Here is a link to the problem with a picture. I think the answerer on cramster did it wrong.
http://www.cramster.com/answers-apr-11/physics/torque-masses-m1-45-kg-m2-70-kg-connected_1275174.aspx#post-1243203-comment-76306
Two masses m1 = 4.5 kg and m2 = 7.0 kg are connected via a piece of rope wound around a pulley as
shown in the figure above. The 4.5 kg mass is sliding along the table surface. The coefficient of
kinetic friction between the 4.5 kg mass and the table surface is μ K=0.40. Assume g=9.8 m/s2. Also assume the rope has no mass and that pulley is frictionless.
The mass m1 is accelerating to the right with acceleration a. The pulley has a moment of inertia of 2 kgm^2 and a radius of 0.05m.
a) Find the acceleration of the mass.
b)Find the tension in the cable attached to m2.
c)Find the energy dissipated by friction after 0.7 seconds.
http://www.cramster.com/answers-apr-11/physics/torque-masses-m1-45-kg-m2-70-kg-connected_1275174.aspx#post-1243203-comment-76306
Two masses m1 = 4.5 kg and m2 = 7.0 kg are connected via a piece of rope wound around a pulley as
shown in the figure above. The 4.5 kg mass is sliding along the table surface. The coefficient of
kinetic friction between the 4.5 kg mass and the table surface is μ K=0.40. Assume g=9.8 m/s2. Also assume the rope has no mass and that pulley is frictionless.
The mass m1 is accelerating to the right with acceleration a. The pulley has a moment of inertia of 2 kgm^2 and a radius of 0.05m.
a) Find the acceleration of the mass.
b)Find the tension in the cable attached to m2.
c)Find the energy dissipated by friction after 0.7 seconds.
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Denote the tension in the horizontal piece of rope by Tx
Denote the tension in the vertical piece of rope by Ty
Both M1 and M2 will accelerate with the same acceleration a (non-elastic rope)
Then we have three equations of motion:
M1 * a = Tx - mu M1 g
M2 * a = M2 g - Ty
I a/R = (Ty - Tx) R
A little rewriting gives
Tx = M1 ( a + mu g)
Ty = M2 ( g - a)
So the third is
I a / R = (M1 - M2) a R + (M2 - mu M1) g R
or
I a = (M1 - M2) a R^2 + (M2 - mu M1) g R^2
and so
a = (M2 - mu M1) g R^2 / ( I - (M1 - M2) R^2 )
If you plug in the values given you will have a.
Then the equation for Ty will give the tension in the rope at M2
Also, if you have a, you will be able to calculate the distance M1 has moved in 0.7 seconds.
It is the frictional force (mu M1 g) that you have to multiply with this distance to know the amount of work the frictional force performed (and that equals the amount of energy dissipated.
Alternatively, calculate the potential energy of the block M2 at the beginning and end of the 0.7 second movement. Calculate also the speed at t=0.7 s. Then you will have initial mechanical energy, final mechanical energy, and hence the missing part will be dissipated. It is very instructive to se these two methods yield the same result.
Denote the tension in the vertical piece of rope by Ty
Both M1 and M2 will accelerate with the same acceleration a (non-elastic rope)
Then we have three equations of motion:
M1 * a = Tx - mu M1 g
M2 * a = M2 g - Ty
I a/R = (Ty - Tx) R
A little rewriting gives
Tx = M1 ( a + mu g)
Ty = M2 ( g - a)
So the third is
I a / R = (M1 - M2) a R + (M2 - mu M1) g R
or
I a = (M1 - M2) a R^2 + (M2 - mu M1) g R^2
and so
a = (M2 - mu M1) g R^2 / ( I - (M1 - M2) R^2 )
If you plug in the values given you will have a.
Then the equation for Ty will give the tension in the rope at M2
Also, if you have a, you will be able to calculate the distance M1 has moved in 0.7 seconds.
It is the frictional force (mu M1 g) that you have to multiply with this distance to know the amount of work the frictional force performed (and that equals the amount of energy dissipated.
Alternatively, calculate the potential energy of the block M2 at the beginning and end of the 0.7 second movement. Calculate also the speed at t=0.7 s. Then you will have initial mechanical energy, final mechanical energy, and hence the missing part will be dissipated. It is very instructive to se these two methods yield the same result.