Max \omega for Stationary Bar on Prism 1 with Coeff. k < cot\alpha

  • Thread starter sArGe99
  • Start date
  • Tags
    Max Prism
In summary, a prism with a bar of mass m placed on it experiences a horizontal acceleration \omega directed to the left. The maximum value of this acceleration for the bar to remain stationary relative to the prism is given by the equation \omega = \frac {g (sin \alpha - k cos \alpha)}{cos \alpha + k sin \alpha}, where k < cot\alpha. The importance of this data is that if this condition is not met, the sign of the expression for acceleration will change, leading to incorrect results. Additionally, the tendency of the bar to move up the prism is due to the prism's acceleration towards it, which introduces additional forces.
  • #1
sArGe99
133
0

Homework Statement



Prism 1 with bar 2 of mass m placed on it gets a horizontal acceleration [tex]\omega[/tex] directed to the left. At what maximum value of this acceleration [tex]\omega[/tex] will the bar be still stationary relative to the prism, if the coefficient of friction between them [tex]k < cot\alpha[/tex]

th_untitled3.jpg


Homework Equations




[tex]mg sin \alpha - m \omega cos \alpha[/tex] = Force trying to move the body along the prism.
k ([tex]m\omega sin \alpha + mg cos \alpha[/tex]) = Frictional force.


The Attempt at a Solution


Since the prism is accelerating to the left, the bar experiences a pseudo force [tex]m\omega [/tex] directed to the right. Resolving it has components
[tex]m \omega cos \alpha [/tex] opposed to [tex]mg sin \alpha [/tex] and the component [tex]m\omega sin \alpha [/tex] acting along the direction of [tex]mg cos \alpha[/tex]

At equilibrium, the frictional force equals the force which acts along the prism.
Equating, [tex]mg sin \alpha - m\omega cos\alpha[/tex] = k ([tex]m\omega sin \alpha + mg cos \alpha[/tex])
Further simplifying,

[tex]\omega = \frac {g (sin \alpha - k cos \alpha)}{cos \alpha + k sin \alpha}[/tex]

This is not the correct answer. Can you please point out where I've made a mistake?
 
Physics news on Phys.org
  • #2
sArGe99 said:
At equilibrium, the frictional force equals the force which acts along the prism.
Equating, [tex]mg sin \alpha - m\omega cos\alpha[/tex] = k ([tex]m\omega sin \alpha + mg cos \alpha[/tex])
Looks like you made a sign error. Which way does the friction force act? Which direction are you calling positive?

(I'd use ΣF = 0 parallel to the wedge, coupled with a consistent sign convention.)
 
  • #3
the maximum value of k is cot(alpha). So have you tried putting that in?
 
  • #4
Doc Al said:
Looks like you made a sign error. Which way does the friction force act? Which direction are you calling positive?

(I'd use ΣF = 0 parallel to the wedge, coupled with a consistent sign convention.)

Oh. Thanks for that correction. I've got the right answer now. :biggrin:
I reckon I can use the sign system in any direction as long as I put the signs correctly?
 
  • #5
Its given [tex]k < cot\alpha[/tex]. What is the importance of this data in this question? I got the correct answer although I didn't have this point in mind. So I'm thinking why its given..
 
Last edited by a moderator:
  • #6
sArGe99 said:
I reckon I can use the sign system in any direction as long as I put the signs correctly?
Absolutely.
sArGe99 said:
Its given [tex]k < cot\alpha[/tex]. What is the importance of this data in this question? I got the correct answer although I didn't have this point in mind. So I'm thinking why its given..
Consider the sign of your expression for the acceleration if that condition were not met. :wink:
 
  • #7
Sign of the expression for acceleration? I don't notice any particular change when [tex]k < cot\alpha[/tex]! :frown: Where am I going wrong?
 
  • #8
sArGe99 said:
Sign of the expression for acceleration? I don't notice any particular change when [tex]k < cot\alpha[/tex]! :frown: Where am I going wrong?
Take a look at your (corrected) expression. What happens to the sign of the denominator when k > cot α ?
 
Last edited:
  • #9
The denominator turns out to be zero when [tex]k= cot\alpha[/tex]. :tongue:
Thanks for the help.
 
Last edited:
  • #10
I see that the tendency of the body 2 is to move up. Why is it like that, usually it always tends to move down, right?
 
  • #11
sArGe99 said:
I see that the tendency of the body 2 is to move up. Why is it like that, usually it always tends to move down, right?
The block would tend to move down the prism, if the prism weren't accelerating towards it. That changes things by introducing additional forces.
 

1. What does "Max \omega for Stationary Bar on Prism 1 with Coeff. k < cot\alpha" mean?

This phrase refers to the maximum angular velocity (represented by the Greek letter omega, \omega) that a stationary bar can withstand when placed on a prism with a coefficient of friction (k) that is less than the cotangent of the angle of the prism (\alpha).

2. Why is it important to know the maximum angular velocity for a stationary bar on a prism?

Knowing the maximum angular velocity is important because it helps determine the stability of the bar on the prism. If the angular velocity exceeds the maximum, the bar may slip or fall off the prism, potentially causing damage or injury.

3. How is the maximum angular velocity calculated?

The maximum angular velocity can be calculated using the formula \omega = \sqrt{\frac{gk}{\mu R}}, where g is the acceleration due to gravity, k is the coefficient of friction, \mu is the static friction coefficient, and R is the radius of the bar.

4. What factors can affect the maximum angular velocity for a stationary bar on a prism?

The maximum angular velocity can be affected by several factors, including the coefficient of friction, the angle of the prism, the weight and shape of the bar, and the surface conditions of both the prism and the bar.

5. How can the maximum angular velocity be increased?

The maximum angular velocity can be increased by increasing the coefficient of friction, using a steeper angle for the prism, decreasing the weight of the bar, or by using materials with better grip or friction on the surfaces of the bar and prism.

Similar threads

  • Introductory Physics Homework Help
3
Replies
97
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
547
  • Introductory Physics Homework Help
Replies
6
Views
121
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
636
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
2
Replies
38
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
Back
Top