/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 84 A local ice hockey team has aske... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A local ice hockey team has asked you to design an apparatus for measuring the speed of the hockey puck after a slap shot. Your design is a 2.00 -m-long, uniform rod pivoted about one end so that it is free to rotate horizontally on the ice without friction. The \(0.800 \mathrm{~kg}\) rod has a light basket at the other end to catch the \(0.163 \mathrm{~kg}\) puck. The puck slides across the ice with velocity \(\overrightarrow{\boldsymbol{v}}\) (perpendicular to the rod), hits the basket, and is caught. After the collision, the rod rotates. If the rod makes one revolution every 0.736 s after the puck is caught, what was the puck's speed just before it hit the rod?

Short Answer

Expert verified
Using the aforementioned steps, students can find the initial speed of the puck before it hit the rod by applying the conservation of angular momentum principle.

Step by step solution

01

Find the Angular Velocity after the Collision

After the collision, the rod rotates and makes one revolution every 0.736s. Hence, the angular velocity is \( \omega = \frac{2\pi}{T} = \frac{2\pi}{0.736 \, s}\) (radians per second).
02

Calculate the Angular Momentum after the Collision

The total angular momentum after the collision is given by the moment of inertia of the system times its angular velocity. The moment of inertia for a rigid rod rotated about one end can be given by \(\frac{1}{3}m r^2\), and the moment of inertia of a particle at the end of the rod can be given by \(m r^2\). Hence, the total moment of inertia for this system (rod + puck), \(I_{\text{total}} = \frac{1}{3} m_{\text{rod}} r^2 + m_{\text{puck}} r^2\). The total angular momentum then is \(L_{\text{after collision}} = I_{\text{total}} \omega\)
03

Determine the Initial Velocity of the Puck

Before the collision, the rod does not move hence has no angular momentum. The puck moves with velocity \(v\) and has hence an angular momentum \(L_{\text{before collision}} = m_{\text{puck}} r v\). However, since net external torque acting on this system is zero, the initial and final angular momentum should be equal, \( L_{\text{after collision}} = L_{\text{before collision}}\). Doing some calculations, we can now solve for \(v = \frac{L_{\text{after collision}}}{ m_{\text{puck}} r }\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Angular Velocity
Imagine an ice skater spinning around; the rate at which they rotate is essential for understanding their motion. Analogously, in the world of physics, particularly when dealing with rotational movement, we use the term angular velocity to describe how fast an object rotates or revolves relative to another point, usually the center of rotation.

Angular velocity, denoted by the symbol \(\omega\), is a vector quantity that represents the rate of change of the angular position of an object. It's measured in radians per second (rad/s). When we look at the hockey puck apparatus problem, after the puck strikes the rod and is caught, the system begins rotating, and we must compute the angular velocity to find out how fast the puck was traveling initially.

By knowing the rod makes one full revolution (which is \(2\pi\) radians) in 0.736 seconds, we can calculate the angular velocity using \( \omega = \frac{2\pi}{T} \) with \( T \) being the period. This is crucial because the angular velocity tells us how quickly the rod and puck are turning as one unit post-collision, which then allows us to delve deeper into the conservation of angular momentum.
Moment of Inertia
The concept of moment of inertia is central to understanding how different objects will rotate, given the same amount of rotational force. Think of it as the rotational equivalent of mass in linear motion. The moment of inertia describes how difficult it is to change an object's rotational speed about an axis.

The moment of inertia depends not just on the mass but also on the distribution of that mass relative to the axis of rotation. For example, a figure skater with arms extended has a larger moment of inertia compared to when their arms are close to their body, which is why skaters spin faster with arms pulled in—because their moment of inertia is smaller.

In the hockey puck exercise, we use the formulas for the moment of inertia of a rod about its end (\(\frac{1}{3}m_{\text{rod}} r^2\)) and a point mass at the end of the rod (\(m_{\text{puck}} r^2\)). When the puck strikes the rod and is caught, it affects the total moment of inertia of the system, which we ultimately use to apply the principle of the conservation of angular momentum to find the puck’s speed.
Conservation of Angular Momentum
In physics, the conservation of angular momentum is a fundamental principle that states if no external torque acts on a system, the total angular momentum of that system remains constant. This is analogous to the conservation of linear momentum, where the total linear momentum is conserved in the absence of external forces.

In our ice hockey exercise, before the puck hits the rod, it has a certain linear momentum which, upon impact, becomes angular momentum. Because there is no external torque (since the pivot is frictionless and there's no other external influences), the angular momentum before and after the collision is the same. So, knowing the angular velocity and moment of inertia after the collision enables us to calculate the initial angular momentum, which is equivalent to that of the puck's just before impact.

Here we used the fact that \( L_{\text{after collision}} = L_{\text{before collision}} \) to solve for the puck's speed. Using this conservation law is immensely powerful as it provides a reliable method to solve complex dynamic problems with rotating systems, such as the hockey puck apparatus.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A uniform solid disk made of wood is horizontal and rotates freely about a vertical axle at its center. The disk has radius \(0.600 \mathrm{~m}\) and mass \(1.60 \mathrm{~kg}\) and is initially at rest. A bullet with mass \(0.0200 \mathrm{~kg}\) is fired horizontally at the disk, strikes the rim of the disk at a point perpendicular to the radius of the disk, and becomes embedded in its rim, a distance of \(0.600 \mathrm{~m}\) from the axle. After being struck by the bullet, the disk rotates at \(4.00 \mathrm{rad} / \mathrm{s}\). What is the horizontal velocity of the bullet just before it strikes the disk?

A machine part has the shape of a solid uniform sphere of mass \(225 \mathrm{~g}\) and diameter \(3.00 \mathrm{~cm}\). It is spinning about a frictionless axle through its center, but at one point on its equator it is scraping against metal, resulting in a friction force of \(0.0200 \mathrm{~N}\) at that point. (a) Find its angular acceleration. (b) How long will it take to decrease its rotational speed by \(22.5 \mathrm{rad} / \mathrm{s} ?\)

A woman with mass \(50 \mathrm{~kg}\) is standing on the rim of a large horizontal disk that is rotating at \(0.80 \mathrm{rev} / \mathrm{s}\) about an axis through its center. The disk has mass \(110 \mathrm{~kg}\) and radius \(4.0 \mathrm{~m} .\) Calculate the magnitude of the total angular momentum of the woman-disk system. (Assume that you can treat the woman as a point.)

A block with mass \(m\) is revolving with linear speed \(v_{1}\) in a circle of radius \(r_{1}\) on a frictionless horizontal surface (see Fig. E10.42). The string is slowly pulled from below until the radius of the circle in which the block is revolving is reduced to \(r_{2}\). (a) Calculate the tension \(T\) in the string as a function of \(r\), the distance of the block from the hole. Your answer will be in terms of the initial velocity \(v_{1}\) and the radius \(r_{1} .\) (b) Use \(W=\int_{r_{1}}^{r_{2}} \overrightarrow{\boldsymbol{T}}(r) \cdot d \overrightarrow{\boldsymbol{r}}\) to calculate the work done by \(\vec{T}\) when \(r\) changes from \(r_{1}\) to \(r_{2}\). (c) Compare the results of part (b) to the change in the kinetic energy of the block.

One force acting on a machine part is \(\overrightarrow{\boldsymbol{F}}=(-5.00 \mathrm{~N}) \hat{\imath}+\) \((4.00 \mathrm{~N}) \hat{\jmath} .\) The vector from the origin to the point where the force is applied is \(\vec{r}=(-0.450 \mathrm{~m}) \hat{\imath}+(0.150 \mathrm{~m}) \hat{\jmath}\). (a) In a sketch, show \(\vec{r}, \vec{F}\), and the origin. (b) Use the right-hand rule to determine the direction of the torque. (c) Calculate the vector torque for an axis at the origin produced by this force. Verify that the direction of the torque is the same as you obtained in part (b).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.