# Rotation: Speed of an object as it slips off a rotating disk

• sksmith6471
In summary, the problem involves a 75g mass placed 75cm from the center of a rotating platform with a uniform angular acceleration of 0.125rad/s. The coefficient of static friction between the mass and the platform is 0.250. By setting up a free body diagram and using the sum of forces in the x and y directions, the speed of the mass can be calculated using the equation v=ωr. After solving for ω, the speed is approximately 1.36 m/s, which is the maximum speed the mass can rotate without slipping. A small error in the answer options has been corrected by the instructor.
sksmith6471

## Homework Statement

A $75g$ mass sits $75cm$ from the center of a rotating platform undergoing a uniform angular acceleration of $0.125rad/s$. The coefficient of static friction between the mass and the platform is $0.250$. What is the speed of the mass when is slides off?

A. 0.889 m/s
B. 1.26 m/s
C. 1.44 m/s
D. 1.58 m/s
E. It will never slide off

I'm also given an image of a disk with a mass on it, and the mass is located at the edge of the disk a distance r from the center.

My variables as far as I can tell:
m=75g
r=.75m
g=9.8m/s2

## Homework Equations

∑Fx=fs=mar
∑Fy=N-mg=0
fssN
ar2r
at=αr

I'm not sure if there are other equations I need in order to solve this problem

## The Attempt at a Solution

I set up a free body diagram using the sum of the forces in the x and y directions:
∑Fx=fs=mar
∑Fy=N-mg=0

I solved for fssmg

I replaced a w/ ω2r to get ∑Fxsmg=mω2r

Then I solved for ω to get $ω=\sqrt{μg/r}=\sqrt{\frac{(.25)(9.8)}{.75}}$. This gave me an answer of $1.807rad/s$ which isn't close to any of the given answers.

I'm also unsure because I didn't actually use the value for the angular acceleration. I thought that by solving the problem how I did I would be able to find the maximum speed the object would be able to rotate on the disk without slipping, and any values greater than that would cause the object to slide and fall off. I'm struggling with this problem and I don't know if I'm even starting correctly.

Hi sksmith6471. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

The answer I get is not listed; it is partway between that of B and C.

Last edited by a moderator:
Wow, thank you. I didn't even notice my units were different than the given answers. I converted my units to m/s using ##v=\omega r##. This gave me 1.356m/s which is between answers B and C. This should be the speed of the object just before it starts to slip.

Did I approach this problem correctly? If the speed was even slightly greater than this value it should start to slip right? These questions are created by our instructor so there may be an error with the answers.

You're both missing the fact that the centripetal acceleration is just one of the components of the acceleration. There is also a tangential acceleration to be calculated. The total acceleration can then be found using Pythagorean theorem and related to the friction force using Newton's 2nd law.

dauto said:
You're both missing the fact that the centripetal acceleration is just one of the components of the acceleration. There is also a tangential acceleration to be calculated. The total acceleration can then be found using Pythagorean theorem and related to the friction force using Newton's 2nd law.
You're right! I originally calculated tangential accelerating force, saw that it was well below tangential friction so dismissed it. I think I can see which option to choose, now.

I'm a bit confused. Are you saying that ∑Fx=fs=mar should actually be ∑Fx=fs=mα? then I take the squareroot ar2+at2 plug it into my force equation and then solve for ω?

Last edited:
sksmith6471 said:
I'm a bit confused. Are you saying that ∑Fx=fs=mar should actually be ∑Fx=fs=mα? then I take the squareroot ar2+at2 plug it into my force equation and then solve for ω?

Yes, that's right.

Ok, I set up my equation: ∑Fx=fs=mα→ μmg=m\sqrt{ar2+at2} (I'm not sure why itex isn't working for me, so i omitted the commands). After some simplification I get ω2=\sqrt{(μg)2-(αr)2}/r.

I plugged in my values at this point and found that:

Then i plugged in :
v=ωr → v=(1.806)(.75)≈1.355m/s.

Edit: I'm also confused as to why (assuming my math is correct) the answer that I just got is approximately the same as in my second post.

Last edited:
Now your calculation seems correct. And you're right, the tangential acceleration had little effect because the angular acceleration is so small. Are you sure you provided the correct data for the problem?

I spoke with my teacher and he said there was a typo in the answer selection. The correct answer is in fact 1.36 m/s and answer B should be 1.36 m/s. Thank you dauto and NascentOxygen.

As a side note: he also mentioned that the acceleration he gave was the tangential acceleration. Luckily the value doesn't make too much of a difference so the answer wasn't affected by much.

## 1. What is rotation?

Rotation is the circular movement of an object around a central axis or point.

## 2. How is the speed of an object affected as it slips off a rotating disk?

The speed of an object as it slips off a rotating disk is affected by the tangential velocity of the object, the distance from the center of rotation, and the angular velocity of the disk. As the object moves farther away from the center, its tangential velocity decreases, causing it to slow down.

## 3. What is tangential velocity?

Tangential velocity is the speed at which an object is moving in a circular path, tangential to the circle at any given point. It is perpendicular to the radial velocity, which is the speed at which the object is moving towards or away from the center of rotation.

## 4. How does the angular velocity of the disk affect the speed of the object?

The angular velocity of the disk directly affects the speed of the object as it slips off. The higher the angular velocity, the faster the object will move along the circular path. However, as the object moves farther from the center of rotation, its tangential velocity decreases, causing it to slow down.

## 5. What is the relationship between the distance from the center of rotation and the speed of the object?

The distance from the center of rotation and the speed of the object have an inverse relationship. As the distance increases, the speed of the object decreases. This is due to the fact that the tangential velocity of the object is directly proportional to the distance from the center of rotation.

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