Question

A stone is suspended from the free end of a wire that is wrapped around the outer rim of a pulley, as shown in the figure (see the figure ). The pulley is a uniform disk with mass 10.4 and radius 51.0 and turns on frictionless bearings. You measure that the stone travels a distance 12.4 during a time interval of 2.80 starting from rest.

Answer 1

h=a•t²/2
a = 2•h/t² =2•12.4/2.8²=3.16 m/s².
According to the Newton’s Second Law for rotation
the torque is the moment of inertia times
an angular acceleration
M=I•ε,
M = T•R.
I=m•R²/2.
ε =a/R.
T•R = (m•R²/2)• (a/R)=
= m•R• a/2.
T= m•a/2.
The tension is
T=10.4•3.16/2=16.4 N.
Forces on the stone are
ma = mg – T,
m = T/(g-a) =16.4/(9.8-3.16)=2.47 kg.