A 1.6 kg mass and a 6 kg mass are connected by a massless string over a pulley that is in the shape of a solid wheel having a radius 0.13 m and mass 5.5 kg. Determine the linear acceleration of the blocks.
m₁a=- m₁g+T₁ …….(1)
- m₂a=T₂- m₂g ……(2)
Iε=(T₂-T₁)R => T₂-T₁ = Iε/R=Ia/R²
If the wheel is disc I=mR²/2 =>
T₂-T₁= Iε/R= mR²a/2R²=ma/2
subtract (2) from (1)
m₁a+ m₂a= - m₁g +T₁- T₂+m₂g.
a(m₁+ m₂) = g(m₂-m₁) - (T₂-T₁)=
= g(m₂-m₁) - ma/2.
a(m₁+ m₂+m/2)= g(m₂-m₁)
a= g(m₂-m₁)/(m₁+ m₂+m/2)
(If the wheel is the hoop I=mR² =>
T₂-T₁= Iε/R= mR²a/R²=ma)
To determine the linear acceleration of the blocks, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.
First, let's consider the forces acting on the 1.6 kg mass and the 6 kg mass separately.
For the 1.6 kg mass:
- The weight force (mg), where g is the acceleration due to gravity (9.8 m/s²) and m is the mass (1.6 kg).
- The tension force in the string (T) directed upwards.
For the 6 kg mass:
- The weight force (mg), where g is the acceleration due to gravity (9.8 m/s²) and m is the mass (6 kg).
- The tension force in the string (T) directed downwards.
Since the masses are connected by a massless string over a pulley, the tension force in the string is the same for both masses.
Now, let's consider the forces acting on the pulley:
- The weight force (mg), where g is the acceleration due to gravity (9.8 m/s²) and m is the mass of the pulley (5.5 kg).
- The tension force in the string (T) acting tangentially along the circumference of the pulley, causing it to rotate.
Since the pulley is considered a solid wheel, there is also a moment of inertia (I) associated with its rotation. The moment of inertia can be calculated using the formula I = 0.5 * m * r^2, where m is the mass of the pulley and r is its radius.
Now, let's calculate the net force acting on each object:
For the 1.6 kg mass:
Net force = T - mg
For the 6 kg mass:
Net force = mg - T
For the pulley:
Net force = T - mg (because the tension force is acting tangentially)
Now, we can set up the equations of motion:
For the 1.6 kg mass:
T - mg = (1.6 kg) * a (Equation 1)
For the 6 kg mass:
mg - T = (6 kg) * a (Equation 2)
For the pulley:
T - mg = (5.5 kg) * a/r (Equation 3)
To solve these equations, we need to eliminate the tension force (T) from the system. We can do this by adding Equation 1 to Equation 2:
T - mg + mg - T = (1.6 kg + 6 kg) * a
0 = (7.6 kg) * a
a = 0 m/s²
Therefore, the linear acceleration of the blocks is 0 m/s².