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Chapter 15: Problem 11

A yo-yo of total mass \(m\) consists of two solid cylinders of radius \(R\), connected by a small spindle of negligible mass and radius \(r\). The top of the string is held motionless while the string unrolls from the spindle. Show that the acceleration of the yo-yo is \(g /\left(1+R^{2} / 2 r^{2}\right)\). [Hint: The acceleration and the tension in the string are unknown. Use \(\tau=\Delta L / \Delta t\) and \(F=m a\) to determine these two unknowns.

Short Answer

Expert verified
The acceleration of the yo-yo is \( a = \frac{g}{1 + \frac{R^2}{2 r^2}} \).

Step by step solution

01

Identify Forces Acting on the Yo-Yo

First, identify the forces acting on the yo-yo. The force due to gravity acting downwards is given by the weight of the yo-yo, which is \(mg\). The tension \(T\) in the string acts upwards. When these forces are considered, the net force \(F_{net}\) acting on the yo-yo can be expressed as:\[ F_{net} = mg - T \]
02

Apply Newton's Second Law for Linear Motion

Using Newton's second law, the net force on the yo-yo can also be written in terms of the linear acceleration \(a\) as:\[ F_{net} = ma \]Substituting the expression for \(F_{net}\) from Step 1, we have:\[ mg - T = ma \]
03

Establish the Torque Equation

Next, consider the torque \(\tau\) acting on the yo-yo about the axis of the spindle. The force causing the torque is the tension \(T\), and it acts at a distance \(r\) (the radius of the spindle) from the axis. Hence, the torque is:\[ \tau = Tr \]The torque is also related to the angular acceleration \(\alpha\) and the moment of inertia \(I\). The moment of inertia of the yo-yo (two cylinders) is \(I = m R^2 / 2\). So, we have:\[ \tau = I \alpha \]
04

Relate Linear and Angular Variables

The relationship between linear acceleration \(a\) and angular acceleration \(\alpha\) is given by:\[ \alpha = \frac{a}{r} \]Substituting for \(\alpha\) in the torque equation from Step 3, we get:\[ Tr = \frac{m R^2}{2} \times \frac{a}{r} \rightarrow Tr = \left(\frac{m R^2 a}{2 r}\right) \]
05

Solve for Tension T

From the torque equation in Step 4, solve for \(T\):\[ T = \frac{m R^2 a}{2 r^2} \]
06

Substitute T into Linear Equation

Substitute the expression for \(T\) from Step 5 into the linear motion equation from Step 2:\[ mg - \frac{m R^2 a}{2 r^2} = ma \]
07

Solve for Acceleration a

Simplifying the equation from Step 6 to solve for \(a\), we have:\[ mg = ma + \frac{m R^2 a}{2 r^2} \]\[ mg = a \left(m + \frac{m R^2}{2 r^2}\right) \]Divide both sides by the expression in parentheses:\[ a = \frac{g}{1 + \frac{R^2}{2 r^2}} \]
08

Conclusion

Thus, we determine that the acceleration of the yo-yo is given by the expression:\[ a = \frac{g}{1 + \frac{R^2}{2 r^2}} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Second Law
Newton's Second Law is fundamental to understanding motion. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). This principle helps us determine how forces cause changes in motion. In the yo-yo exercise, we're dealing with both vertical forces鈥攁 combination of gravitational force and tension from the string.
Newton's Second Law allows us to express this relationship as:
  • In terms of force: \( F_{net} = mg - T \).
  • Here, \( mg \) is the gravitational force, pulling down the yo-yo, and \( T \) is the tension in the string, acting upwards.
This net force results in linear acceleration ( a ). Thus, understanding this law is crucial to solving for the yo-yo鈥檚 acceleration.
Torque and Angular Acceleration
Torque is all about rotations. It tells us how much a force acting at a distance from the pivot can rotate an object. Think of opening a door: the farther away you push from the hinge, the easier it is to swing open. The torque (\tau ) for the yo-yo is about the axis of the spindle, calculated using the tension in the string and the spindle's radius:
  • Formula for torque: \( \tau = T r \)
  • Where \( T \) is the force or tension and \( r \) is the radius of the spindle.
But, torque also relates to the angular acceleration (\alpha ). Use the moment of inertia to make that connection:
  • Torque relates to angular acceleration through moment of inertia: \( \tau = I \alpha \).
Thus, torque connects the force applied and its effectiveness in producing rotational movement.
Moment of Inertia
Moment of Inertia (I ) is like rotational mass. It describes how much torque is needed for a certain angular acceleration. It's the rotational equivalent of mass in linear motion. For the yo-yo, composed of two cylinders, the moment of inertia is:
  • Moment of Inertia formula: \( I = \frac{m R^2}{2} \)
  • Here, \( R \) is the radius of the cylinders, and \( m \) is the mass.
Thus, the yo-yo鈥檚 design鈥攖wo large cylinders鈥攁ffects its rotational movement. Larger radius and mass result in larger moment of inertia, meaning it requires more torque to achieve the same angular acceleration as a smaller object.
Linear and Angular Relationships
Linear and angular quantities can be directly related. Understanding this relationship lets us transition between the two, offering insights into how linear movements relate to rotations. For the yo-yo, the crucial relationship is:
  • Angular acceleration (\alpha ) and linear acceleration ( a ): \( \alpha = \frac{a}{r} \).
  • Here, \( r \) is the radius of the spindle.
This equation shows that as the yo-yo accelerates linearly, it also rotates. The smaller the spindle's radius, the greater the angular acceleration for the same linear acceleration. Thus, understanding these connections helps solve for unknowns, linking motion in straight lines to rotations seamlessly.

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Most popular questions from this chapter

An object thrown straight up in the air is momentarily at rest when it reaches the top of its motion. Does that mean that it is in equilibrium at that point? Explain.

A ball is connected by a string to a vertical post. The ball is set in horizontal motion so that it starts winding the string around the post. Assume that the motion is confined to a horizontal plane, i.e., ignore gravity. Michelle and Astrid are trying to predict the final velocity of the ball when it reaches the post. Michelle says that according to conservation of angular momentum, the ball has to speed up as it approaches the post. Astrid says that according to conservation of energy, the ball has to keep a constant speed. Who is right? [Hint: How is this different from the case where you whirl a rock in a circle on a string and gradually reel in the string?]

A uniform ladder of mass \(m\) and length \(L\) leans against a smooth wall, making an angle \(\theta\) with respect to the ground. The dirt exerts a normal force and a frictional force on the ladder, producing a force vector with magnitude \(F_{1}\) at an angle \(\phi\) with respect to the ground. Since the wall is smooth, it exerts only a normal force on the ladder; let its magnitude be \(F_{2}\). (a) Explain why \(\phi\) must be greater than \(\theta\). No math is needed. (b) Choose any numerical values you like for \(m\) and \(L\), and show that the ladder can be in equilibrium (zero torque and zero total force vector) for \(\theta=45.00^{\circ}\) and \(\phi=63.43^{\circ}\).

The sun turns on its axis once every \(26.0\) days. Its mass is \(2.0 \times 10^{30} \mathrm{~kg}\) and its radius is \(7.0 \times 10^{8} \mathrm{~m}\). Assume it is a rigid sphere of uniform density. (a) What is the sun's angular momentum? In a few billion years, astrophysicists predict that the sun will use up all its sources of nuclear energy, and will collapse into a ball of exotic, dense matter known as a white dwarf. Assume that its radius becomes \(5.8 \times 10^{6} \mathrm{~m}\) (similar to the size of the Earth.) Assume it does not lose any mass between now and then. (Don't be fooled by the photo, which makes it look like nearly all of the star was thrown off by the explosion. The visually prominent gas cloud is actually thinner than the best laboratory vacuum ever produced on earth. Certainly a little bit of mass is actually lost, but it is not at all unreasonable to make an approximation of zero loss of mass as we are doing.) (b) What will its angular momentum be? (c) How long will it take to turn once on its axis?

Penguins are playful animals. Tux the Penguin invents a new game using a natural circular depression in the ice. He waddles at top speed toward the crater, aiming off to the side, and then hops into the air and lands on his belly just inside its lip. He then bellysurfs, moving in a circle around the rim. The ice is frictionless, so his speed is constant. Is Tux's angular momentum zero, or nonzero? What about the total torque acting on him? Take the center of the crater to be the axis. Explain your answers.
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