In the figure, block 1 has mass m_1 = 464 g, block 2 has mass m_2 = 571 g, and t
ID: 1570166 • Letter: I
Question
In the figure, block 1 has mass m_1 = 464 g, block 2 has mass m_2 = 571 g, and the pulley is on a frictionless horizontal axle and has radius R = 5.16 cm. When released from rest, block 2 falls 70.3 cm in 4.52 s without the cord slipping on the pulley. (a) What is the magnitude of the acceleration of the blocks? What are (b) tension T_2 (the tension force on the block 2) and (c) tension T_1 (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. (a) Number Units (b) Number Units (c) Number Units (d) Number Units (e) Number UnitsExplanation / Answer
Assume block 1 is on a flat, frictionless table and m2 drops vertically
Caculate acceleration: 0.703 = 0.5*a*4.52^2 => a = 1.406/20.43 = 0.069m/s^2 <------- a)
Since m1 is attached to m2, m1 also accelerates at 0.069 which means the force pulling m1 must be: 0.464*0.069 = 0.0319N
Is T1 the tension in the rope pulling block 1,T1 = 0.0319N <--------- c)
If block 2 were stationary, T2 would be 0.571*g = 5.6N
T2 must be less because block 2 is accelerating downward at 0.069m/s^2.
If block 2 were accelerating downward at 9.81m/s^2, T2 would = 0. Thus
T2 = 0.571*[9.81-.069] = 5.56 N <------------- b)
The difference in T1 and T2 is the force rotating the pulley
5.56 - 0.0319 = 5.53N
We know that Torque = I*alpha = F*R where I is the rotational moment of inertia, alpha is the angular acceleration and the radius R = 0.0516m
F*R = 5.53*0.0516 = 0.285 = I*alpha
We know the acceleration of the edge of the pulley a = 0.069m/s^2
We know a = R*alpha => alpha = a/R = 0.069/0.0516 = 1.34 rad/s^2 <---------- d)
e) I = F.R/alpha
I = 0.285/1.34=0.213 (kg. m2)