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.

1. Using power = torque x angular velocity only works if the values are constant, and they aren't. Otherwise you have to integrate the torque (which should have units of N-m, not N/m) with the angular displacement to find the work done. (Work = torque x angular displacement) or integrate with the velocity, to get the power. For constant acceleration, this will yield the average speed during that phase, not the final speed.

 

2. Or use the work (from above) and then apply P = W/t

 

3. You found the kinetic energy using (1/2)Iw^2, which is also the work done. Now apply P = W/t

 

 

Check the three methods - they should all give the same answer. (And it's not 75.4 W, assuming your other values are correct)

Ok thanks for that.

 

Am I right in saying by either using the Kinetic energy rotational / time = 18J/3secs = 6 watts

 

or by using Torque x angular displacement = 2 N-m x 9rads = 18 = workdone, then P=W/t again to get 6 watts.

Ok thanks for that.

 

Am I right in saying by either using the Kinetic energy rotational / time = 18J/3secs = 6 watts

 

or by using Torque x angular displacement = 2 N-m x 9rads = 18 = workdone, then P=W/t again to get 6 watts.

 

Right. or torque x average speed = 2 N-m x 3 rad/s = 6 watts