Consider the system shown in the diagram. The pulley is a uniform cylinder with
ID: 1432367 • 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 ? = 30 degrees . 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
Tension in string above m2 = T2
Tension in string with m1 = T1
moment of inertia of pulley = I
I = mR^2
(T2 - T1)R = I(o) = Ia/R [o is angular acc of the pulley, R is radius of pulley, a is acc of the blocks]
m2g - T2 = m2a
T1 - m1*g*sin(alpha) = m1a
=>m2 - m1*g*sin(alpha) = a(I/R^2 + m1 + m2)
=> a = [m2g - m1*g*sin(alpha)]/[mp + m1 + m2] = [1 - 2.5*sin(30)]*9.8/[2.5+1+0.6] = -0.5975 m/s/s
a) acc of m1 : a = 0.5975 m/s/s down the incline
acc of m2 : a = 0.5975 m/s/s down
angular acc of pulley = a/R = 14.939 rad/s^2
T1 = -2.5*0.5975 + 2.5*9.8*sin(30) = 10.75 N
T2 = 9.8 + 0.5975 = 10.3975 N
b) v = u + at
1 = 0.5975*t
t = 1.673 s
to find the position
position 1 = 60cm above the gound
position 2 = x cm above the ground
Total distance travelled = (60-x)/sin(30) = d
Now d = 0.5*a*t^2 = 0.5*0.0597*1.673^2 = 1.066cm
1.066*sin(30) = 60 - x
0.5332 = 60 - x
x = 59.4667cm