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Homework Help: Block Around a Track

  1. Nov 29, 2014 #1
    1. The problem statement, all variables and given/known data
    A small block of mass m = 0.50 kg is fired with an initial speed of v0 = 4.0 m/s along a horizontal section of frictionless track, as shown in the top portion of Figure P7.58. The block then moves along the frictionless, semicircular, vertical tracks of radius R = 1.5 m. (a) Determine the force exerted by the track on the block at points A and B. (b) The bottom of the track consists of a section (L = 0.40 m) with friction. Determine the coefficient of kinetic friction between the block and that portion of the bottom track if the block just makes it to point C on the first trip. (Hint: If the block just makes it to point C, the force of contact exerted by the track on the block at that point is zero.)


    2. Relevant equations

    ## a_c = m v^2 / r ##, ## F = ma ##

    3. The attempt at a solution

    (a) We begin by using CoE with the level at B defined as 0. Right before sliding onto the track, the total energy is ## \frac{1}{2} m 4^2 + m g 3 = m(8+3g) ##

    At point A, the total energy is ## \frac{1}{2} m v^2 + mg(1.5) = m(\frac{v^2}{2} + 1.5g) ##. Energy is conserved, so ## 2(8+3g-1.5g) = v^2 \Rightarrow v^2 = 45.4 ## (Assuming g = 9.8).

    Since the only force at point A is the one causing the centripetal acceleration, the force of track on block is ## \frac{mv^2}{r} = 15.13 ## N.

    At point B, the total energy is ## \frac{1}{2} mv^2 ##. Energy conserved, so ## 2 (8+3g) = v^2 = 74.8 ##. At the bottom, however, the weight of the object takes role, and net force = ## \frac{mv^2}{r} - mg = 20.03 ## N. (Not sure it's supposed to be + mg or -mg, but I think it should be mg. Any ideas on this would be appreciated)

    Is this right? If so, I will proceed to post my part B solution. No answer attached with this one, so not sure.


    (b) At the top, normal force is 0, so mg is the only force contributing to the centripetal force. Therefore, ## mg = mv^2 / r \Rightarrow v^2 = 14.7 ##. Conservation of energy with the point at the bottom of the semicircle and point C gives ## \frac{1}{2} m 14.7 + 3mg = \frac{1}{2} mv^2 \Rightarrow v^2 = 73.5 ## at the bottom.

    Now, by the Work-Energy theorem, the work done by friction is equal to the change in energy. Since no change in potentional energy, all change is in kinetic energy, so ## W_fric = 1/2*m*(74.8-73.5) = 0.325 ##. In addition, ## W = Fd##, so ##0.325 = 0.4F \Rightarrow F = 0.8125 ##. In addition, ##F = \mu * mg \Rightarrow \mu = 0.17 ##.

    Is all of this correct, or did I make a mistake somewhere? (Again, can't find the same problem anywhere online)


    Last edited: Nov 29, 2014
  2. jcsd
  3. Nov 29, 2014 #2


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    The force from the track on the block has to be such that the resultant force keeps the block in the circular orbit. What is the resultant force and what magnitude should it have?

    Your reasoning otherwise seems essentially ok - but - You really should not insert values into your equations until the final step. Inserting numbers really obscures the physics and makes it difficult to check limiting cases. Furthermore, your equations are missing units which makes things even more difficult to read (and technically even wrong). Without units, your expressions do not mean anything. For example, v^2 = 45.4 does not tell you anything - v could be measured in cm/hour, lightyears/Myr, m/s, or km/day.
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