Plank being pulled across two cylinders

  • Thread starter Thread starter bvschaefer
  • Start date Start date
  • Tags Tags
    Cylinders
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 6K views
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: 1,115
Last edited: