Question

A large reel of paper is installed at the end of a paper machine.

2. A large reel of paper installed at the end of a paper machine. d1 = 2m, Tension on paper = 1000N. At t0, J = 1000kgm^2 and the speed ofReel#1 is 60rpm. Assume no loss. a. Calculate the motor power and torque at t0. b. Suppose the paper was torn by an accident at to and the reels are disconnected. The motor was running until it was tripped (motor torque = 0) after 1 sec. Calculate the speed of Reel#1 0.5 sec after the accident (in rpm). c. Sketch speed (in rpm), torque. and power from t0 to t0+ 2 sec.

Answer 1

a)

On a rotational system the torque is:

$\tau =F\times l$

Since the force and radius are perpendicular:

$\tau =1000N\cdot 1m=1000\, Nm$

The power is:

$P=\tau \cdot \omega =1000\, Nm\cdot 6.28\, rad/s=6280\, W$

b)

We are looking for the angular velocity at t=0.5 sec. The equation that describes the angular velocity is:

$\omega =\omega _{0}-\alpha t$

To find the angular acceleration:

$\tau =I\alpha$

$\alpha =\frac{\tau }{I}=\frac{1000}{1000}=1\, \frac{rad}{s^{2}}$

Since the speed is decreasing the acceleration is negative:

$\omega =\omega _{0}+\alpha t=60-4.77=55,23\, rpm$

remember 0.5 rad/s=4.77 rpm

c)

Applying the equation: $\omega =\omega _{0}-\alpha t$ The graph of the angular velocity until it stops (12.57 sec)

w (rad/s) 60 t (s) 0 12.57

The power decreases steadily according to the equation:

$P=\tau \cdot \omega$ , When the torque goes to zero (1 sec) the power also goes to zero

P (W) 6280

The torque remains constant until it is tripped (t=1 sec)

torque (Nm) 1000 Nm