Can someone help me with this rotational dynamics problem.

  • #51
I ended up getting a=.901m/s^2
 
  • #52
SammyS
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I ended up getting a=.901m/s^2
[STRIKE]I didn't get that[/STRIKE].

I'll re-check what I got, but I have already gone over my answer a few times.

Added in Edit:

Your result here is good!
 
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  • #53
I'm sorry I meant .934 m/s^2 I was practicing this problem with different quantities and accidentally wrote the answer one of the others.
 
  • #54
SammyS
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I'm sorry I meant .934 m/s^2 I was practicing this problem with different quantities and accidentally wrote the answer one of the others.
I didn't get that either.

When you got a = 0.552 m/s2 your moment of inertia was 4 times what it should have been, so when you had Fp = 6a, that was also 4 times the size it should have been.
 
  • #55
ehild
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See attached picture. The acceleration of the plank is ap. F-2Fp=Map (1)

The motion of the cylinder is translation of the centre of mass and rotation about the CM.
The CM accelerates with acm. The friction between cylinder and plank accelerates the CM of the cylinders forward, and accelerates forward rotation; the rotational resistance Fg between ground and cylinder drives translation forwards, but hinders rotation. .
So Fp+Fg=macm. (2)

r(Fp-Fg)=Iβ (β is the angular acceleration)

The cylinders do not slip: βr=0.5 m.
The moment of inertia about the CM is I=0.5 mr2.
The torque equation becomes:

Fp-Fg=0.5 macm.(3)

Adding equations (2) and(3):

2Fp=1.5macm (4)


The plank does not slip on the cylinders, so its velocity is the same as the topmost point of the cylinders, which is twice the velocity of the CM:
acm=0.5 ap.

Now you have the equations (1) and (4)

F-2Fp=Map
2Fp=1.5map/2

Add them:

F=Map+0.75map.

ehild
 

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  • #56
SammyS
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I ended up getting a=.901m/s^2
OK ...

I went over my calculations yet again.

The answer is a = 6.4/7.1 = 0.9014... .

However you came up with this for a previous post (post #51) of yours, this answer looks good.

I apologize for my repeatedly poor arithmetic.
 
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