Rotational mechanics. cylinder over a plank

  • #1
vinay.ryder13
4
0

Homework Statement


a solid cylinder of mass m and radius R and a plank of mass M are placed on a smooth horizontal surface. there is sufficient friction between the cylinder and the plank's surface to prevent sliding of the cylinder. the fixed pulley is smooth. if the plank is pulled with a horizontal force F, the acceleration of center of the cylinder is? (Refer the figure *dia.bmp*),(take tension in the string as T and friction between the cylinder & plank as f).

View attachment dia.bmp



2. The attempt at a solution

1. started off with the force equation of the plank => F- ( T + f)= Ma ----equation(a)

2. equation of motion of cylinder => T-f = macm -----equation(1)

3. rotational equation of cylinder => fR = Iα
fR = mR2/2 * α
f = macm/2

substituting in equation(1) and solving => T = 3macm/2

we know, acm = a/2 ( doubtful about the requisite)

so substituting the results in equation(a),

F - 2macm= 2Macm

=> acm= F/2(M+m).

but the solution given the referred book is acm = F/3m+M.

have i done some mistake is writing the equations or at some other point? please help! thanks.
 

Answers and Replies

  • #2
kushan
256
0
Hello there ,
You are doing fine , just a small mistake :smile: is stopping you from getting the right answer .

The accelration of cylinder from grounds point of view would be twice of the accelration of block M.
T+f=2ma

and F+f-T=Ma
try to understand these and then solve again
:cool:
 
  • #3
vinay.ryder13
4
0
@kushan - but, drawing fbd for plank is giving the eqn F-(f+T)=Ma. That is pricking me. Please can u verify it by an fbd? :)
 
  • #4
vinay.ryder13
4
0
If F-(f+T)=Ma is correct then the cylinder's eqn would also change.
 
  • #5
kushan
256
0
Vinay ,
The friction is acting along the direction of F on the Cylinder
But [BOLD] according to newtons law , friction in opp. direction should act on the slab [/BOLD]
 
  • #6
vinay.ryder13
4
0
Oh...i get it now. Thanks! :)
 

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