Rotational Dynamics of rope

  • Thread starter Swatch
  • Start date
  • #1
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.
 

Answers and Replies

  • #2
OlderDan
Science Advisor
Homework Helper
3,021
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
mukundpa
Homework Helper
524
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
89
0
Yes I forgot the acceleration. Got the right answer. Thanks
 
  • #5
89
0
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
Doc Al
Mentor
44,958
1,223
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
89
0
Thanks again. :biggrin:
 
  • #8
23
0
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.
 

Related Threads on Rotational Dynamics of rope

  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
2
Replies
32
Views
5K
  • Last Post
Replies
16
Views
4K
  • Last Post
Replies
5
Views
531
  • Last Post
Replies
3
Views
12K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
24
Views
1K
  • Last Post
Replies
19
Views
8K
Replies
9
Views
248
Top