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Chapter 10: Problem 19

In another race, a solid sphere and a thin ring roll without slipping from rest down a ramp that makes angle \(\theta\) with the horizontal. Find the ratio of their accelerations, \(a_{\text {ring }} / a_{\text {sphere }}\)

Short Answer

Expert verified
Answer: The ratio of accelerations of a solid sphere and a thin ring rolling without slipping down a ramp at angle θ with the horizontal is 7/10.

Step by step solution

01

1. Write down the expressions of the forces acting on the objects

The gravitational force \(F_g\) acts vertically downwards on both the sphere and the ring. It can be decomposed into two components; one along the incline (\(F_\parallel\) which causes the object to roll) and one perpendicular to the incline (\(F_\perp\) which doesn't contribute to the motion). We will find \(F_\parallel\) for both objects. For both objects, we have: \(F_g = mg\) where \(m\) is the mass of the object (sphere or ring) and \(g\) is the acceleration due to gravity. The component of gravitational force along the incline, in terms of \(\theta\) is given by: \(F_\parallel = mg\sin\theta\)
02

2. Set up equations of motion

For rolling without slipping, we need to equate the translational and rotational motion. We use Newton's second law for translational motion and the torque equation for rotational motion: For translational motion, we have: $$F_\parallel = ma$$ For rotational motion, we have: $$I\alpha=R \cdot F_\parallel$$ where \(I\) is the moment of inertia of the object, \(\alpha\) is the angular acceleration, and \(R\) is the radius. Since the objects are rolling without slipping, we can relate the linear and angular acceleration using the formula: $$a=R\alpha$$
03

3. Calculate the moment of inertia for each object

Now, we need to find the moment of inertia, \(I\), for each object. The moment of inertia depends on the distribution of mass in the object. For a solid sphere of mass \(m\) and radius \(R\), its moment of inertia about an axis through the center is given by: \(I_{\text {sphere }}=\frac{2}{5}mR^2\) For a thin ring of mass \(m\) and radius \(R\), its moment of inertia about an axis perpendicular to the plane of the ring and passing through the center is given by: \(I_{\text {ring }}=mR^2\)
04

4. Solve the equations for the accelerometer and find the ratio

Now we need to solve the equations of motion for both objects to find the accelerations for each object: For the sphere: $$\begin{aligned} m a_{\text {sphere }} &=m g \sin \theta \\ \frac{2}{5} m R^2 \alpha_{\text {sphere }} &=R \cdot a_{\text {sphere }} \\ a_{\text{sphere}} &= \frac{5}{7}g\sin\theta \end{aligned}$$ For the ring: $$\begin{aligned} m a_{\text {ring }} &=m g \sin \theta \\ m R^2 \alpha_{\text {ring }} &=R \cdot a_{\text {ring }} \\ a_{\text{ring}} &= \frac{1}{2}g\sin\theta \end{aligned}$$ Now we can find the desired ratio: $$\frac{a_{\text {ring }} }{a_{\text {sphere }}}=\frac{\frac{1}{2} g \sin \theta}{\frac{5}{7} g \sin \theta}=\boxed{\frac{7}{10}}$$

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Most popular questions from this chapter

Determine the moment of inertia for three children weighing \(60.0 \mathrm{lb}, 45.0 \mathrm{lb}\) and \(80.0 \mathrm{lb}\) sitting at different points on the edge of a rotating merry-go-round, which has a radius of \(12.0 \mathrm{ft}\).

A wheel with \(c=\frac{4}{9}\), a mass of \(40.0 \mathrm{~kg}\), and a rim radius of \(30.0 \mathrm{~cm}\) is mounted vertically on a horizontal axis. A 2.00 -kg mass is suspended from the wheel by a rope wound around the rim. Find the angular acceleration of the wheel when the mass is released.

A cylinder with mass \(M\) and radius \(R\) is rolling without slipping through a distance \(s\) along an inclined plane that makes an angle \(\theta\) with respect to the horizontal. Calculate the work done by (a) gravity, (b) the normal force, and (c) the frictional force.

In a tire-throwing competition, a man holding a \(23.5-\mathrm{kg}\) car tire quickly swings the tire through three full turns and releases it, much like a discus thrower. The tire starts from rest and is then accelerated in a circular path. The orbital radius \(r\) for the tire's center of mass is \(1.10 \mathrm{~m},\) and the path is horizontal to the ground. The figure shows a top view of the tire's circular path, and the dot at the center marks the rotation axis. The man applies a constant torque of \(20.0 \mathrm{~N} \mathrm{~m}\) to accelerate the tire at a constant angular acceleration. Assume that all of the tire's mass is at a radius \(R=0.35 \mathrm{~m}\) from its center. a) What is the time, \(t_{\text {throw }}\) required for the tire to complete three full revolutions? b) What is the final linear speed of the tire's center of mass (after three full revolutions)? c) If, instead of assuming that all of the mass of the tire is at a distance \(0.35 \mathrm{~m}\) from its center, you treat the tire as a hollow disk of inner radius \(0.30 \mathrm{~m}\) and outer radius \(0.40 \mathrm{~m}\), how does this change your answers to parts (a) and (b)?

A figure skater draws her arms in during a final spin. Since angular momentum is conserved, her angular velocity will increase. Is her rotational kinetic energy conserved during this process? If not, where does the extra energy come from or go to?
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