PHYSICS!!

A person thought it would be a good idea to change a tire on the side of a hill. The tire has a width
of W = 0.30 m and a radius of R = 0.40 m. The inside of the wheel is empty, but the wheel is solid
from a radius of R/2 to R with uniform density. The density of the material is 2000 kg/m3
.
(a) What is the mass, weight and moment of inertia of the wheel?
(b) If the wheel starts from rest and rolls down a 4.00 m ramp, 30.0
o
relative to the horizontal, what
is the wheel’s speed at the bottom of the ramp?
(c) What average torque do you need to apply to stop it at the bottom of the ramp in one revolution?

asked by Vanessa
  1. first mass of rubber
    = 2000 * .3 * ∫ 2 pi r dr
    from r = .2 to r = .4

    = 600 (2 pi) (1/2)(.16-.04)

    = 226 kg

    226 * 9.81 = 2219 Newtons weight

    I = 2000*.3 *∫ 2 pi r^3 dr
    = 3770 (1/4)(.4^4-.2^4)
    = 22.6
    ===================

    posted by Damon
  2. change in potential energy = m g h
    = 2219*4*sin 30 = 4438 Joules

    = (1/2)mv^2 + (1/2)I w^2
    but w = v/r = v/.4 = 2.5 v
    so
    4438 = (1/2)(226)v^2 + (1/2)(22.6)(6.25 v^2)
    or
    4438 = 184 v^2
    so
    v = 24.2 m/s
    and w = 60.4 radians/s

    Torque * angle = change in Ke
    T * 2 pi = (1/2) I w^2

    posted by Damon

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