When the motor in the figure below raises the m = 1010kg mass, it produces a tension of 1.04E+4N in the cable on the right side of the pulley. The pulley has a moment of inertia of 78.9kgm^2 and a radius of 0.736m. The cable rides over the pulley without slipping. Determine the acceleration of the m = 1010kg mass. Use g=9.81m/s^2.
Draw free-body diagrams of the mass and the pulley. Do not assume that the tension in the cable is the same on both sides of the pulley.
(the diagram has a pulley and the motor is on the right side and it raising the mass which is on the left side of the pulley)
To determine the acceleration of the 1010 kg mass, we need to apply Newton's second law of motion. In order to do that, let's first draw free-body diagrams for both the mass and the pulley.
First, let's consider the free-body diagram of the mass:
1. Draw a dot to represent the mass.
2. Draw an arrow pointing upwards from the dot to represent the weight of the mass (mg), where m = 1010 kg and g = 9.81 m/s^2.
3. Draw another arrow pointing to the right from the dot to represent the tension in the cable on the left side of the pulley. Note that the tension here is unknown, so we'll label it as T1.
4. Finally, draw a arrow pointing to the left from the dot to represent the acceleration of the mass. We'll label this as a.
Next, let's consider the free-body diagram of the pulley:
1. Draw a circle to represent the pulley.
2. Draw an arrow pointing downwards from the circle to represent the weight of the pulley (mg), where m is the mass of the pulley and g = 9.81 m/s^2.
3. Draw two arrows on either side of the pulley, pointing in opposite directions. These represent the tensions in the cable on the left and right sides of the pulley. We'll label these as T1 and T2, respectively.
Now, we can apply Newton's second law of motion to the mass:
ΣF = ma
In the vertical direction:
T1 - mg = 0 (since the mass is not moving up or down)
Therefore, T1 = mg.
In the horizontal direction:
T2 - T1 = ma
We can substitute the known values into the equation:
T2 - mg = ma
We know that the tension in the cable on the right side of the pulley is 1.04E+4 N, so T2 = 1.04E+4 N.
Substituting these values into the equation:
1.04E+4 N - (1010 kg)(9.81 m/s^2) = (1010 kg) a
Now, we can solve for the acceleration (a):
a = (1.04E+4 N - (1010 kg)(9.81 m/s^2)) / (1010 kg)
a ≈ 0.220 m/s^2
Therefore, the acceleration of the 1010 kg mass is approximately 0.220 m/s^2.