Question

(b) Write an expression for the net external torque acting on the system of the two blocks + wheel + string about the wheel's axle. What are its rotation sense and vector direction? [NOTE: The string is part of the system, so the string tension forces provide internal torques, not external torques.]

Answer 1

τext =(M*R*g)*(m₂ - m₁) / (m₁ + m₂)

The wheel is rotating clockwise. Its vector direction is - k which is perpendicular to the plane of the wheel.

Step-by-step explanation:

Given info

m₁ = mass of block 1

m₂ = mass of block 2

m₁ < m₂

M = Mass of the wheel

R = Radius of the wheel

I = Moment of inertia of the wheel = M*R²

We assume that the wheel is rotating clockwise since m₁ < m₂

In order to get an expression for the net external torque acting on the system we can apply the following equation

τext = I*α

where α is the angular acceleration of the wheel which can be found as follows

at = R*α ⇒ α = at / R

then we have to compute the acceleration of the system (at), using the Newton's 2nd Law

Block 1:

∑Fy = m₁*at (↑+) ⇒ T - m₁*g = m₁*at (I)

Block 2:

∑Fy = m₂*at (↓+) ⇒ - T + m₂*g = m₂*at (II)

If we apply

(I) + (II) ⇒ at = g*(m₂ - m₁) / (m₁ + m₂)

Now, we get α:

α = at / R = (g*(m₂ - m₁) / (m₁ + m₂)) / R

⇒ α = (g / R)*(m₂ - m₁) / (m₁ + m₂)

Finally

τext = I*α = (M*R²)*((g / R)*(m₂ - m₁) / (m₁ + m₂))

⇒ τext =(M*R*g)*(m₂ - m₁) / (m₁ + m₂)

The wheel is rotating clockwise. Its vector direction is - k which is perpendicular to the plane of the wheel.