testing header
Math Goodies is a free math help portal for students, teachers, and parents.
Free Math
Interactive Math Goodies Software

Buy Math Goodies Software
testing left nav
Math Forums @ Math Goodies
Math Forums @ Math Goodies
Home | Profile | Register | Active Topics | Members | Search | FAQ
Save Password
Forgot your Password?

 All Forums
 Homework Help Forums
 Miscellaneous Math Topics
 New Topic  Reply to Topic
 Printer Friendly
Author Previous Topic Topic Next Topic  

Junior Member

4 Posts

Posted - 09/20/2008 :  09:24:14  Show Profile  Reply with Quote
A 40cm diameter 50kg, solid pulley wheel is used to drive a conveyor belt system on a production line. The belt is very light and its mass can be taken as zero. The pulley wheel is accelerated from rest at 2 rad/s^2 for 3 s and then rotated at constant velocity for a further 10s before being decellarted uniformly back to rest in 2s.

1 How far does the belt move during the whole 15s
2 What power is required to accelrate the pulley wheel during the acceleration phase i.e the initial 3 s

For question 1i did the following (If i've went wrong please tell me!)

Multiplied the diameter by pi to get the circumfrence. 1256.64mm
Then used the formula theta = (omega0 + omega)t all divided by 2 to get my angular displacement for the accelration in 3 secs. - 9 Rads
Then used the formula theta = (omega0 + omega)t all divided by 2 to get my angular displacement for the decceleration in 2 secs - 6 Rads
For the 10 secs of constanst velocity i got 60 rads as the spped was 6 rads per second.

I added all three together and divided by 2pi to get the number of revolutions - 11.94
I then multipled the number of revolutions by the circumfrence of the wheel - 15000mm (If thats right I'll be amazed!)

I thought i had a fair idea how to do number 2 however After looking through my books I thought I had it all sorted, then i reread a part of the distance learning package, It said that if the mass is not all at the same radius then J=mk^2 where K is the radius of gyration. It then says for any questions given the radius o gyration will be given. Am I right in saying that I have not been given enough information in this question to answer it correctly?
Go to Top of Page

Advanced Member

5634 Posts

Posted - 09/20/2008 :  11:49:59  Show Profile  Reply with Quote
75 radians of angular displacement is correct.

are you familiar with the formula for arc length?

s = r

s = (20 cm)(75 radians) = 1500 cm = 15 m

power = work/time = (change in rotational kinetic energy)/time

rotational kinetic energy = (1/2)Iw

where I = rotational inertia of the pulley (usually (1/2)Mr since a pulley is normally a solid disk ... for your pulley, I = (1/2)(50 kg)(0.2 m) = 1 kg-m)
w (omega) = angular speed

P = [(1/2)(1)(6) - 0]/3 = (18 J)/(3 s) = 6 watts
Go to Top of Page
  Previous Topic Topic Next Topic  
 New Topic  Reply to Topic
 Printer Friendly
Jump To:
Math Forums @ Math Goodies © 2000-2004 Snitz Communications Go To Top Of Page
This page was generated in 0.09 seconds. Snitz Forums 2000
testing footer
About Us | Contact Us | Advertise with Us | Facebook | Blog | Recommend This Page

Copyright © 1998-2014 Mrs. Glosser's Math Goodies. All Rights Reserved.

A Hotchalk/Glam Partner Site - Last Modified 22 Oct 2014