physics2- No-slipFrictionlessThe ball is a solid sphere of radius 4.0cm and mass 0.14kg. the half-pipe has one side with sufficientstatic friction that we can assume no sliding will occur there. The ball is released from rest at a height of0.77m above the bottom of the half-pipe on the no-slip side. What is its angular speed when it is on thefrictionless side of the track? How high does it rise on the frictionless side? 4) Describe what physics Law or Laws you will use to solve each part of the problem.5) Set up and solve for the first question. Find the angular velocity of the ball when it is on thefrictionless side.

Answered: 4) Describe what physics Law or Laws…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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physics2-

No-slip
Frictionless
The ball is a solid sphere of radius 4.0cm and mass 0.14kg. the half-pipe has one side with sufficient
static friction that we can assume no sliding will occur there. The ball is released from rest at a height of
0.77m above the bottom of the half-pipe on the no-slip side. What is its angular speed when it is on the
frictionless side of the track? How high does it rise on the frictionless side?

4) Describe what physics Law or Laws you will use to solve each part of the problem.
5) Set up and solve for the first question. Find the angular velocity of the ball when it is on the
frictionless side.

Expert Solution

Step 1

To solve the first part of this problem, we have to use the law of conservation of energy. According to which, energy is neither be created nor be destroyed but can be transferred from one form to another.

Here the system consists of a half-pipe and a ball, When a ball is at a height h, then it ai associated with potential energy. when it is on the frictionless side of the pipe, this potential energy is converted into rotational kinetic energy.

$P o t e n t i a l e n e r g y = R o t a t i o n a l k i n e t i c e n e r g y m g h = \frac{1}{2} I ω^{2} . . . . . . . . 1 w h e r e m - m a s s o f t h e b a l l I - m o m e n t o f i n e r t i a o f t h e b a l l = m r^{2} ω - a n g u l a r v e l o c i t y . F r o m 1, w e c a n h a v e ω = \sqrt{\frac{2 m g h}{I}} ω = \sqrt{\frac{2 \times 0.14 \times 9.8 \times 0.77}{0.14 \times {(4 \times 10^{- 2})}^{2}}} ω = 97.121$

Angular velocity of the ball is 97.121 rad/s

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