Solid disk radius of gyration

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1. A solid disc of radius r, is rolling down a variable incline (a ramp). Show that the acceleration of the centre of mass, C is given by:

a=g[ sinθ – F/Ncosθ ].
I have done this as shown below.

N is the normal reaction and F is friction.
N = mgcosθ
F = µN = µmgcosθ

mgsinθ – F = ma
mgsinθ – µmgcosθ = ma
a = g[ sinθ – µcosθ ]

but µ = F/N

a=g[ sinθ – F/Ncosθ ]

2. Determine an expression for the value of F/N where the only unknowns are the angle θ, the radius r and radius of gyration k.

I have tried this question and can’t get the right answer and need some help please.
I know
T = Iα
I = mk2 also for a solid disk I = 0.5mr2
α = a/r
m=N/gcosθ
Can anyone help with this question please?
 

Answers and Replies

  • #2
Doc Al
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doner said:
1. A solid disc of radius r, is rolling down a variable incline (a ramp). Show that the acceleration of the centre of mass, C is given by:

a=g[ sinθ – F/Ncosθ ].
I have done this as shown below.

N is the normal reaction and F is friction.
N = mgcosθ
F = µN = µmgcosθ

mgsinθ – F = ma
mgsinθ – µmgcosθ = ma
a = g[ sinθ – µcosθ ]

but µ = F/N

a=g[ sinθ – F/Ncosθ ]
I find this question quite strange, since you should have no trouble finding the acceleration directly without using F/N. In any case, while your answer is correct, the method is not. You assume that friction equals µN, but this is not true in general. Remember this is static friction, so F is less than (or possibly equal to) µN.

But you don't need to use µN at all; just stick to:
N = mgcosθ
mgsinθ – F = ma​
and combine these two.

2. Determine an expression for the value of F/N where the only unknowns are the angle θ, the radius r and radius of gyration k.

I have tried this question and can’t get the right answer and need some help please.
I know
T = Iα
I = mk2 also for a solid disk I = 0.5mr2
α = a/r
m=N/gcosθ
Forget the radius of gyration; you don't need it. Combine the torque equation (T = Iα) with the force equation (mgsinθ – F = ma) and you can solve for the acceleration. And then find F/N.
 
  • #3
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i have combined it but i get F/N = 1/3 TAN@
But the question wants an expression with r and k in it also.
 
  • #4
Doc Al
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If the disk is uniform, that's the correct answer. The only thing that I can think of is to pretend that you don't know if the disk is uniform or not. Then you can write the rotational inertia in terms of the radius of gyration and solve for the acceleration, then F/N. Then you'd have r and k in your answer.
 
  • #5
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Why don't you try asking your lecturer, maybe she can help!!!
 

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