In the figure, block 1 has mass m1 = 450 g, block 2 has mass m2 = 570 g, and the
ID: 1459919 • Letter: I
Question
In the figure, block 1 has mass m1 = 450 g, block 2 has mass m2 = 570 g, and the pulley is on a frictionless horizontal axle and has radius R = 4.6 cm. When released from rest, block 2 falls 73 cm in 5.3 s without the cord slipping on the pulley. (a) What is the magnitude of the acceleration of the blocks? What are (b) tension T2 (the tension force on the block 2) and (c) tension T1 (the tension force on the block 1)? (d) What is the magnitude of the pulley?s angular acceleration? (e) What is its rotational inertia? Caution: Try to avoid rounding off answers along the way to the solution. Use g = 9.81 m/s^2.Explanation / Answer
(a) Block travels 73 cm in 5.3sseconds therfore the acceleration is
s = ut+1/2at2
73 = 0.5*a*5.32
a = 5.1975 cm/s
(b) When we draw the FBD for m2 and write the euation
let T2 is the tension in the string.
T2 = m2g - m2a = (0.57)(9.81 - (5.1975/100)) = 5.562 N
(c) Tension in the block 1
T1 = m1g+ m1a = (0.45)(9.81 + 0.05197) = 4.437 N
(d) We knwo that
Torque = I*angular acceleration
and also
a = (m2-m1)g / (m1+m2 + (I/R2)) where I is the moment of inertia of the pulleyand R is the radius of the pulley.
Putting all the values we get
I = 3.326 kg-cm2
Therefore
(T2 - T1)*R = I*angular acceleration
angular acceleration = 155.57 rad/s2