Cylinder rolling without slipping on a truck

[LCSphysicist]

Homework Statement

A section of steel pipe of large diameter and relatively thin wall is mounted as shown on a flat-bed truck. The driver of the truck, not realizing that the pipe has not been lashed in place, starts up the truck with a constant acceleration of 0.5 g. As a result, the pipe rolls backward (relative to the truck bed) without slipping, and falls to the ground. The length of the truck bed is 5 m.
(a) With what horizontal velocity does the pipe strike the ground?
(b) What is its angular velocity at this instant?
(c) How far does it skid before beginning to roll without slipping, if the coefficient of friction between the pipe and ground is 0.3?
(d) What is its linear velocity when its motion changes to rolling without slipping?

Relevant Equations

t = rf
t = Ia

I don't know how to start it

Is the truck who make the cylinder roll, initially? If yes, how? Since the truck force would pass by the center of the cylinder.

 

[etotheipi]

Is the truck who make the cylinder roll, initially? If yes, how? Since the truck force would pass by the center of the cylinder.

It'll be because the truck 'leaves behind' the cylinder, in a vague sense. It might be easiest to do the first part of the question in the accelerating reference frame of the truck, so that the cylinder feels an inertial body force of magnitude ##0.5mg## to the right (and friction!). We are told it rolls without slipping on the truck, so that should let you also eliminate ##\ddot{\theta}## whilst on the truck. In addition to the velocity w.r.t. the lab when it finally leaves the truck.

 

[LCSphysicist]

It'll be because the truck 'leaves behind' the cylinder, in a vague sense. It might be easiest to do the first part of the question in the accelerating reference frame of the truck, so that the cylinder feels an inertial body force of magnitude ##0.5mg## to the right. We are told it rolls without slipping on the truck, so that should let you calculate ##\ddot{\theta}## whilst on the truck. In addition to the velocity w.r.t. the lab when it finally leaves the truck.

I thought in attack the problem by this way, but I arrived at so many contradictions that i gave up.
First i didn't know if this force inertial would produce any torque in the body, when i takes for granted, i didn't know in which point would the force acting on the body.
So i gave up this ideia and try attack just with friction force the responsible for the torque, so i imediate see a problem:
f is friction
-mA is the inertial force

Rotation:
|f*r| = |I*x''/r|
I = mr²
|f| = |mx''|

and:
f + m(-a) = mx''
No matter how direction i choose to the friction force, both will be absurd and so it need to be disconsidered.

So, I have no idea: S

 

[etotheipi]

The situation in the truck is not dissimilar to something rolling down a hill. You have an inertial body force of magnitude ##0.5mg## acting to the right through the centre of mass (producing no torque about the CM), and a frictional force of magnitude ##f## acting to the left, which does product a torque about the centre of mass.

The frictional force in this case is necessary, otherwise ##a_x = r\ddot{\theta}## (the rolling condition) could not be satisfied.

 

[LCSphysicist]

Where is the error? :C

 

[etotheipi]

I would say your working is pretty much correct, save a sign error; if ##A## is the magnitude of the acceleration of the truck and ##\ddot{x}## is the acceleration of the pipe w.r.t. the truck,
$$mA - f = m\ddot{x}$$ $$fr = mr^2 \alpha = \frac{mr^2 \ddot{x}}{r} \implies f = m\ddot{x}$$ So now substituting back into the first relation gives $$mA = 2m\ddot{x} \implies \ddot{x} = \frac{A}{2}$$

 

[LCSphysicist]

I would say your working is pretty much correct, save a sign error; if ##A## is the magnitude of the acceleration of the truck and ##\ddot{x}## is the acceleration of the pipe w.r.t. the truck,
$$mA - f = m\ddot{x}$$ $$fr = mr^2 \alpha = \frac{mr^2 \ddot{x}}{r} \implies f = m\ddot{x}$$ So now substituting back into the first relation gives $$mA = 2m\ddot{x} \implies \ddot{x} = \frac{A}{2}$$

Wow,damn signs! XD
thxx

 

[etotheipi]

Wow,damn signs! XD
thxx

Yeah, I think it's easy to make errors like that because it's not obvious how the senses of the rotational & linear motion are connected. Probably best just to step back and think about it, as in "if the cylinder is accelerating to the right (##\ddot{x} > 0##) then the clockwise acceleration has to also be positive".

 

[Halc]

While most parts of the question can be answered with the details given (and they assume the pipe doesn't bounce or anything), part b asks for angular velocity of the pipe, which cannot be known without its radius, not specified.