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The block and spool problem (dynamics of rigid bodies)

  • #1

Homework Statement


A block and spool are each pulled by a string across a frictionless surface. The string is wound around the spool such that it may unwind as it is pulled, the string attached to the block is at the center of mass. The block and spool have the same mass and the strings (massless) pull each with the same constant force.

Is the magnitude of the acceleration of the block greater than, less than, or equal to that of the spool? Explain.
Do the block and spool cross the finish line at the same time?

Homework Equations


conceptual?

The Attempt at a Solution


Less than[/B]
For spool:
net torque = Fnet*r = I*α = T*r
^(foregoing sin(theta) because not applicable)
I=(1/2)Mr^2 <-- for solid cylinder
Tr=(1/2)M(r^2)α
T=(1/2)M(r)α
rα= a_tangential
a_tangential = (2T/M)

For block:
Fnet= ma= T
a= T/m
m=M
T/M<2T/M
I think I might be using the equations wrong because this conclusion doesn't add up to the conceptual answer?
Things I understand:
The work of the hand for the spool is greater than that for the block because it has to apply the force over a longer distance, which follows KE= 0.5mv^2 + 0.5I(ω)^2 for the spool.

Things that are tripping me up:
Shouldn't the center of mass of the spool and the center of mass of the block have the same acceleration because they have the same mass and the same net force?
Conceptually, I'm thinking that they cross at the same time, but my math doesn't add up?
 

Answers and Replies

  • #2
TSny
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Welcome to PF!
T=(1/2)M(r)α
rα= a_tangential
a_tangential = (2T/M)
Can you explain why you interpret rα as the "acceleration of the spool"?
 
  • #3
Welcome to PF!


Can you explain why you interpret rα as the "acceleration of the spool"?
(a_string= ) a_tangential = rα (where alpha is the angular acceleration of the rotating object and r is the radius)
is the no slip condition for rolling objects and pulleys. I assumed that you could treat the string around the spool as you would a pulley such that the tangential acceleration of the spool at the point of contact between the string and spool is equal to the acceleration of the string.
I should also note that the spool is situated such that its base is on the ground and it will not roll.
 
  • #4
TSny
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(a_string= ) a_tangential = rα (where alpha is the angular acceleration of the rotating object and r is the radius)
is the no slip condition for rolling objects
Does the "no slip condition" apply to the spool in this problem? The spool is on a frictionless surface.
[EDIT: Or, by "no slip", are you are referring to the string not slipping on the spool rather than the spool not slipping on the table?]
 
  • #5
Does the "no slip condition" apply to the spool in this problem? The spool is on a frictionless surface.
My thinking is that the no slip condition applies to the string around the spool like it would a pulley. ie the string is pulling the spool and the string is not slipping around the spool, so the acceleration and velocity at the point of contact would be the same. If the tangential acceleration for the spool is equal to the acceleration of the string at the point of contact then the acceleration of the center of mass should also equal that. If this isn't right, then I don't know if it applies or not.

Edit: Yes, exactly.
 
  • #6
TSny
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##r \alpha## gives the tangential acceleration of a point on the rim of the spool relative to the center of mass of the spool. The linear acceleration of a point of the string as it unwinds from the spool will be ##r \alpha## relative to the center of mass, but not relative to the table.
 
  • #7
TSny
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If the tangential acceleration for the spool is equal to the acceleration of the string at the point of contact then the acceleration of the center of mass should also equal that.
Yes, the tangential acceleration of a point on the rim of the spool will equal the acceleration of the string at the point of contact. However, why should the acceleration of the center of mass equal the acceleration of the string?
 
  • #8
##r \alpha## gives the tangential acceleration of a point on the rim of the spool relative to the center of mass of the spool. The linear acceleration of a point of the string as it unwinds from the spool will be ##r \alpha## relative to the center of mass, but not relative to the table.
Isn't there an exception for accelerating reference frames of rolling without slipping wherein Newton's second law is upheld (net torque = Iα)?
 
  • #9
TSny
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Isn't there an exception for accelerating reference frames of rolling without slipping wherein Newton's second law is upheld (net torque = Iα)?
Yes. This equation is upheld in the center of mass reference frame (even if the center of mass is accelerating). So ##\sum \tau_c = I_c \alpha## where the ##c## implies that the torques and moment of inertia are calculated relative to the center of mass.
 
  • #10
Yes, the tangential acceleration of a point on the rim of the spool will equal the acceleration of the string at the point of contact. However, why should the acceleration of the center of mass equal the acceleration of the string?
Doesn't it have to for the situation to be physically possible?
 
  • #11
TSny
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Your equation ##Tr = I \alpha## is correct as long as ##I## is the moment of inertia of the spool relative to the center of mass. But care is needed in getting from ##\alpha## to the acceleration of the center of the spool relative to the table. In fact, in this problem it is easier to get the acceleration of the center of mass of the spool directly from F = ma. Then, if you want, you can get the acceleration of the string relative to the table.
 
  • #12
Your equation ##Tr = I \alpha## is correct as long as ##I## is the moment of inertia of the spool relative to the center of mass. But care is needed in getting from ##\alpha## to the acceleration of the center of the spool relative to the table. In fact, in this problem it is easier to get the acceleration of the center of mass of the spool directly from F = ma. Then, if you want, you can get the acceleration of the string relative to the table.
So I should't even consider torque and instead just consider the transnational force of tension?
 
  • #13
rcgldr
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Shouldn't the center of mass of the spool and the center of mass of the block have the same acceleration because they have the same mass and the same net force?
Linear acceleration (at the center of mass) = force / mass, regardless of the point of application (the same linear acceleration for the same force applied either at the center of mass or at the edge of spool).

Conceptually, I'm thinking that they cross at the same time, but my math doesn't add up?
Since the force is the same, then the string has to accelerate faster in the spool case in order to generate the same force. You've already figured out that the surface of the spool accelerates at 2a with respect to the center of mass of the spool. That means the upper surface and the string accelerate at 2a+a = 3a, and the bottom surface accelerates at -2a+a = -a, with respect to the frictionless surface.
 
Last edited:
  • #14
TSny
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So I should't even consider torque and instead just consider the transnational force of tension?
Yes, if all you need is the acceleration of the center of the spool.
 

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