Question

The block of weight 113 lb is pushed with a force of P = 31 lb on top of two cylindrical rollers with a weight of 42 lb and a radius of R = 1.57 ft. Determine the block's speed after it has been moved 2.08 ft to the left. The block is originally at rest. No slipping occurs.

Answer 1

Answer:

The block's speed will be of 5,1 ft/s

Explanation:

For this problem solve it is used the work and energy theorem which state:

$W=dK$

Where W is the net work done and dK the change in the kinetic energy between in the time after and before the force is applied. The net work can expressed as:

$W=F.d$

Here F and d are applied force and the displacement. While the kinetc energy for the i body:  

$Ki=1/2*m_{1}V_{1}^2+1/2*I_{1}w_{1}^2$

Where mi,Ii Vi, wi are the mass, the moement of inertia, the linear and the angular velocity for the i body respectively.  Therefore to calcule the total change in kinetic energy we take into account all the bodies involves, in this case the block and the two rollers. Assuming fixed rollers but can rotate, and the block with linear movement, the final kinect energy is:

$Kf=1/2*m_{b}V_{b}^2+1/2*I_{r1}w_{r1}^2+1/2*I_{r2}w_{r2}^2$

Note since at the beginning the bodies system is at rest hence the initial kinetic energy is null. In the above equation the subscripts b, r1 and r2 relate to the block and the two rollers.

For solid roller the moment of inertia can be calculated as:

$I_{r}=1/2*m_{r}R_{r}$

Where mr and Rr are the mass and radius rollers. Replacing the moment of inertia in the kinetic energy:

$Kf=1/2*m_{b}V_{b}^2+1/4*m_{r1}R_{r1}w_{r1}^2+1/4*m_{r2}R_{r2}w_{r2}^2$

As at the point of contact there is no displacement between the surfaces of the block and the rollers, in that point the linear velocities of this bodies are the same. So can expressed the linear velocity of the block as function of the rotating speed and the rollers radius:

$V_{b}=w_{r}R$

Entering latest in the new kinetic energy expression:

$Kf=1/2*m_{b}.(w_{r}.R)^2+1/4*m_{r}Rw_{r}^2+1/4*m_{r}Rw_{r}^2$

By doing this the kinetic energy expressed only as function of the roller rotating speed. As the roller rotatings speed are the same, and the masses and the radius are known, the the kinetic energy can expressed as:

$Kf=1/2*113\ lb.\1.57\ ft\ w^2+1/4*42\ lb.\ 1.57\ ft\ w^2+1/4*42\ lb.\ 1.57\ ft\ w^2$

$Kf=191\ lb.ft^2 w^2$

Before calculating the work, we will obtain the force in the proper units:

$F=30\ lbf\ *\ 32.16\frac{lb.ft/s}{lbf}$

$F=964.8\ lb.ft/s^2$

Then calculate the work:

$W=964.8\ lb.ft/s\ *\ 2.08\ ft$

$W=2006.8\ lb.ft^2/s^2$

By equating the work and kinetic energy equations and solving for the rotating speed:

$2008.8\ lb.ft^2/s^2=190\ lb.ft^2 w^2$

$w=\sqrt(\frac{2006.8\ lb.ft^2/s^2}{190\ lb.ft^2})$

$w=3.24 1/s$

And finally multiplying the roller rotating speed by the rollers radius, the block speed result:

$V=1.57 ft\ * 3.24 1/s$

$V=5.1 ft/s$