Question

A hoop of \(2.00-\mathrm{m}\) circumference is rolling down an inclined plane of length \(10.0 \mathrm{m}\) in a time of \(10.0 \mathrm{s} .\) It started out from rest. (a) What is its angular velocity when it arrives at the bottom? (b) If the mass of the hoop, concentrated at the rim, is \(1.50 \mathrm{kg},\) what is the angular momentum of the hoop when it reaches the bottom of the incline? (c) What force(s) supplied the net torque to change the hoop's angular momentum? Explain. [Hint: Use a rotation axis through the hoop's center. \(]\) (d) What is the magnitude of this force?

Short answer

Based on the given information, the hoop's angular velocity at the bottom of the inclined plane is \(\frac{\pi\sqrt{2}}{2} \, rad/s\). The angular momentum of the hoop at the bottom of the plane is found to be \(\frac{3\sqrt{2}}{4} \, kg \cdot m^2/s\). The gravitational force acting on the hoop and the normal force exerted by the plane on the hoop are responsible for the net torque. The magnitude of the force responsible for the net torque is \(\frac{3.00}{5\pi} \, N\).

Step-by-step solution

(a) Determine angular velocity of the hoop at the bottom of the plane

To find the angular velocity (\(\omega\)) of the hoop at the bottom of the inclined plane, we will first find the linear velocity (\(v\)). We can use the equation \(v^2 = u^2 + 2as\), where \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the distance traveled. The initial velocity (\(u\)) of the hoop is \(0\) since it starts from rest. The acceleration of the hoop (\(a\)) can be calculated by \(a=\frac{s}{t^2}\), where \(s\) is the length of the inclined plane (\(10.0 \,m\)) and \(t\) is the time (\(10.0 \,s\)). 1. Calculate acceleration: \(a = \frac{10.0 \,m}{(10.0 \,s)^2} = 0.10 \, m/s^2\). 2. Calculate linear velocity: \(v^2 = u^2 + 2as = 0^2 + 2(0.10)(10.0) = 2 \, m^2/s^2\), so \(v = \sqrt{2} \, m/s\). 3. Calculate angular velocity: The relationship between linear velocity (\(v\)) and angular velocity (\(\omega\)) can be expressed as \(v = r\omega\), where \(r\) is the radius of the hoop. Since the circumference of the hoop is \(2.00 \,m\), the radius can be computed as \(r = \frac{2.00 \,m}{2\pi}\). Thus, the angular velocity (\(\omega\)) can be found using the formula: \(\omega = \frac{v}{r} = \frac{\sqrt{2} \, m/s}{\frac{2.00 \,m}{2\pi}} = \frac{\pi\sqrt{2}}{2} \, rad/s\).

(b) Find the angular momentum of the hoop at the bottom of the plane

We can compute the angular momentum (L) using the formula: \(L = I\omega\), where \(I\) is the moment of inertia of the hoop and \(\omega\) is the angular velocity. For a hoop, \(I = Mr^2\), and since the mass of the hoop \(M = 1.50 \, kg\) and the radius \(r = \frac{2.00 \,m}{2\pi}\), the moment of inertia can be determined. 1. Calculate the moment of inertia: \(I = Mr^2 = (1.50 \, kg)\left(\frac{2.00 \, m}{2\pi}\right)^2 = \frac{3.00}{2\pi^2} \, kg\cdot m^2\). 2. Calculate the angular momentum: \(L = I\omega = \frac{3.00}{2\pi^2} \, kg\cdot m^2 \cdot \frac{\pi\sqrt{2}}{2} \, rad/s = \frac{3\pi\sqrt{2}}{4\pi} \, kg \cdot m^2/s = \frac{3\sqrt{2}}{4} \, kg \cdot m^2/s\).

(c) Identify the force(s) responsible for the net torque

The force(s) responsible for the net torque on the hoop are the gravitational force acting on the hoop, which pulls it down the inclined plane, and the normal force exerted by the plane on the hoop perpendicular to the plane. It is important to note that since we are considering a rotation axis through the hoop's center, these forces cause net torque.

(d) Determine the magnitude of force responsible for the net torque

The total force acting on the hoop can be decomposed into two components: one, perpendicular to the inclined plane (normal force), and the other, parallel to the inclined plane (tangential force). Since only the tangential component contributes to the torque, we'll only focus on that component. Given that we know the angular velocity of the hoop (\(\omega = \frac{\pi\sqrt{2}}{2} \, rad/s\)) and its mass (\(M = 1.50 \, kg\)), the tangential force (\(F_t\)) can be found using the equation \(F_t = MAR\), where \(A = r\alpha\) is the tangential acceleration and \(\alpha = \frac{a}{r}\) is the angular acceleration, given acceleration \(a = 0.10 \, m/s^2\) and radius \(r = \frac{2.00 \, m}{2\pi}\). 1. Calculate angular acceleration: \(\alpha = \frac{a}{r} = \frac{0.10 \, m/s^2}{\frac{2.00 \,m}{2\pi}} = \frac{\pi}{10} \, rad/s^2\). 2. Calculate tangential acceleration: \(A = r\alpha = \frac{2.00 \, m}{2\pi} \cdot \frac{\pi}{10} = \frac{1}{5} \, m/s^2\). 3. Calculate tangential force: \(F_t = MAR = (1.50 \, kg) \cdot \frac{1}{5} \, m/s^2 \cdot \frac{2.00 \, m}{2\pi} = \frac{3.00}{5\pi} \, N\). So the magnitude of the force responsible for the net torque is \(\frac{3.00}{5\pi} \, N\).