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Chapter 8: Problem 55

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

Expert verified
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

01

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: $$\tau_{net} = 2.0 \cdot 10.0 - 2.0 \cdot 10.0 = 0 N\cdot m$$
02

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: $$I = \frac{1}{2}(180)(2.0)^2 = 360 kg\cdot m^2$$
03

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: $$\alpha = 0 rad/s^2$$
04

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 \(4.0s\).

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Most popular questions from this chapter

A planet moves around the Sun in an elliptical orbit (see Fig. 8.39 ). (a) Show that the external torque acting on the planet about an axis through the Sun is zero. (b) since the torque is zero, the planet's angular momentum is constant. Write an expression for the planet's angular momentum in terms of its mass \(m,\) its distance \(r\) from the Sun, and its angular velocity \(\omega\) (c) Given \(r\) and a, how much area is swept out during a short time At? [Hint: Think of the area as a fraction of the area of a circle, like a slice of pie; if \(\Delta t\) is short enough, the radius of the orbit during that time is nearly constant.] (d) Show that the area swept out per unit time is constant. You have just proved Kepler's second law!
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A mechanic turns a wrench using a force of \(25 \mathrm{N}\) at a distance of \(16 \mathrm{cm}\) from the rotation axis. The force is perpendicular to the wrench handle. What magnitude torque does she apply to the wrench?
Derive the rotational form of Newton's second law as follows. Consider a rigid object that consists of a large number \(N\) of particles. Let \(F_{i}, m_{i},\) and \(r_{i}\) represent the tangential component of the net force acting on the ith particle, the mass of that particle, and the particle's distance from the axis of rotation, respectively. (a) Use Newton's second law to find \(a_{i}\), the particle's tangential acceleration. (b) Find the torque acting on this particle. (c) Replace \(a_{i}\) with an equivalent expression in terms of the angular acceleration \(\alpha\) (d) Sum the torques due to all the particles and show that $$\sum_{i=1}^{N} \tau_{i}=I \alpha$$
A sign is supported by a uniform horizontal boom of length \(3.00 \mathrm{m}\) and weight \(80.0 \mathrm{N} .\) A cable, inclined at an angle of \(35^{\circ}\) with the boom, is attached at a distance of \(2.38 \mathrm{m}\) from the hinge at the wall. The weight of the sign is \(120.0 \mathrm{N} .\) What is the tension in the cable and what are the horizontal and vertical forces \(F_{x}\) and \(F_{y}\) exerted on the boom by the hinge? Comment on the magnitude of \(F_{y}\).
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