Another friction problem (rotational this time)

In summary, the conversation discusses a homework problem involving a spinning record and a coin. The linear speed of the coin is calculated to be 0.59m/s and the speed at which the coin would slide off is determined to be 0.84m/s. Adding another coin on top does not affect these results. A diagram is also drawn to show the linear velocity and instantaneous acceleration of the coin.
  • #1
groundhog92
7
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
 
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  • #2
The centripetal acceleration of the object is inward isn't it?

Otherwise your numbers look OK.
 
  • #3
for simplicity.

It seems like you have solved the problem correctly. Your calculations and equations all make sense and are consistent. Good job! For part d, your diagram and description of the linear velocity and instantaneous acceleration are correct. The linear velocity is tangent to the circular trajectory and the instantaneous acceleration is perpendicular to the velocity and directed towards the center of the record. Keep up the good work!
 

FAQ: Another friction problem (rotational this time)

What is rotational friction and how is it different from regular friction?

Rotational friction is a type of friction that occurs when an object rotates or rolls over a surface. It is different from regular friction in that it is caused by the contact between two surfaces in motion, rather than just two stationary surfaces rubbing against each other.

How is rotational friction measured?

Rotational friction is typically measured using a unit called torque, which is a measure of the force required to rotate an object. This can be measured using a torque sensor or by calculating the torque based on the weight and distance of the object.

What factors affect the amount of rotational friction?

The amount of rotational friction can be affected by several factors, including the weight and shape of the object, the type and texture of the surface it is rolling on, and the speed at which it is rotating. In general, heavier and rougher objects will have more rotational friction.

How does rotational friction impact the motion of an object?

Rotational friction can slow down the motion of an object by converting its kinetic energy into heat. This can cause the object to roll or rotate at a slower speed, or even come to a complete stop if the friction is strong enough.

Can rotational friction be reduced or eliminated?

While it is not possible to completely eliminate rotational friction, it can be reduced through various methods such as using lubricants, changing the surface or shape of the object, or reducing the weight of the object. However, some amount of rotational friction is necessary for objects to roll and rotate in a controlled manner.

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