Plank being pulled across two cylinders

  • Thread starter Thread starter bvschaefer
  • Start date Start date
  • Tags Tags
    Cylinders
Click For Summary

Homework Help Overview

The problem involves a plank being pulled across two cylindrical rollers, focusing on the dynamics of the system, including forces, accelerations, and torques. It is situated within the context of classical mechanics, specifically dealing with translational and rotational motion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law and the parallel axis theorem to analyze the system. There are attempts to derive equations for torque and acceleration, as well as considerations of the forces acting on both the plank and the cylinders.

Discussion Status

Some participants have provided guidance on the need for torque equations and the relationship between angular and linear acceleration. Others have noted similarities to previous discussions, suggesting a potential for deeper exploration of the problem's mechanics. The conversation is ongoing, with various aspects of the problem being examined.

Contextual Notes

Participants are navigating the complexities of the problem, including the roles of friction and the conditions of rolling without slipping. There is a focus on the need for clear definitions of forces and motions involved, as well as the implications of the setup on the equations governing the system.

bvschaefer
Messages
1
Reaction score
0

Homework Statement



A plank with a mass M = 6.30 kg rides on top of two identical, solid, cylindrical rollers that have R = 5.30 cm and m = 2.00 kg. The plank is pulled by a constant horizontal force Farrowbold of magnitude 5.00 N applied to the end of the plank and perpendicular to the axes of the cylinders (which are parallel). The cylinders roll without slipping on a flat surface. There is also no slipping between the cylinders and the plank.

(a) Find the initial acceleration of the plank when the rollers are equidistant from edge of the plank

(b)Find the acceleration of the rollers at this moment

(c)What friction forces are acting at this moment? (Fg and Fp)

Homework Equations



F = ma
T = Iα
T = F x R

The Attempt at a Solution


Using parallel axis theorem, I have:
I = Icm + MR2
I = .5(2.00kg)(.053)2 + (2.00kg)(.053)2
I = 0.00841
where point of rotation is at the ground.

From here, I know I have to use a torque equation to solve for acceleration but am stuck, any help is appreciated!
 
Physics news on Phys.org
welcome to pf!

hi bvschaefer! welcome to pf! :smile:
bvschaefer said:
… I know I have to use a torque equation to solve for acceleration but am stuck, any help is appreciated!

yes, you need the torque of F about the bottom point …

what is the difficulty with that? :confused:

(btw, you will also need an equation relating α and a)
 
bvschaefer said:

Homework Statement



A plank with a mass M = 6.30 kg rides on top of two identical, solid, cylindrical rollers that have R = 5.30 cm and m = 2.00 kg. The plank is pulled by a constant horizontal force Farrowbold of magnitude 5.00 N applied to the end of the plank and perpendicular to the axes of the cylinders (which are parallel). The cylinders roll without slipping on a flat surface. There is also no slipping between the cylinders and the plank.

Consider the motion of the plank and the cylinder separately: The plank performs translation, the cylinders roll, which is translation of their centre of mass and rotation about the centre of mass.

Write up Newton's second low for the translational motions, collecting the forces exerted both on the plank and on the cylinders.

The friction Fp between the plank and cylinders point in opposite direction as the pulling force in case of the plank, but the friction of magnitude Fp drives the cylinders forward. The rotational resistances between the ground and the cylinder, (Fg) points also forwards.
Both forces of friction exert some torque on the cylinders


The cylinders roll without slipping that means the translational speed of their CM is equal to ωR, the acceleration of the CM is aCM=βR, (β is the angular acceleration of the forward rotation).
The plank does not slip on the cylinders: so its velocity is the same as the linear velocity of the topmost point of the cylinder. What is it compared to the velocity of the CM?

You have two equation for the cylinders: one for the translation of the CM, and one for the rotation. What are they?

ehild
 

Attachments

  • plank.JPG
    plank.JPG
    6.7 KB · Views: 1,094
Last edited:

Similar threads

  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
6K
  • · Replies 55 ·
2
Replies
55
Views
10K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K