What Happens When a Slipping Sphere Begins to Roll Without Slipping?

Click For Summary
SUMMARY

The discussion focuses on the dynamics of a uniform density sphere transitioning from slipping to rolling without slipping. The sphere has an initial angular speed of 10 revolutions per second, a radius of 0.08 meters, and a mass of 7.3 kg. The coefficient of friction is 0.21, which plays a critical role in determining the energy lost due to friction and the time taken to achieve pure rolling motion. Key equations include the rotational inertia (I = 2/5mr²) and kinetic energy formulas for both rotational and translational motion.

PREREQUISITES
  • Understanding of rotational dynamics and inertia
  • Familiarity with kinetic energy equations
  • Knowledge of friction and its effects on motion
  • Basic principles of angular and linear acceleration
NEXT STEPS
  • Calculate the energy lost due to friction during the transition to rolling
  • Determine the time taken for the sphere to reach pure rolling motion
  • Explore the relationship between angular velocity and linear velocity using the equation v = rw
  • Investigate the effects of varying the coefficient of friction on rolling motion
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in the mechanics of rolling motion and energy conservation principles in dynamics.

jbphys303
Messages
1
Reaction score
0

Homework Statement



A uniform density sphere is released such that it has an angular speed of 10 rev/sec and no initial linear velocity. The angular velocity vector is perfectly perpendicular to the linear momentum vector. Initially the ball slips as it moves along the surface, but after time t pure rolling without slipping begins.

The coefficient of friction between the sphere and the surface is 0.21.
The radius of the ball is 0.08m
mass sphere= 7.3 kg
w= 10 rev/sec

#1: How fast is the ball rolling at time t?

#2: How long did it take to reach this speed?

#3: How much energy was lost between time t=0 and t=t?


Homework Equations



Rotational Inertia (I)=2/5mr^2
kinetic energy (rotational)= 1/2Iw^2
ke= 1/2mv^2
v=rw



The Attempt at a Solution



I think I can figure it out once I know how much energy is lost by friction
So far I set up the following equation
1/2Iw^2-(energy lost by friction)= 1/2Iw(final)^2+ 1/2mv^2
Any help would be appreciated
 
Last edited:
Physics news on Phys.org
The motion of the ball is composed of translation of its CM and of rotation around the CM. The kinetic friction points forward, accelerates translation and its torque with respect to the CM decelerates rotation. At the time when v=rw, static friction takes place and the ball rolls.

Use both equation for linear acceleration ma = F and for angular acceleration I*dw/dt=-RF. Solve for v and w. Find the time when wR=v.

ehild
 

Similar threads

  • · Replies 42 ·
2
Replies
42
Views
4K
Replies
4
Views
3K
  • · Replies 60 ·
3
Replies
60
Views
5K
  • · Replies 21 ·
Replies
21
Views
5K
  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 35 ·
2
Replies
35
Views
5K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
2K
  • · Replies 97 ·
4
Replies
97
Views
6K