Consider the system shown in the diagram. The pulley is a uniform cylinder with
ID: 1544111 • Letter: C
Question
Consider the system shown in the diagram. The pulley is a uniform cylinder with mass m_3 = 0.400 kg and radius R = 4.00 cm, the other two masses are m_1 = 2.00 kg and m_2 = 1.00 kg, and alpha = 25.0 degree. Assume the rope is massless, there is no slipping of the rope on the pulley, and there is no friction between and the incline. (a) What is the acceleration of m_1 and m_2 (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 m_2 starts from rest 50.0 cm above the ground, where is it when it has speed 0.500 m/s?Explanation / Answer
writing equations of motion for m1
m1*g*sin(alpha) - T1 = m1*a
Tension in the string connected to m1 is T1 = m1*g*sin(alpha) - m1*a
T1 = (2*9.81*sin(25)) - (2*a)
T1 = 8.3 - 2*a
for mass m2
T2 - m2*g = m2*a
T2 = m2*(a+g) = 1*(a+9.8)
Net Torque acting is Tnet = (T2-T1)*R = I*alpha
I is the moemnt of inertia of the pulley = 0.5*m*R^2 = 0.5*0.4*0.04^2 = 3.2*10^-4 kg-m^2
alpha is the angular accelaration of the pulley = a/R = a/(0.04)
Tnet = (T2-T1)*R = 3.2*10^-4*(a/0.04)
((a+9.8)-(8.3+2*a))*0.04 = 3.2*10^-4*(a/0.04)
accelaration is a = 1.25 m/s^2 downward along the incline
angular accelaration is alpha = a/R = 1.25/0.04 = 31.25 rad/s^2
Tension along the rope connected to m1 is T1 = 8.3 - 2*a = 8.3-(2*1.25) = 5.8 N
Tension along the rope connected to m2 is T2 = 1*(a+9.8) = 1.25+9.8 = 11.05 N
b) initial speed is Vo = 0 m/s
final speed is V =0.5 m/sec
accelaration is a = 1.25 m/s^2
Using kinematic equations
V^2-Vo^2 = 2*a*S
0.5^2 - 0^2 = 2*1.25*S
S = 0.1 m = 10 cm
then 50-10 = 40 cm above the ground