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Accelerating, Rolling Cylinder

  • #1
68
1

Homework Statement


A solid cylinder (I = 1/2 mr^2) is being pulled by a force applied at 40 degrees above the horizontal to the axle and directed through the center of the roller. Roller moves across a flat, horizontal surface. Cylinder has radius of 0.60 m, and mass of 250 kg. Assume roller rolls with just enough friction to keep from slipping, determine the amount of force needed to give the roller a horizontal acceleration of 0.50 m/s^2

Homework Equations


Angular Acc. = a / r
Torque = I (Ang. Acc.) : F Cos(40) (r) = (1/2 mr^2) a/r

The Attempt at a Solution


Is the last equation I wrote sufficient to solve problem? Or do you need to do sum of forces in x direction equal to horizontal acceleration times mass? If so, does static friction point in direction opposite of motion or in same direction of motion (i.e. to prevent the rolling disk from slipping in place)?
 

Answers and Replies

  • #2
19,954
4,108

Homework Statement


A solid cylinder (I = 1/2 mr^2) is being pulled by a force applied at 40 degrees above the horizontal to the axle and directed through the center of the roller. Roller moves across a flat, horizontal surface. Cylinder has radius of 0.60 m, and mass of 250 kg. Assume roller rolls with just enough friction to keep from slipping, determine the amount of force needed to give the roller a horizontal acceleration of 0.50 m/s^2

Homework Equations


Angular Acc. = a / r
Torque = I (Ang. Acc.) : F Cos(40) (r) = (1/2 mr^2) a/r

The Attempt at a Solution


Is the last equation I wrote sufficient to solve problem? Or do you need to do sum of forces in x direction equal to horizontal acceleration times mass? If so, does static friction point in direction opposite of motion or in same direction of motion (i.e. to prevent the rolling disk from slipping in place)?
What is your assessment? Have you drawn a free body diagram yet?

Chet
 
  • #3
68
1
I"m stuck with free-body diagram as I'm not sure direction of friction force.

Can I choose the point of contact with the ground as the point of rotation and then just use the equation for torque (as I did in my first post) applied to the pulling force?
 
  • #4
19,954
4,108
I"m stuck with free-body diagram as I'm not sure direction of friction force.

Can I choose the point of contact with the ground as the point of rotation and then just use the equation for torque (as I did in my first post) applied to the pulling force?
What does this say about the direction of the friction force? (If you're at a loss, take a guess).
How does the linear velocity of the cylinder relate to the angular velocity of the cylinder, if static friction is not exceeded?

Chet
 
  • #5
68
1
I understand that if cylinder is not slipping then v (center of cylinder) = w * r. Not sure how to interpret your first statement (what does what say?)
Would appreciate feedback on whether just assuming the point of contact with ground is the point of rotation is a reasonable approach.
Thanks.
J
 
  • #6
19,954
4,108
I understand that if cylinder is not slipping then v (center of cylinder) = w * r. Not sure how to interpret your first statement (what does what say?)
Would appreciate feedback on whether just assuming the point of contact with ground is the point of rotation is a reasonable approach.
Thanks.
J
My first question was, "to get the cylinder rotating in the direction that it does rotate, does the ground have to exert a tangential force on the cylinder in the rearward direction or in the forward direction? This is the static frictional force that the ground exerts on the cylinder. You are going to need to do both a rotational torque balance and a linear force balance.

The point of contact with the ground is the point that the tangential force from the ground is applied. The rotation takes place around the axis of the cylinder.

Chet
 
  • #7
68
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OK, thanks. I think I"m getting a hold on this. (It was confusing that in some cases--e.g., when force comes from a string wrapped around cylinder and exiting tangentially on the top--the force of static friction could be in direction of motion.)
Conceptually, I just don't get why linear force balance is computed in the "standard" way. How can something that's rolling have F = MA applied to the center of mass using the same equations that apply to a box being dragged across a frictionless surface? (Again, I understand how to mathematically get the answer with two equations--I have a hard time intuitively accepting this as true.)
 
  • #8
19,954
4,108
OK, thanks. I think I"m getting a hold on this. (It was confusing that in some cases--e.g., when force comes from a string wrapped around cylinder and exiting tangentially on the top--the force of static friction could be in direction of motion.)
Conceptually, I just don't get why linear force balance is computed in the "standard" way. How can something that's rolling have F = MA applied to the center of mass using the same equations that apply to a box being dragged across a frictionless surface? (Again, I understand how to mathematically get the answer with two equations--I have a hard time intuitively accepting this as true.)
OK Jon4444. I can see where it can be confusing. Let's see if I can help. I think what you are saying is, "If I have a translating, rotating body, different parts of the body are accelerating at different rates and in different directions. Why is it still valid to apply the usual force balance to the motion of the center of mass?"

If you have a single point particle, I assume you have no problem applying F = Ma. Now suppose you have an array of point particles (say 100) that are joined to their nearest neighbors by massless springs. The springs are very stiff. The entire array can be translating and rotating. Suppose that some of the particles at the outer region of the array are also subjected, not only to the spring forces from its neighbors, but also to external forces (that can be considered acting on the combined array). Now, suppose you write an F = Ma force balance on each and every individual particle of the array. No problem, right? Next, suppose you sum up all these individual force balances. All the forces in the springs will cancel out with one another in the summation, and the Ma side of the equation will be just the sum of the masses of the particles times their individual accelerations. How does this sum relate to the acceleration of the center of mass of the array, and to the total combined mass of the particles? The left side of the equation will just be equal to the external forces applied to the array.

Chet
 
  • #9
68
1
Thanks again Chet--that's a helpful analogy.
 
  • #10
311
23
In rolling without slipping cases, you can give the cylinder an effective mass (which is greater than the actual mass)
Give the cylinder an arbitrary velocity (say 10 m/s)
Calculate the rotating and linear KE's
The effective mass = actual mass * ( 1 + ( KE rotating / KE linear ) )
 

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