In the figure, block 1 has mass m 1 = 460 g, block 2 has mass m 2 = 550 g, and t
ID: 1451928 • Letter: I
Question
In the figure, block 1 has mass m1 = 460 g, block 2 has mass m2 = 550 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 5.1 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
As the pulley has mass and radius is given, it will have moment of inertia. So, the tension on both the strings will be different, even though the blocks will have same acceleration.
If you use the relation s=(1/2)at*t
you will get the acc. as 0.056132256 m/s^2
The equation for 0.550kg block is
0.550 * a = 0.550 g - T2 ( where T2 is the tension in that string and a is the acceleration i.e, 0.06
T2 = 5.364627259 N
The equation for 0.46kg block is
0.46 * 0.056132256 = T1 - 0.46 g ( T1 is the tension in the other part of the string)
T1 = 4.538420838 N
also, a= r *(p) where p is the angular acceleration of the pulley.
therefore, p = a/r = 0.056132256/0.046 = 1.220266435 ( take all the units in S.I. system)
These tensions provide a torque to the pulley. As it rotates in clockwise direction ( if the 0.550 kg block is on your right ), there is a net clockwise torque acting.
torque = force * distance of the force from the axis
= ( T2-T1 ) * r
but torque is also equal to ( I * p ) where I is the moment of inertia and p is the angular acceleration.
you now know ( T2-T1) = 0.826206421 N
r = 0.046m
p = 1.220266435
substitute
and get
I which is 0.031145243 kg-m^2