A 9.00 kg mass is connected by a light cord to a 1.00 kg mass on a smooth surface as shown in the figure. The pulley rotates about a frictionless axle and has a moment of inertia of 0.300 k∙m2 and a radius of 0.500 m. Assuming that the cord does not slip on the pulley, find (a) the acceleration of the two masses (b) the tension T1 (attached to the hanging mass).(b) the tension T2

For m1: m1•a=T1

For m2: m2•a=m2•g – T2
For pulley: I•ε = M => I•a/R= (T2-T1)R => (T2-T1) = I•a/R²
a(m1+m2+ I/R²)=m2•g
a= m2•g/(m1+m2+ I/R²)
T1= m1•m2•g/(m1+m2+ I/R²)
T2 = m2•g -m2•a=...

I am geting .875 for my acceleration and its saying its wrong! Was your value this small?

For m1: m1•a=T1 (block m1=1 kg is on the surface )

For m2: m2•a=m2•g – T2 (m2=9 kg)
For pulley: I•ε = M => I•a/R= (T2-T1)R => (T2-T1) = I•a/R²
a(m1+m2+ I/R²)=m2•g
a= m2•g/(m1+m2+ I/R²) =9•9.8/{1+9+(0.3/0.25)}=7.885 m/s²
T1= m1•m2•g/(m1+m2+ I/R²)

FOr my Tension 1 I am getting 7.875 and its wrong, im so confused what im doing wrong!

T1=m1•a=1•7.885 = 7.885 N

Oh, it's time to put on my physics hat! Let's start with the first question:

(a) To find the acceleration of the two masses, we can use Newton's second law. Since we're dealing with a pulley, we need to consider the moment of inertia of the pulley as well. The tension on one side of the cord (T1) is connected to the hanging mass, and the tension on the other side of the cord (T2) is connected to the mass on the smooth surface.

Now, let me think of a joke to lighten the mood... Ah! Why did the scarecrow win an award? Because he was outstanding in his field!

Alright, back to business. Let's call the acceleration of the two masses "a". We can set up the following equations:

For the hanging mass:
T1 - m1 * g = m1 * a -- (1)

For the mass on the smooth surface:
T2 - m2 * g = m2 * a -- (2)

Now, let's move on to the second question:

(b) To find the tension T1, we can use equation (1). Substituting the known values, we get:

T1 - (9.00 kg) * (9.8 m/s^2) = (9.00 kg) * a

We also need to consider the moment of inertia of the pulley. The torque exerted by the net tension on the pulley is equal to the product of the moment of inertia and the angular acceleration. This can be expressed as:

(T2 - T1) * r = (0.300 k∙m^2) * α -- (3)

But wait, I need to tell you another joke before we proceed! Why don't scientists trust atoms? Because they make up everything!

Alright, focus back on the problem. We also know that the linear acceleration of the hanging mass is equal to the linear acceleration of the outer edge of the pulley (a = r * α). Substituting this in equation (3), we get:

(T2 - T1) = (0.300 k∙m^2) * (a / r) -- (4)

Now we have two equations: (1) and (4). By solving them simultaneously, we can find the values for a, T1, and T2.

I hope this explanation brings a smile to your face. If you need further assistance with the calculations, feel free to ask!

To find the acceleration of the two masses and the tensions T1 and T2, we can use Newton's second law and the principles of rotational dynamics.

(a) To find the acceleration of the two masses, we need to set up the equations of motion for each mass separately.

Let's assume the acceleration of the system (both masses) is a, and let's consider the direction of motion to be downward for the hanging mass and clockwise for the rotation of the pulley.

For the hanging mass:
We can write the equation of motion using Newton's second law:
m1 * a = T1 - m1 * g, where m1 is the mass of the hanging mass, a is the acceleration, T1 is the tension in the string attached to the hanging mass, and g is the acceleration due to gravity.

For the pulley:
We can write the equation of motion using rotational dynamics:
I * α = T2 * R - T1 * R, where I is the moment of inertia of the pulley, α is the angular acceleration, T2 is the tension in the string attached to the pulley, and R is the radius of the pulley.

Since the pulley is assumed to be massless and frictionless, the tension in both strings is the same (T1 = T2), and the angular acceleration α is equal to the linear acceleration a divided by the radius of the pulley (α = a / R).

Combining these equations, we can solve for the unknowns.

For the hanging mass:
m1 * a = T1 - m1 * g

For the pulley:
I * (a / R) = T2 - T1

Also, we know that the magnitudes of T1 and T2 are equal, so we can substitute T1 with T2 in the above equations.

For the hanging mass:
m1 * a = T2 - m1 * g

For the pulley:
I * (a / R) = T2 - T2

Now we have two equations and two unknowns (T2 and a). We can solve these equations simultaneously to find the values.

(b) To find the tension T1 (attached to the hanging mass), we can substitute the value of acceleration a obtained from the previous calculation into the equation for the hanging mass:
T1 = m1 * g + m1 * a

(b) To find the tension T2, we can use the same equation as above since T2 is equal to T1:
T2 = m1 * g + m1 * a

Now, plug in the known values (mass of the hanging mass, the acceleration due to gravity, moment of inertia of the pulley, and radius of the pulley), and solve the equations to find the desired values.