Another friction problem (rotational this time)

  • Thread starter Thread starter groundhog92
  • Start date Start date
  • Tags Tags
    Friction Time
Click For Summary
SUMMARY

The discussion centers on calculating the linear speed of a coin placed on a spinning record and determining the conditions under which it would slide off due to friction. The linear speed of the coin, calculated using the formula for linear velocity, is 0.59 m/s. The critical speed at which the coin would slide off, factoring in a static friction coefficient of 0.5, is determined to be 0.84 m/s. Adding another coin on top does not affect the sliding speed, as the masses cancel out in the calculations.

PREREQUISITES
  • Understanding of centripetal force and its formula: F=(mv^2)/r
  • Knowledge of static friction and its coefficient: Ff=umg
  • Familiarity with calculating linear velocity using circumference and time
  • Basic principles of rotational motion and acceleration
NEXT STEPS
  • Explore the effects of varying the coefficient of friction on rotational motion
  • Learn about the implications of adding mass in rotational dynamics
  • Study the relationship between angular velocity and linear velocity
  • Investigate the role of gravitational acceleration in circular motion calculations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and rotational dynamics, as well as educators looking for practical examples of friction and motion in circular paths.

groundhog92
Messages
7
Reaction score
0

Homework Statement


Once again I just need someone to check my work, I think I have it correct. If a coin of mass .005kg is placed on a spinning record .14m from the center and it takes the coin 1.5 sec to complete a revolution then a) what is the linear speed of the coin. b)If the coefficient of static friction is .5 then at what speed would the coin slide off c) If you were to add another coin directly atop the first how would the affect your results in part b) and d) draw and label the linear velocity and instantaneous acceleration in a diagram.


Homework Equations


F=(mv^2)/r
Ff=umg u=coefficient of friction
Circumference/ Time = linear velocity

The Attempt at a Solution


part a
.88/1.5 = .59m/s

part b
(mv^2)/r=umg
(v^2)/r=ug
v=sqrt(rug)
v=.84m/s

part c
It wouldn't change because the masses canceled out

part d
This one I was a bit confused about. I drew the linear velocity tangent to the coins circular trajectory. I drew the instantaneous acceleration perpendicular to that going out, through the coin, from the center of the record.

Note: In the calculations for part b gravity was 10 instead of 9.81
 
Physics news on Phys.org
The centripetal acceleration of the object is inward isn't it?

Otherwise your numbers look OK.
 

Similar threads

Replies
3
Views
2K
  • · Replies 60 ·
3
Replies
60
Views
5K
  • · Replies 42 ·
2
Replies
42
Views
4K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
1
Views
1K
  • · Replies 97 ·
4
Replies
97
Views
6K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
13
Views
2K
Replies
4
Views
2K