Question

Two children standing on opposite sides of a merry-goround (see Fig. 8.5 ) are trying to rotate it. They each push in opposite directions with forces of magnitude \(10.0 \mathrm{N} .\) (a) If the merry-go-round has a mass of $180 \mathrm{kg}\( and a radius of \)2.0 \mathrm{m},$ what is the angular acceleration of the merry-go-round? (Assume the merry-go-round is a uniform disk.) (b) How fast is the merry-go-round rotating after \(4.0 \mathrm{s} ?\)

Short answer

Answer: The angular acceleration of the merry-go-round is $0 rad/s^2$, and its angular velocity after 4.0 seconds is $0 rad/s$.

Step-by-step solution

Calculate net torque acting on the merry-go-round

We are given that the two children apply equal and opposite forces of \(10.0 N\) on the merry-go-round. Let the distance of each child from the center of the merry-go-round be \(r\). Since the force applied by both children is along the radial direction, we can compute the net torque on the merry-go-round using the formula: $$\tau_{net} = r\cdot F_{1} - r \cdot F_{2}$$ Given \(F_{1} = F_{2} = 10.0 N\) and \(r = 2.0 m\), we can calculate the net torque: $$\tau_{net} = 2.0 \cdot 10.0 - 2.0 \cdot 10.0 = 0 N\cdot m$$

Compute the moment of inertia

We are asked to assume that the merry-go-round is a uniform disk. The moment of inertia for a uniform disk of mass \(M\) and radius \(R\) is given by: $$I = \frac{1}{2}MR^2$$ The mass of the merry-go-round is \(180 kg\), and the radius is \(2.0 m\). Plugging these values, we have: $$I = \frac{1}{2}(180)(2.0)^2 = 360 kg\cdot m^2$$

Calculate the angular acceleration

Using Newton's second law for rotational motion, we know torque is related to angular acceleration and moment of inertia by the equation: $$\tau_{net} = I \cdot \alpha$$ Here, \(\alpha\) is the angular acceleration. Plugging the values for \(\tau_{net}\) and \(I\), we get: $$0 N\cdot m = (360 kg\cdot m^2) \cdot \alpha$$ Since the net torque acting on the merry-go-round is zero, its angular acceleration is also zero: $$\alpha = 0 rad/s^2$$

Calculate angular velocity after \(4.0 s\)

To determine the angular velocity \(\omega\) after \(4.0s\), we can use the kinematic equation for angular velocity: $$\omega = \omega_0 + \alpha \cdot t$$ Here, \(\omega_0\) is the initial angular velocity of the merry-go-round, and \(t\) is the time elapsed. We are not given the initial angular velocity \(\omega_0\), so we will assume it to be zero, which means the merry-go-round starts from rest. Since the angular acceleration is zero, the angular velocity remains constant: $$\omega = 0 + 0 \cdot 4.0 = 0 rad/s$$ Thus, the merry-go-round is not rotating after \(4.0s\).