Consider the system shown in the diagram. The pulley is a uniform cylinder with
ID: 1430822 • Letter: C
Question
Consider the system shown in the diagram. The pulley is a uniform cylinder with mass m3 = 0.60 kg and radius R = 4.0 cm, the other two masses are m1 = 2.5 kg and m2 = 1.0 kg, and ? = 30o . Assume the rope is massless, there is no slipping of the rope on the pulley, and there is no friction between m1 and the incline.
(a) What is the acceleration of m1 and m2 (both magnitude and direction)? What is the angular acceleration of the pulley? What are the tensions in the rope (both to the left of the pulley and to the right of the pulley?
(b) If m2 starts from rest 60 cm above the ground, where is it when it has speed 1.0 m/s?
m2Explanation / Answer
the system moves with acceleration a
m1 moves down
for m1
m1*g*sin30 - T1 = m1*a
(2.5*9.8*sin30)-T1 = (1.25*a)..........(1)
for m2
T2 - m2*g = m2*a
T2 - (1*9.8) = 1*a ...........(2)
for m3
net torque = I*angular acceleration
(T1 - T2)*R = I*a/R
I = (1/2)*m3*R^2
(T1-T2) = (1/2)*m3*a
T1 - T2 = (1/2)*0.6*a .............(3)
solving 1 , 2 , 3
T1 = 11.04 N
T2 = 10.7 N
a = 0.96 m/s^2
for m1
acceleration = a = 0.96 m/s^2
direction down the incline
for m2
acceleration = a = 0.96 m/s^2
direction up wards
for pulley
angular accelration = a/R = 0.96/0.04 = 24 m/s^2
tension
left rope T1 = 11.04 N
right rope = T2 = 10.76 N
(b)
from equation of motion
vf^2 - vi^2 = 2*ay*dy
vf^2 - vi = 2*-a*sin30*(yf-yi)
1^2 - 0 = -2*0.96*(yf-0.6)
f = 7.9 cm from ground