force of gravity down slope
= m g sin 30 = .5 m g
force of friction up slope = FF
use FF to get angular acceleration
FF r = I alpha
so
alpha = FF r/I = FF r/(2/5 mr^2)
alpha = 5FF/2mr
FF = 2 m r alpha/5
a = r alpha so alpha = a/r
and F = m a
so
m a = .5 m g - FF
m a = .5 m g - 2 m r alpha/5
m a = .5 m g - 2 m a/5 = .1 mg
Caramba !
a = g/10
CHECK MY ARITHMETIC !!!
so
v = g t /10
d = (1/2) (g/10) t^2
Consider a solid ball of mass m = 50g and is placed on a long flat slope that makes an angle of 30° to the horizontal. The ball was initially at rest and was then released such that it rolled down the slope without slipping. With a detailed explanation of your method determine the speed of the ball after it had rolled 4.0m down the slope. You may assume acceleration due to gravity is 10.0ms^-1 and that the moment of inertia of a solid sphere is given by the formula l=2/5mr^2.
5 answers
ay that makes no sense cuz. Fix up look sharp bruv.
Damon, the problem I am trying to solve is explaining it. I don't understand how I'm meant to provide such a detailed explanation as most of the method makes up the description of the answer?
Ayyyyy wag1 Anonymous my mans making no sense, come lecture Thursday 9am and we'll fix up and look sharp G
Deadline's gone babe.
1 answer