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Block sliding down a plane

  1. Apr 9, 2014 #1
    1. The problem statement, all variables and given/known data

    A block with mass m = 5.00kg slides down a surface inclined 36.9 ∘ to the horizontal (the figure (Figure 1) ). The coefficient of kinetic friction is 0.27. A string attached to the block is wrapped around a flywheel on a fixed axis at O. The flywheel has mass 23.0kg and moment of inertia 0.500 kg⋅m2 with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.200m from that axis.

    YF-10-55.jpg

    What is the acceleration of the block down the plane?

    2. Relevant equations

    Moment of Inertia

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


    3. The attempt at a solution
    R = radius of wheel = [itex] I = \frac{MR^2}{2} => \sqrt(\frac{1}{M})=R [/itex]

    r = radius to point P (where the force acts as a torque)

    [itex]F_{net x } = mgsin\theta-\mu_kmgcos\theta-T=ma[/itex]

    [itex] \tau = rT = I\alpha = \frac{Ia}{R}[/itex]

    [itex] T = \frac{Ia}{rR} [/itex]

    [itex] mg(sin\theta - \mu_kcos\theta)=ma+\frac{Ia}{rR}[/itex]

    [itex] mg(sin\theta - \mu_kcos\theta)\frac{1}{m+\frac{I}{rR}}=a[/itex]

    Somehow this is giving me the wrong answer, what did I do wrong?

    EDIT: it seems that when you set R=r it'll give you the right answer. This does not make sense, the radius of the Cylinder is not equivalent to the radius of the torque (which is defined as .2 in the problem). They differ, if only slightly, however mastering physics tells me I'm wrong.
     
    Last edited: Apr 10, 2014
  2. jcsd
  3. Apr 10, 2014 #2
    The RHS should have 'r' instead of 'rR' .
     
  4. Apr 10, 2014 #3
    I don't think it should, I divided through by r to arrive at [itex]T=\frac{Ia}{rR}[/itex]
    the division takes place in the line above the one you are citing
     
  5. Apr 10, 2014 #4
    Okay...I was reading 'a' as 'α' .

    You are right in figuring out that R=r in your work. Basically R i.e radius features only in the MI (MR2/2) .In the torque equation as well as the no slip condition ,we need to consider 'r'.

    The relationship between 'a' and 'α' is a=αr and not a =αR .The tangential acceleration of the tip of the pulley where the rope leaves it is αr which is in turn equal to the acceleration of the block.
     
  6. Apr 10, 2014 #5
    Ohhh, yes you're absolutely right. What was the point in throwing in the weight of the pulley then, extraneous information?
     
  7. Apr 10, 2014 #6
    To check your understanding whether you can pick up the relevant piece of information from the problem :smile:
     
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