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Question

two blocks, of mass m1 = 250 g and m2 = 700 g, are connected by a massless cord that is wrapped around a uniform disk of mass M = 500 g and radius R = 12.0 cm. The disk can rotate without friction about a fixed horizontal axis through its center; the cord cannot slip on the disk. The system is released from rest.

a) Find the magnitude of the acceleration of the blocks.
=.......m/s^2
(b) Find the tension T1 in the cord at the left.
=.......N
(c) Find the tension T2 in the cord at the right.
=........N

Answer 1

Look at FBDs of each component of the system using downward motion of the 700 g block as positive

The 700g block

0.7*9.81-T2=0.7*a

The 250 g block
T1-0.25*9.81=0.25*a

now the disk

angular motion is analogous to translational in that the sum of torques is equal to the moment of inertia times angular acceleration.
Angular acceleration is related to translational as
alpha=a/R

0.12*(T2-T1)=I*a/R

I for a uniform disk is
I=m*R^2/2

so
0.12*(T2-T1)=.5*.5*.12*a

Since there is no slipping in the system all of the a's for each component are equal

The three equations are

0.12*(T2-T1)=.5*.5*.12*a
T1-0.25*9.81=0.25*a
0.7*9.81-T2=0.7*a

solve for the three variables

(T2-T1)=a/4
T1=(a+9.81)/4
0.7*(9.81-a)=T2

0.7*(9.81-a)-(a+9.81)/4=a/4

1.8*9.81/4.8=a

a=3.68 m/s^2
T1=3.37 N
T2=4.29 N

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