Test 3 sol - the fulcrum. If we imagine that all of the...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Solutions to Test Three 1)The tangential speed of a spinning object can be found by thinking of the  circumference of the wheel as the “distance” traveled in each rotation; the tangential  speed is therefore the number of circumferences per second, or V T  = (2 pi) (20 m) x (1circumference / 50 sec)  V T  = 2.512 m/s 2)You must sum the torques in this problem, and the torque from the measuring stick is  tricky.  Its mass lies all along its length, not in one location at one distance from the  fulcrum like the rock.  The key is to only worry about the stick's center of mass and  the torque it contributes.   The center of mass of the stick will be at its center, positioned one meter to the right of 
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: the fulcrum. If we imagine that all of the stick's mass is concentrated there, we'll have an identical problem from a physics standpoint, and it is one we can solve: T = F x R T total = 0 T total = T rock + T stick's Center of Mass T total = (M rock G) x (-1 m) + (M stick G) x (1 m) = 0 M rock = M stick = 30 kg 3)If the person's usual weight on Earth is 600 N, then by F = MA, F = M A 600 N = M (10 m/s^2) M = 60 kg Now, using F = MA again with A = 26.7 m/s^2 for Jupiter, F = (60 kg) (26.7 m/s^2) = 1602 Newtons...
View Full Document

This note was uploaded on 02/13/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

Page1 / 2

Test 3 sol - the fulcrum. If we imagine that all of the...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online