A frictionless pulley has the shape of a uniform solid disk of mass 5.00 k g and
ID: 3897525 • Letter: A
Question
A frictionless pulley has the shape of a uniform solid disk of mass 5.00kg and radius 22.0cm . A 1.10kgstone is attached to a very light wire that is wrapped around the rim of the pulley , and the stone is released from rest. As it falls down, the wire unwinds without stretching or slipping, causing the pulley to rotate. How far must the stone fall so that the pulley has 8.50J of kinetic energy?
To see if your results are reasonable, you can compare the final velocity of the stone as it falls down unwinding the wire from the pulley, to the velocity the stone would have if falling the same distance while unconnected to the pulley. What is the velocity of an untethered stone after falling 1.13m from rest?
Explanation / Answer
let M = 5 kg
m = 1.1 kg
r = 22 cm = 0.22 m
let T is the tension in the string.
and alfa is the angular acceleration of disk and a is the acceleration of the falling block.
net force acting on fallinf block,
Fnet = m*g - T
m*a = m*g - T
==> T = m*g - m*a --(1)
net torque acting on disk = T*r*sin(90)
I*alfa = T*r
(M*r^2/2)*(a/r) = T*r
===> T = M*a/2 --(2)
from equations 1 and 2
m*g - m*a = M*a/2
m*g = a(m + M/2)
a = m*g/(m+M/2)
= 1.1*9.8/(1.1 + 5/2)
= 2.994 m/s^2
let d is the distnace travelled
v^2-u^2 = 2*a*d
v = sqrt(2*a*d)
KE = 0.5*m*v^2
= m*a*d (kinetic energy gained by falling object)
loss of potential energy = gain in kinetic enrgy
m*g*d = m*a*d + 8.5
d*(m*g - m*a) = 8.5
d = 8.5/(m*g - m*a)
= 8.5/(1.1*9.8 - 1.1*2.994)
= 1.135 m
we know,
v^2- u ^2 = 2*a*d
here, u = 0
so v = sqrt(2*a*d)
= sqrt(2*2.994*1.3)
= 2.79 m/s <<<<<<<<<<=------Answer