A 2.01kg stone is tied to a thin, light wire wrapped around the outer edge of th
ID: 1264061 • Letter: A
Question
A 2.01kg stone is tied to a thin, light wire wrapped around the outer edge of the uniform 8.19kgcylindrical pulley shown in the figure below (Figure 1) . The inner diameter of the pulley is 59.2cm , while the outer diameter is 1.47m . The system is released from rest, and there is no friction at the axle of the pulley.
1) What is the rotational inertia of the pulley?
2) Find the acceleration of the stone.
3) Find the tension in the wire.
A 2.01kg stone is tied to a thin, light wire wrapped around the outer edge of the uniform 8.19kgcylindrical pulley shown in the figure below (Figure 1) . The inner diameter of the pulley is 59.2cm , while the outer diameter is 1.47m . The system is released from rest, and there is no friction at the axle of the pulley. 1) What is the rotational inertia of the pulley? 2) Find the acceleration of the stone. 3) Find the tension in the wire. 4) Find the angular acceleration of the pulley.Explanation / Answer
m = 2.01 kg
M = 8.19 kg
R1 = 59.2 cm = 0.592 m
R2 = 147 cm = 1.47 m
I = M*(R1^2+R2^2)/2
= 8.19*(0.592^2+1.47^2)/2
= 10.28 kg.m^2
let T is tension in the string and a is the acceleration of the falling block
m*a = m*g - T
T = m*(g-a) ---(1)
Torque acting on pulley
T*R2 = I*alfa
T*R2 = I*a/R2
T = I*a/R2^2 --(2)
from equations 1 and 2
m*(g-a) = I*a/R2^2
m*g - m*a = Ia/R2^2
a*(m + I/R2^2) = m*g
a = m*g/(m + I/R2^2)
= 2.01*9.8/(2.01+ 10.28/1.47^2)
= 2.913 m/s^2
b) T = 2.01*(9.8-2.913) = 13.84 N
c) alfa = a/R2 = 2.913/1.47= 1.98 rad/s^2
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