Question Involving angular and linear motion and friction.

In summary, Dave has hit a brick wall while trying to work out how to solve a question for a science assignment. He needs help from someone who is more experienced in the subject. Greg has replied to Dave and is willing to help him out.
  • #1
Dave_George
7
0
Hello to all,

Im new here, just found the site and wishing I had stumbled across this a few months back when I started my Electrical engineering course.

I hope this is in the correct section, I did read through where to put this but apologies if I have placed this in the wrong section...

I have a distinction level question in my science assignment which is causing me some headache, any help, guidance or pointers in the right direction would be much appreciated! I am not looking for answers but looking for ways too those answers if possible..

Q) A solid flat disc is rotating horizontally. If a mass of 2KG is placed 0.15M from the centre of rotation and assuming a coefficient of friction of 0.3, determine;

i)The maximum angular velocity that the disc can be rotated at without the block moving

ii)The linear velocity of the block if the angular velocity in (i) is exceeded.

iii)If the block mass was increased to 4KG what effects would this have on the max angular veolcity calculated in (i)...


I have some idea of linear and angular motion but we only briefly touched on combining the two into questions like these, we were told to "Have a go" at this question if we could, problem of course is that if you can't do it you can't hit the distinction grade...With a few pointers i think i could work this out, can anyone help??

many thanks, hope to see you in the forums

David
 
Physics news on Phys.org
  • #2
Hi Dave...would I be correct in assuming that you have had a go at this question but somewhere in your working you've hit a brick wall that you can't seem to get past (either through some subtle error or using a wrong method)?...could you post your working (or even your ideas) so that we can see what bit is stopping you from making progress?
 
Last edited:
  • #3
Hi Greg, thanks for your reply.

I have indeed hit a bit of a brick wall, I think its through not knowing which method to use to work this out, possibly we haven't covered it enough...but I have some handouts and will try and explain where I am currently as far as idea's...

I can see that the values I have been given are what's required to work this out, the size and mass of the rotating disc are not important, and that this seems to be centered around the centripetal force which I think is F = m w2 r

There is obviously a force acting downwards on the block which I think is F=mg

So I think the answer to i is that the maximum velocity the disc can rotate is when the centripetal force is greater than the force acting downwards?

Force downwards is F=mg therefore F=2x9.81 = 19.62n/m2

Centripetal force is F=m w2 r but I have no value for angular velocity (w2)?

Can I therefore assume a value of say 1rpm (0.105 rad/s) and work out what the centripetal force would be for that and then work up until that force became higher? 1rpm would be F=2x0.105sqx0.15 = 0.0033075n/msq


Am I along the right lines or totally off course?

Any advice would be greatly appreciated. Also for the radius do I assume that 0.15m is the value as I have not been given the size of the disc?

suffice to say I am not sure that the above is the right way to work this out, hope you can help!

best regards

David
 
  • #4
Welcome to PF, Dave!

1. You are not asked to find the angular velocity in general, you are asked to find the maximum angular velocity the disk can have if the block is not to move relative to the disk.

2. Remember that the force of static friction F is always less than maximal static friction, 0.3mg

3. F must provide the centripetal acceleration of the block; hence we have the inequality:
[tex]m\omega^{2}r=F\leq{0.3}mg[/tex]

r is the distance of the block to the centre of the rotation.
Use this inequality to determine [itex]\omega_{max}[/itex]
 
  • #5
arildno said:
Welcome to PF, Dave!

1. You are not asked to find the angular velocity in general, you are asked to find the maximum angular velocity the disk can have if the block is not to move relative to the disk.

2. Remember that the force of static friction F is always less than maximal static friction, 0.3mg

3. F must provide the centripetal acceleration of the block; hence we have the inequality:
[tex]m\omega^{2}r=F\leq{0.3}mg[/tex]

r is the distance of the block to the centre of the rotation.
Use this inequality to determine [itex]\omega_{max}[/itex]


Hi arildno, many thanks for your reply, I have studied it but have to admit you have lost me a bit...!

I can sort of see what you mean, just some of the terminology confusing me perhaps??:confused:

Do appreciate your time in helping me:smile:

best regards

David
 
  • #6
Where did you get lost to begin with?
 
  • #7
Hmm to put it in slightly different terms...for part i)...assume that the angular rotation is miniscule...the block would not move...what force is preventing the block from moving?...what would cause the value of this force to increase?...and finally, what is the greatest value of this force?

If we know what the greatest value of this force is then we can say that if the object is not going to move then the value of this force must always be equal to the centripetal force...and so at such point that this force is at it's greatest value, what is implied?

Have tried to leave some of it to you :smile:
 
Last edited:
  • #8
GregA said:
Hmm to put it in slightly different terms...for part i)...assume that the angular rotation is miniscule...the block would not move...what force is preventing the block from moving?...what would cause the value of this force to increase?...and finally, what is the greatest value of this force?

If we know what the greatest value of this force is then we can say that if the object is not going to move then the value of this force must always be equal to the centripetal force...and so at such point that this force is at it's greatest value, what is implied?

Have tried to leave some of it to you :smile:

Cheers for the reply again, reading what you have written above...

the force preventing the block from moving must surely be "mg" and also the 0.3 coefficient of friction?

The thing that would cause the value of this force to increase is the weight of the block increasing perhaps?

The greatest value of this force? I am not sure on this one, (Im starting to feel quite stupid now...got top marks in all my other science assignments, but this question has me stumped!)

am i along the right lines with the above?

regards

David
 
  • #9
Dave_George said:
Cheers for the reply again, reading what you have written above...

the force preventing the block from moving must surely be "mg" and also the 0.3 coefficient of friction?

The thing that would cause the value of this force to increase is the weight of the block increasing perhaps?

The greatest value of this force? I am not sure on this one, (Im starting to feel quite stupid now...got top marks in all my other science assignments, but this question has me stumped!)

am i along the right lines with the above?

regards

David

the force preventing the block from moving must surely be "mg" and also the 0.3 coefficient of friction?
this is close but [tex] \mu [/tex] should not so much be combined with mg as it should be combined with R (where R = normal reaction)...though in this scenario your answer is a simply a shortcut using [tex] R=mgcos \theta [/tex] ...on an incline your way of thinking would cause problems.

The thing that would cause the value of this force to increase is the weight of the block increasing perhaps?
Hmm...imagine that you put a 20kg bar-bell on top of a rotating surface and at the speed it's rotating it has no tendency to slide off...if you put another another 9 similar weights on top of it can you imagine this 200kg mass of weights starting to move towards the centre?:wink: (the only force acting on the object is friction...and this force is trying to prevent the object from going anwhere (relative to the bit of surface it's sitting on!...relative to you the friction is making that object go round in circles)) by using F=ma you get:

[tex]\inline \mu mg =ma [/tex] (using your shortcut here)...can you see something that both sides of this equation share in common?...what does this imply?...finally with regards to angular rotation is there any way of expressing "a" in a form that is more useful?
 
Last edited:
  • #10
GregA said:
the force preventing the block from moving must surely be "mg" and also the 0.3 coefficient of friction?
this is close but [tex] \mu [/tex] should not so much be combined with mg as it should be combined with R (where R = normal reaction)...though in this scenario your answer is a simply a shortcut using [tex] R=mgcos \theta [/tex] ...on an incline your way of thinking would cause problems.

The thing that would cause the value of this force to increase is the weight of the block increasing perhaps?
Hmm...imagine that you put a 20kg bar-bell on top of a rotating surface and at the speed it's rotating it has no tendency to slide off...if you put another another 9 similar weights on top of it can you imagine this 200kg mass of weights starting to move towards the centre?:wink: (the only force acting on the object is friction...and this force is trying to prevent the object from going anwhere (relative to the bit of surface it's sitting on!...relative to you the friction is making that object go round in circles)) by using F=ma you get:

[tex]\inline \mu mg =ma [/tex] (using your shortcut here)...can you see something that both sides of this equation share in common?...what does this imply?...finally with regards to angular rotation is there any way of expressing "a" in a form that is more useful?

Both sides of the equation share M in common? I am not sure what this implies to be honest! And with regards to angular rotation when you write "a" do you mean acceleration?

Sorry if it seems like I am just not getting this, however this question is so unlike the other ones we have done I just can't seem to make sense of it.

Appreciate you trying to point me in the right direction though

regards

Dave
 
  • #11
yes..."a" = acceleration..and it can be expressed as something else more helpful...Thing is I don't know how much more info I can just give you without running the risk of stealing any satisfaction you could get from arriving at an answer.
 
  • #12
GregA said:
yes..."a" = acceleration..and it can be expressed as something else more helpful...Thing is I don't know how much more info I can just give you without running the risk of stealing any satisfaction you could get from arriving at an answer.

Im not sure how to express "a" as something more helpfull GregA, don't worry about it the only satisfaction I am going to get from this question is when I burn my question paper!

Im going to email my tutor and request we do an example that is similar to this next time I am in college, I have no examples to look back on and even a web search to find similar problems has proven fruitless.

regards

Dave
 
  • #13
with regards to angular rotation...the centrepetal acceleration can be expressed either as [tex] r \omega^2 [/tex] or [tex] \frac {v^2}{r} [/tex] (where r = radius of the circle being described)
The first of these is two is the better choice based on the fact that you know [tex] r [/tex] and with a bit of manipulation will leave you can arrive at an expression that leaves only one variable [tex] \omega^2 [/tex]...the other two parts of the question just require a slight bit of thinking :smile:
 
Last edited:
  • #14
GregA said:
with regards to angular rotation...the centrepetal acceleration can be expressed either as [tex] r \omega^2 [/tex] or [tex] \frac {v^2}{r} [/tex] (where r = radius of the circle being described)
The first of these is two is the better choice based on the fact that you know [tex] r [/tex] and with a bit of manipulation will leave you can arrive at an expression that leaves only one variable [tex] \omega^2 [/tex]...the other two parts of the question just require a slight bit of thinking :smile:

So your saying a=rw2 ? (I don't know how to put the correct text format for the formulae) Which can then be transposed to make w2 the subject I suppose. Where does the acceleration come into this though Greg when I was asked to find the max angular velocity?

I can sort of see that if i can do the first part the other two parts should slot into place...

cheers

David
 
  • #15
Think this way. If the disc rotates, the mass is centripetally accelerating (until it starts to slide). (GregA showed how to relate the centripetal acceleration to the angular velocity.) Now, for any object to accelerate, Newton's 2nd law tells us that a force must be exerted on the object. In this problem, the only force available to provide the centripetal force is the force of static friction. Figure out the maximum force that static friction can exert. Given this, figure out the maximum possible acceleration. Then you'll be able to find the maximum angular velocity.
 

1. How does friction affect the motion of an object?

Friction is a force that opposes the motion of an object. It acts in the opposite direction of the object's motion and can cause it to slow down or come to a stop.

2. What is the difference between angular and linear motion?

Linear motion involves movement in a straight line, while angular motion involves rotation around an axis. Angular motion also includes concepts such as angular velocity and acceleration.

3. How do you calculate the force of friction?

The force of friction can be calculated using the formula F = μN, where μ is the coefficient of friction and N is the normal force acting on the object.

4. What factors affect the coefficient of friction?

The coefficient of friction can be affected by the nature of the surfaces in contact, the force pressing the surfaces together, and the presence of any lubricants or substances between the surfaces.

5. How does friction affect the acceleration of an object?

Friction can reduce the acceleration of an object by opposing its motion. The amount of friction present can also affect the rate at which an object accelerates or decelerates.

Similar threads

  • Introductory Physics Homework Help
Replies
25
Views
1K
  • Introductory Physics Homework Help
2
Replies
62
Views
9K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
2
Replies
38
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
866
  • Introductory Physics Homework Help
Replies
2
Views
5K
  • Introductory Physics Homework Help
Replies
6
Views
4K
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
Back
Top