The system shown below is released from rest. The mb = 35 kg block is 2 m above the ledge. The pulley is a uniform disk with a radius of 10 cm and mass M = 5.0 kg. Assume that the string does not slip on the pulley.

(a) Find the speed of the 35 kg block just before it hits the ledge.

(b) Find the angular speed of the pulley at that time.

(c) Find the tensions in the strings.
I having trouble with this part I was able to get part a and b.
I suppose to use T1-T2=(1/2*M)a

To solve part (c) and find the tensions in the strings, you can use the equation of motion for the system. The equation to use is:

T1 - T2 = (1/2)M * a

Here, T1 is the tension in the string on the side of the 35 kg block, T2 is the tension in the string on the side of the pulley, M is the mass of the pulley (5.0 kg), and a is the acceleration of the system.

To solve for the acceleration, you can use Newton's second law:

ΣF = ma

Considering the forces acting on the system, you have:

1. The tension T1 acting upwards on the 35 kg block.
2. The tension T2 acting downwards on the pulley.
3. The weight of the 35 kg block acting downwards.
4. The weight of the pulley acting downwards.

Taking the positive direction upwards, we can write the equation as:

T1 - T2 - (35 kg) * g - (5.0 kg) * g = (35 kg + 5.0 kg) * a

Where g is the acceleration due to gravity (9.8 m/s^2).

Now, you can substitute the values and solve for the tensions T1 and T2.

To find the tensions in the strings (T1 and T2), we can use the equation of motion for the pulley. The equation is:

T1 - T2 = Iα

where T1 and T2 are the tensions in the strings, I is the moment of inertia of the pulley, and α is the angular acceleration of the pulley.

We can solve this equation by substituting the expression for α in terms of a (linear acceleration):

α = a / r

where r is the radius of the pulley. We can also use the relationship between the linear acceleration a and the angular acceleration α:

a = rα

Substituting these expressions into the equation of motion, we have:

T1 - T2 = (I / r) * (a / r)

The moment of inertia of a disk is given by I = (1/2) * M * r^2, where M is the mass of the disk and r is its radius. Substituting this expression into the equation, we get:

T1 - T2 = ((1/2) * M * r^2 / r^2) * a

Simplifying the equation, we have:

T1 - T2 = (1/2) * M * a

Now we can substitute the expression for acceleration a from part (a), which is a = g / 3, into the equation:

T1 - T2 = (1/2) * M * (g / 3)

where g is the acceleration due to gravity. Substituting the given values:

T1 - T2 = (1/2) * 5.0 kg * (9.8 m/s^2 / 3)

T1 - T2 = 8.17 N

So, the tension in string 2 (T2) is 8.17 N less than the tension in string 1 (T1).

(T1 - T2) = I*alpha = (1/2)M*R^2*(a/R)

(1/2) M*a.

I agree with you. M is the pulley mass.

You need the accleration of the masses, a.

That equals the angular acceleration (which you say you have got by doing b), multiplied by R.