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15.
Problem 11.94 (HRW) We are given that two blocks, mass
Solution: Click For PDF Version We will solve this problem by using the Newton’s laws for
translational and rotational motions. As the cord does not slip and the disk
rotates because of torque on it, the relation between the linear acceleration of
the blocks,
We will draw free-body diagrams for the blocks and that for
the disk and apply the laws of motion for setting up the equations of motion
connecting tensions
We will analyse this problem further and demonstrate conservation of energy. We will compute the change in potential energy of the system as the masses move from and show that at any instant the change in the potential energy is equal to the sum of kinetic energy of the two masses and the rotational energy of the disk. As
From the free-body diagrams of masses
The third equation of motion is obtained by equating the net torque on the
disk,
These three equations can be algebraically solved for finding expressions for
The solutions are
Substituting
We will now verify the law of conservation of energy. Let us observe the system at a time when the block
Therefore, the total kinetic energy of the two blocks will be
The angular velocity
Therefore, the rotational energy of the pulley will be
The total motion energy of the blocks and the pulley will be the sum of these two expressions and is
Substituting the expression for
Therefore, the change in potential energy of the system is equal to the change in its motion energy. This is the law of conservation of energy.
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(400g)
and mass
(600g),
are connected by a massless cord that is wrapped around a disk of mass
(500g)
and radius
(12cm).
The disk can rotate without friction about fixed horizontal axis through its
centre. The cord cannot slip on the disk. We have to find the magnitude of the
acceleration of the blocks, the tension
in the
cord at the left and the tension
in the
cord at the right. 
, and the
angular acceleration,
, of the
disk is
.
and the
two masses are connected by an unstretchable massless cord, the magnitude of
acceleration
be the
moment of inertia of the pulley. As it is a disk of mass
and
radius,
,
. 
, to
, the
rotational counterpart of the Newton’s second law of motion. It is
in terms
of
,
,
and the acceleration due to gravity
.
, we find
,
and
.
beginning
from rest. At that instant the block
would have
risen by height
. As the
blocks move with constant acceleration,
of the
pulley at that instant will be
.
.
.
that we
have found by solving the equations of motion, we find that at that instant the
total motion energy of the system is
, which
is the change in the potential energy of the two blocks.