Consider the system shown in the diagram. The pulley is a uniform cylinder with
ID: 1323357 • Letter: C
Question
Consider the system shown in the diagram. The pulley is a uniform cylinder with mass m3 =0.80 kg and radius R = 6.0 cm, the other two masses are m1 = 2.5 kg and m2 = 1.5 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? 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, how fast is it going when it hits the ground?
Consider the system shown in the diagram. The pulley is a uniform cylinder with mass m3 =0.80 kg and radius R = 6.0 cm, the other two masses are m1 = 2.5 kg and m2 = 1.5 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? 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, how fast is it going when it hits the ground?Explanation / Answer
a)
Here , Let the accleration of block is a
then , Using second law of motion ,
a = Fnet / m
a = (1.5 - 2.5sin(30))*g/(1.5 + 2.5 + 0.5*.80*0.06^2/.06^2)
a = 0.56 m/s^2
the acceleration of m1 and m2 is 0.56 m/s^2
angular acceleraion = a/r
angular acceleraion = 0.56/.06
angular accleration is 9.28 rad/s^2
and for tension
m2g - T = m2a
1.5 * 9.8 - T = 1.5 * 0.56
T = 13.86 N
the tension in the string is 13.9 N
b)
Now , Using third equation of motion ,
v = sqrt(2gh)
v = sqrt(2*9.8*0.60)
v = 1.08 m/s
the speed of m2 when it hits the ground is 1.08 m/s