A pulley of moment of inertia \(10.0\, \text{kg} \cdot \text{m}^2\) is mounted on a wall as shown in the following figure. Light strings are wrapped around two circumferences of the pulley and weights are attached. Find the net torque and the angular acceleration of the pulley. Assume the following data: \(r_1 = 100\, \text{cm}, r_2 = 40\, \text{cm}, m_1 = 1.0\, \text{kg}, m_2 = 2.0\, \text{kg}\). Show complete logical work for any credit.**Diagram Explanation:**The diagram includes a large pulley with two different radii marked as \(r_1\) and \(r_2\). Two masses, \(m_1\) and \(m_2\), are attached to strings that wrap around the respective circumferences defined by \(r_1\) and \(r_2\). The forces exerted by the masses create torques at different radii on the pulley, influencing its angular acceleration.

We can answer the question by using the formula for torque and Newton's law for rotational motion. The detailed steps are given below.The Torque τ due to a force "F" about some point O is given by τ→=r→×F→=rFsinθ n^ Here, r is the distance from the point O to the point of application of force, and θ is the angle between r→ and F→. The direction of torque, n^ is given by the right hand thumb rule. If you curl your righ hand fingers from the vector r→ towards vector F→, the direction of thumb points along the vector τ→ The following figure will illustrate this In this figure, torque points out of the page. We will apply this formula to find the net torque on the pulley.In the given problem, the force is due to weight of the two masses. The following diagram shows the direction of these forces. Force due to mass 1 is given by F1=W1=m1g and it acts at a distance of r1 from the center of the pulley. Force due to mass 2 is given by F2=W2=m2g and it acts at a distance of r2 from the center of the pulley. Let's consider the direction pointing out of the page as positive, and direction pointing into the page as negative. Now, torque due to weight of mass 1 is given by τ1=r1F1sinθ=r1m1gsinθ We are given that, r1=100cm=1 m and m1=1 kg , and g=9.8 m/s2 is a constant Since force acts perpendicular to radius, we have θ=90° and sinθ=1 Substituting these values in the above expression, we get τ1=1×1×9.8×1=9.8 Nm By right hand thumb rule, this torque points out of the page. So we will give positive value to it. Now, torque due to weight of mass 2 is given by τ2=r2F2sinθ=r2m2gsinθ We are given that, r2=40cm=0.4 m and m2=2 kg Since force acts perpendicular to radius, we again have θ=90° and sinθ=1 Substituting these values in the above expression, we get τ2=0.4×2×9.8×1=7.84 Nm By right hand thumb rule, this torque points into the page. So we will give negative value to it. Soτ2=-7.84 Nm The net torque τ is given by τ=τ1+τ2=9.8-7.84=1.96 Nm Since it has positive value, it points out of the page. Therefore, net torque on the pulley is 1.96 Nm and it points out of the page.Newton's law for rotational motion is given by τ=Iα Here, τ is the torque I is the moment of Inertial α is the angular acceleration We are given that the moment of Inertia of pulley is 10 kg m2. The net torque on the pulley was found above. So substituting these values, we can find the angular acceleration of the pulley τ=Iα1.96=10×αα=1.9610α=0.196 Therefore, angular acceleration of the pulley is 0.196 rad/s2.

Answered: A pulley of moment of inertia 10.0…