Question

(b) A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless horizontal axle. A light inextensible cord is wrapped around the wheel and supports a hanging object of mass m. Show that the angular acceleration of the wheel α is such that:

α =
(g/R) /1 +( I/mR^2)
,
where g is the acceleration due to gravity at the surface of the Earth

Answer 1

The angular acceleration (α) of the wheel is found by applying torque from the hanging mass via the light inextensible cord and using the rotational form of Newton's second law, resulting in the expression α = (g/R) / (1 + ( I/mR²)).

Step-by-step explanation:

To show that the angular acceleration (α) of the wheel is α = (g/R) / (1 + ( I/mR²)), where g is the acceleration due to gravity, R is the radius, I is the moment of inertia, and m is the mass of the hanging object, we use the relationship between torque, moment of inertia, and angular acceleration given by net τ = Iα. We also consider Newton's second law for rotational motion in the form τ = mr²α.

Applying a force through the light inextensible cord, we generate a torque (τ) on the wheel: τ = m*g*R, since torque is the product of force and the radius, and the force here is m*g (the weight of the hanging mass).

Using the rotational form of Newton's second law, we plug this torque into the equation τ = Iα. This gives us Iα = m*g*R. Solving for α, we divide by I: α = m*g*R / I.
To express this in the desired form, we introduce the unitless quantity (I/mR²) into the denominator: α = (g/R) / (1 + (I/mR²)).

Thus, we have shown that the angular acceleration of the wheel is indeed α = (g/R) / (1 + ( I/mR²)).