In the figure, block 1 has mass m 1 = 440 g, block 2 has mass m 2 = 520 g, and t
ID: 1350277 • Letter: I
Question
In the figure, block 1 has mass m1 = 440 g, block 2 has mass m2 = 520 g, and the pulley is on a frictionless horizontal axle and has radius R = 4.7 cm. When released from rest, block 2 falls 73 cm in 4.7 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/s2.
Explanation / Answer
s=ut+1/2 at^2
for block 2, u=0 => a= 0.73/(1/2* 4.7^2) = 0.066 m/s^2
a) So, the acceleration of Masses=0.066 m/s^2
b) Equations individually for masses=
IF T1 is the tension for block1 and T2 for block2,
M2g-T2=M2a------(1)
T1-M1g-M1a------(2)
Substituting masses and acceleration, acceleration is same for both blocks as they are tied together.
=> equation 1 => T2= 0.52( 9.8-0.066)=5.06N
c) => Equation 2 => T1= 0.44( 0.066+9.8)=4.34N
d) The torques on the pulley are due to the two tensions (T1 and T2) torquing it in opposite directions.
Counter-clockwise torque: (T1)R
Clockwise torque: (T2)R
Net torque: (T2)R (T1)R = (T2-T1)R
Net torque = I
(T2-T1)R = I
Remember that , pulley's angular acceleration= = a/R,=0.066/0.047=1.404 rad/s^2
so:
(T2-T1)R = I(a/R) <--- EQUATION 3
Substituting,
Rotational inertia for pulley =I= (5.06-4.34)*0.047/(1.404)=0.024 kg.m^2