Block #1 (mass 2.0 kg) is attached to a rope. The rope stretches around a pulley
ID: 1460350 • Letter: B
Question
Block #1 (mass 2.0 kg) is attached to a rope. The rope stretches around a pulley (which has a radius of 4.0 cm and a mass of 0.25 kg) and is attached to block #2 (mass 5.0 kg) on the other side. There is enough friction between the rope and the pulley wheel so that when the rope moves, it rotates the pulley without slipping. Assume the pulley wheel is a uniform cylinder. Released from rest, the system begins to move with a constant acceleration. What is (a) the acceleration of the masses, (b) the torque on the section of the rope attached to block #1, and (c) the torque on the section of the rope attached to block #2?Explanation / Answer
let acceleration of the masses is a m/s^2, with m2 moving downward and m1 moving upward
let tension in the section of the rope connected to mass 1 is T1.
let tension in the section of the rope connected to mass 2 is T2.
then for mass m2, force balance equation:
m2*g-T2=m2*a
==>T2+m2*a=m2*g
==>T2+5*a=5*9.8
==>T2+5*a=49
==>T2=49-5*a...(1)
for mass m1, force balance equation:
T1-m1*g=m1*a
==>T1-m1*a=m1*g
==>T1-2*a=2*9.8
==>T1-2*a=19.6
==>T1=19.6+2*a...(2)
for the pulley, torque balance equation:
radius of the pulley*(T2-T1)=moment of inertia*angular acceleration
==>radius of the pulley*(T2-T1)=0.5*mass of pulley*radius of pulley^2*linear acceleration/radius
==>T2-T1=0.5*0.25*0.04^2*a/0.04^2
==>T2-T1=0.125*a...(3)
using equation 1 and equation 2 in equation 3,
49-5*a-19.6-2*a=0.125*a
==>29.4=7.125*a
==>a=4.1263 m/s^2
using the value of a in equation 1, we get T2=49-5*4.1263=28.369 N
using the value of a in equation 2, we get T1=19.6+2*4.1263=27.853 N
hence the final answers are:
a) acceleration of the masses is 4.1263 m/s^2
b) torque on the section of the rope attached to block 1 is 27.853 N*0.04=1.1141 N.m
c)torque on the section of the rope attached to block 2 is 28.369*0.04=1.1347 N.m