and weight of a rope affect its rotational dynamics?

The shape and weight of a rope can greatly impact its rotational dynamics. A thicker and heavier rope will require more torque to rotate and will have a larger moment of inertia, while a thinner and lighter rope will be easier to rotate and have a smaller moment of inertia.

3. What are some real-life applications of rotational dynamics of rope?

Rotational dynamics of rope has various practical applications such as in rock climbing, sailing, and amusement park rides. It is also used in engineering and construction, such as in the design of cranes and pulley systems.

4. What is the difference between translational and rotational motion of a rope?

Translational motion refers to the movement of an object in a straight line, while rotational motion involves rotation around a fixed axis. In the case of a rope, translational motion would be the movement of the rope from one point to another, while rotational motion would be the spinning of the rope around a fixed point.

5. How can we calculate the tension in a rotating rope?

To calculate the tension in a rotating rope, we can use the equation T=Iα, where T is the tension, I is the moment of inertia, and α is the angular acceleration. We can also use the equation T=mrω^2, where m is the mass of the rope, r is the radius, and ω is the angular velocity. In both cases, the tension will depend on the mass and rotational parameters of the rope.

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