Plank being pulled across two cylinders

[bvschaefer]

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? (F_g and F_p)

Homework Equations

F = ma
T = Iα
T = F x R

The Attempt at a Solution

Using parallel axis theorem, I have:
I = I_cm + 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!

 

[tiny-tim]

welcome to pf!

hi bvschaefer! welcome to pf!

… 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? ​

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

 

[haruspex]

Apart from the numbers, this looks the same as https://www.physicsforums.com/showthread.php?t=651174

 

[ehild]

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 a_CM=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

 

[Nickie]

(a) To find the initial acceleration of the plank, we can use Newton's second law, F = ma, where F is the net force acting on the plank and m is its mass. In this case, the only force acting on the plank is the horizontal force Farrowbold, so we have:

F = ma
5.00 N = (6.30 kg)a
a = 0.794 m/s^2

(b) To find the acceleration of the rollers, we can use the relationship between torque and angular acceleration, T = Iα, where T is the net torque acting on the rollers, I is the moment of inertia of the rollers, and α is the angular acceleration. In this case, the net torque is due to the friction force acting between the rollers and the ground, and we can assume that the friction force is equal in magnitude to the horizontal force Farrowbold pulling the plank. Therefore, we have:

T = Farrowbold(R + R) = 2FarrowboldR
I = 0.00841 kg m^2 (calculated in part a)
α = a/R = 0.794 m/s^2 / 0.053 m = 15.0 rad/s^2

(c) The friction forces acting at this moment are the static friction force Fg between the rollers and the ground, and the friction force Fp between the plank and the rollers. Since the rollers are not slipping on the ground, the net torque acting on them is zero, which means that the friction force Fg must be equal in magnitude and opposite in direction to the horizontal force Farrowbold pulling the plank. Therefore, we have:

Fg = Farrowbold = 5.00 N

The friction force Fp between the plank and the rollers can be calculated using the relationship T = FpR, where T is the net torque acting on the plank (due to the horizontal force Farrowbold) and R is the radius of the rollers. Therefore, we have:

T = FarrowboldR = FpR
Fp = Farrowbold = 5.00 N