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Kinematics and energy conservation in rotational motion

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A light string that is attached to a large block of mass $4m$ passes over a pulley with negligible rotational inertia and is wrapped around a vertical pole of radius $r$, as shown above. The system is released from rest, and as the block descends the string unwinds and the vertical pole with its attached apparatus rotates. The apparatus consists of a horizontal rod of length 2L, with a small block of mass m attached at each end. The
rotational inertia of the pole and the rod are negligible.

(a) Determine the rotational inertia of the rod-and-block apparatus attached to the top of the pole.
(b) Determine the downward acceleration of the large block.
(c) When the large block has descended a distance D, how does the instantaneous total kinetic energy of the three blocks compare with the value 4mgD ? Check the appropriate space below.
Greater than 4mgD
Equal to 4mgD
____ Less than 4mgD

Justify your answer.

a) The answer is simply $I=2mL^{2}$.

b) Using $4mg-T=4ma$ and $I\alpha = I\alpha/r =Tr$ we can find $a=2gr^{2}/(2r^{2}+L^{2})$.

c) One way is to use conservation of energy to say that the total kinetic energy of the three blocks should equal the change in potential energy $4mgD$. I also tried to use energy conservation and kinematics. The total kinetic energy of the three blocks seems to be $1/2 mv^{2}+1/2mv^{2}+4mv^{2}= 5mv^{2}$ (assuming all masses have the same linear velocity).

Then since I know the acceleration from b) I can find that $v=2gr^{2}/(2r^{2}+L^{2})t$ and so $D=gr^{2}/(2r^{2}+L^{2})t^{2}$. Solving for $t$ and replacing into $v$ I can find the velocity at distance $D$ and then plug into the kinetic energy $5mv^{2}$. Doing this however I dont get $4gmD$. Is my assumption that the linear velocities of all blocks is the same wrong? Or what is wrong with the reasoning?

asked Mar 3, 2017 in Physics Problems by member192 (280 points)
edited Mar 3, 2017 by member192

1 Answer

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Best answer

Your assumption is wrong. You need to apply circle geometry to work out how the distance fallen by the $4m$ block is related to the distance moved around the circle by the two blocks $m$. These distances are in the ratio $r/L$. The velocities are in the same ratio.

answered Mar 3, 2017 by sammy gerbil (28,448 points)
selected Mar 4, 2017 by member192
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