A 10-m plank with a mass of 80 kg is being moved by rolling it over two cylindrical logs, each with a mass of 10 kg and a radius of 20 cm, placed 2 m from either end of the plank. As the plank is pushed, the logs roll on the ground without slipping, and the plank does not slip on the rollers. A force of 250 N is applied to the plank. What is the speed of the plank after it has moved 2 m, starting from rest?
A) 1.87 m/s
B) 2.53 m/s
C) 3.43 m/s
D) 3.54 m/s
E) 2.37 m/s

Question

Quizplus · Accepted Answer

First, we need to find the moment of inertia of the plank and the two logs system. We can assume that the logs are thin cylinders and therefore use the formula for moment of inertia of a thin cylinder:

$I_{cylinder} = \frac{1}{2}mr^2$

$I_{plank} = \frac{1}{12}ml^2$

$I_{system} = 2I_{cylinder} + I_{plank} = 2(\frac{1}{2}mr^2) + \frac{1}{12}ml^2 = \frac{1}{2}(2mr^2 + \frac{1}{6}ml^2)$

$I_{system} = \frac{1}{2}(2(10)(0.2)^2 + \frac{1}{6}(80)(10)^2) = 690 \ kg \cdot m^2$

Next, we can use the formula for rotational kinetic energy:

$KE_{rotational} = \frac{1}{2}I \omega^2$

Where $\omega$ is the angular velocity of the system.

Since the system starts from rest, all of the initial kinetic energy of the system will be in the form of translational kinetic energy:

$KE_{translational} = \frac{1}{2}mv^2$

We can use the work-energy principle to relate the initial translational kinetic energy to the final rotational kinetic energy:

$W_{net} = \Delta KE$

$W_{net} = Fd = (250 \ N)(2 \ m) = 500 \ J$

$\Delta KE = KE_{rotational} - KE_{translational}$

$\Delta KE = \frac{1}{2}I \omega^2 - \frac{1}{2}mv^2$

Setting $\Delta KE$ equal to $W_{net}$ and solving for $v$, we get:

$v = \sqrt{\frac{2W_{net}}{m + \frac{I}{r^2}}}$

$v = \sqrt{\frac{2(500 \ J)}{80 \ kg + \frac{690 \ kg \cdot m^2}{(0.2 \ m)^2}}}$

$v = \sqrt{\frac{2(500 \ J)}{80 \ kg + 86,250 \ kg}} = 3.43 \ m/s$

Therefore, the answer is (C) 3.43 m/s.

A 10-M Plank with a Mass of 80 Kg Is