Block-pulley system and rotational dynamics

[Kennedy]

Homework Statement

Two blocks, m1 = 1.0 kg and m2 = 2.0 kg are connected by a light string as shown in the figure. If the radius of the pulley is 1.0 m and its moment of inertia is of the system is 5.0 kg*m^2, the acceleration of the system is (expressed as a fraction of gravitational acceleration g):

Homework Equations

I believe that Newton's Second Law is the only relevant equation (F = ma).

The Attempt at a Solution

I have the acceleration to be in the direction of m2, because that's the heavier block. So, the m2g - T1 = m2a. Solving for T1 that leads to m2g - m2a = T1. On the other side of the pulley we have T2 - m1g = m1a. Solving for T2, we get T2 = m1a + m1g. Now the pulley, T2(radius) - T1(radius) = τ. We know that τ = Iα, and α = a/r.

Now, (m1a +m1g)(r) - (m2g -m2a)(r) = 5(a/r)

How do I solve this in terms of g?

 

[Abhishek kumar]

Homework Statement

Two blocks, m1 = 1.0 kg and m2 = 2.0 kg are connected by a light string as shown in the figure. If the radius of the pulley is 1.0 m and its moment of inertia is of the system is 5.0 kg*m^2, the acceleration of the system is (expressed as a fraction of gravitational acceleration g):

Homework Equations

I believe that Newton's Second Law is the only relevant equation (F = ma).

The Attempt at a Solution

I have the acceleration to be in the direction of m2, because that's the heavier block. So, the m2g - T1 = m2a. Solving for T1 that leads to m2g - m2a = T1. On the other side of the pulley we have T2 - m1g = m1a. Solving for T2, we get T2 = m1a + m1g. Now the pulley, T2(radius) - T1(radius) = τ. We know that τ = Iα, and α = a/r.

Now, (m1a +m1g)(r) - (m2g -m2a)(r) = 5(a/r)

How do I solve this in terms of g?

Which mass is m1 and which is m2? tension produce by m1 is t1? Then write equation accordingly

 

[Kennedy]

Which mass is m1 and which is m2? tension produce by m1 is t1? Then write equation accordingly

The tension in the string attached to m1 is T1, and the tension in the string attached to m2 is T2.

 

[Kennedy]

The tension in the string attached to m1 is T1, and the tension in the string attached to m2 is T2.

m1 is the 1 kg mass, and m2 is the 2 kg mass

 

[Abhishek kumar]

The tension in the string attached to m1 is T1, and the tension in the string attached to m2 is T2.

The tension in the string attached to m1 is T1, and the tension in the string attached to m2 is T2.

m1 is the 1 kg mass, and m2 is the 2 kg mass

Ok then t1-m1g=m1a(say equation 1)
m2g-t2=m2a(say equation 2)
and for rotational motion
t2-t1=Ia/r^2(say equation 3)
Then put the value t1 and t2 from equation 1and 2 in equation 3 you will get the acceleration

 

[haruspex]

, (m1a +m1g)(r) - (m2g -m2a)(r) = 5(a/r)

How do I solve this in terms of g

I do not see any difficulty. Just collect all the a terms on one side and all the g terms on the other. You have the values for r amd the masses.