Rotational Mechanics (Conveyor belt)

1. Jan 19, 2007

paul9619

1. The problem statement, all variables and given/known data

A 0.4m diameter, 50KG, solid pulley wheel is used to derive a conveyor belt system on a production line. The belt is very light and it's mass can be taken as zero. The pulley wheel is accelerated from rest at 2 rads/s^2 for 3 secs, then rotated at a constant velocity for a further 10 secs, before being decelerated uniformly back to rest in 2 secs.

1) How far does the belt move during the whole 15 secs?
2) What power is required to accelerate the pulley wheel during the acceleration phase (i.e) during initial 3 secs.

2. Relevant equations

(a) w(angular velocity) = Wo (Intial angular velocity) + & (angular acceleration) x t (time)
(b) @(Pheta) = Wo x t + 1/2&t^2

3. The attempt at a solution

I split the calculations into 3 parts Part A - initial acceleration phase (3 secs). Part B - Constant velocity phase (10 secs). Part C - deacceleration phase.

For part A I used the formula (a) to work out the angular velocity at 6 rads/s. I then used formula (b) to find the angular displacement of 9 rads.

For part B I used the formula (b) to work out the angular displacement as 60 rads. (no acceleration just an intial velocity)

For part C I rearranged formula (a) to find the angular deceleration which was -3 rad/s^2. i then put this value into formula (b) to get 6 rads.

I then added up all the angular displacements to give me 9+60+6 = 75 rads. I then converted this to revolutions, so 75/2Pi to give 11.94. The distance travelled is then the circumference of the wheel (Pi x d) multiplied by the revolutions which I worked out at 15 meters. Thats my answer for question 1.

For question 2 I am now a little confused. I have done it this way.

using the following formulas

Radius of Gyration (k) = 0.707 x radius = 0.707 x 0.2 = 0.1414
Moment if Inertia (J) = mk^2 = 50KG x (0.1414)^2 = 0.999698

I have Torque = J x angular acceleration = 0.999698 x 2 rads/s^2 = 2 N/m

Power = Torque x angular velocity = 2N/m x 6 rads/s = 12 watts.

Now I also have a companion who has attempted this question and he tried it this way:

KE rot = 1/2Jw^2= 17.99 Joules.

He says Torque = KErot/angular displacement = 18/9 = 2 N/m

The power = Torque x speed = 2 x (6 x 2Pi) = 75.4 watts

I am confused as to what the actual answer should be for part 2. Any pointers would be much appreciated.

2. Jan 20, 2007

OlderDan

The angular velocity is not constant. The instantaneous power is changing during the acceleration phase. You cannot use torque times final velocity to find the average power. The average power is the total work done on the wheel divided by the time interval it takes to do the work. The work done is the change in kinetic energy of the wheel.