Rotation of a uniform rigid disc about a fixed smooth axis

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  • #1
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Homework Statement


A uniform circular disc has mass M and diameter AB of length 4a. The disc rotates in a vertical plane about a fixed smooth axis perpendicular to the disc through the point D of AB where AD=a. The disc is released from rest with AB horizontal. (See attached diagram)

(a) Calculate the component parallel to AB of the force on the axis when AB makes an angle θ with the downward vertical.

(b) Calculate the component perpendicular to AB of the force on the axis when AB makes an angle θ with the downward vertical

Homework Equations


KE = (1/2)Iω2
I of a disc about axis perpendicular to disc through centre is (1/2)mr2
Transverse component of acc. = r(dω/dt)
Radial component of acc. = rω2

The Attempt at a Solution


I have done (a) okay with no problems and got the correct answer. I used a similar approach for part (b) but got the incorrect answer. For part (b):

The moment of inertia is (1/2)M(2a)2 + Ma2 = 3Ma2
(using the parallel axis theorem).

Using conservation of energy:

(1/2)(3Ma22 = Mgacosθ
(3/2)aω2 = gcosθ

Differentiating w.r.t. θ gives:

3aω(dω/dθ) = -gsinθ
3aω(dω/dt)(dt/dθ) = -gsinθ
(dω/dt) = -(g/3a)sinθ *

Resolving forces: Mgsinθ - Y = Ma(dω/dt)

Substituting in *: Mgsinθ - Y = -(Mag/3a)sinθ

Y = Mgsinθ + (1/3)Mgsinθ = (4/3)Mgsinθ

The answer is supposed to be (2/3)Mgsinθ. I can't see what I have done wrong. Any help would be greatly appreciated!
 

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Answers and Replies

  • #2
Resolving forces: Mgsinθ - Y = Ma(dω/dt)

I don't like what I see here. You have the angular acceleration, not the linear acceleration.
Maybe rather use

[itex]\tau[/itex] = I [itex]\alpha[/itex] ?
 
  • #3
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I don't understand what is wrong with resolving forces, I have used the transverse component of acceleration for circular motion which is radius*(dω/dt) and the forces acting on the disc which are parallel to that acceleration vector (as in the diagram)?

However if I use τ = I(dω/dt) = 3Ma2(dω/dt)

aMgsinθ = 3Ma2(dω/dt)
gsinθ = 3a(dω/dt)
(dω/dt) = (g/3a)sinθ

Before, I got the negative of this - if I substitute this into my force equation however I do get the correct answer of (2/3)Mgsinθ.

But you said that that equation is wrong - how else can I get an expression involving Y though?
 
  • #4
Ok, Y does not have a torque.
Then rather try using the tangential acceleration in your equation:
Resolving forces: Mgsinθ - Y = Ma(dω/dt)
You can see that this equation of yours is dimensionally wrong.
 
  • #5
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Why?

LHS: (kg*m*s-2) - (N) = N
RHS: kg*m*(s-1)(s-1) = kgms-2 = N
 
  • #6
Sorry, yes, you did use the transverse component of the acceleration.
You could setup

[itex]\sum \tau[/itex] = I [itex]\alpha[/itex]

about another point along AB.
 
Last edited:
  • #7
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I believe the radian is not included in dimensional analysis.

For example in τ = Iα, if you include the radian you would have:

LHS = Nm
RHS = kg*m2*rad*s-2 = (kg*m*s-2)*m*rad = Nm*rad

so you must ignore the 'unit' radian.
 
  • #9
haruspex
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Resolving forces: Mgsinθ - Y = Ma(dω/dt)
Because of the way theta is measured, dω/dt is the acceleration anticlockwise (in your diagram). So the above equation should read Y - Mgsinθ = Ma(dω/dt).
 
  • #10
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Oh, and by the same logic for τ = Iα,

aMgsinθ = -3Ma2(dω/dt)

which, along with the force equation, gives the correct answer.

Thank you for your help!
 

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