MHB How Does Angular Velocity Affect a Block on a Rotating Wedge?

[Ciaran]

Hi there,

I've come across the following question and drawn a free body diagram:
A wedge with face inclined at an angle θ to the horizontal is fixed to a rotating
turntable. A block of mass m rests on the inclined plane and the coefficient of
static friction between the block and the wedge is µ. The block is to remain at
position R from the centre of rotation of the turntable.
a)If the general acceleration vector in planar polar coordinates is given by (see attachment), show that the acceleration of the block is a = −Rω^2 rˆ, where ω is the angular velocity of the turntable.
b) Find the components of the block’s acceleration parallel and vertical to the
inclined plane.
c)Find the minimum angular velocity ω to keep the block from sliding down
the face of the wedge

Now, I know I have to use cylindrical coordinates with z being constant due to no vertical motion.I can also show the expression for the acceleration using a free body diagram, but not using the expression attached. However, I get the gist of the expression attached as I see that omega squared is (d(theta)/ dt )^2, it's just actually using it to answer part a). And for the rest of the question, I feel I need to use the expression for part a) so am not confident on how to proceed.

I've done quite a few problems involving spherical coordinates and the like, but this one has really stumped me! Any help would be much appreciated

View attachment 4136

 

[I like Serena]

Hi there,

I've come across the following question and drawn a free body diagram:
A wedge with face inclined at an angle θ to the horizontal is fixed to a rotating
turntable. A block of mass m rests on the inclined plane and the coefficient of
static friction between the block and the wedge is µ. The block is to remain at
position R from the centre of rotation of the turntable.
a)If the general acceleration vector in planar polar coordinates is given by (see attachment), show that the acceleration of the block is a = −Rω^2 rˆ, where ω is the angular velocity of the turntable.
b) Find the components of the block’s acceleration parallel and vertical to the
inclined plane.
c)Find the minimum angular velocity ω to keep the block from sliding down
the face of the wedge

Now, I know I have to use cylindrical coordinates with z being constant due to no vertical motion.I can also show the expression for the acceleration using a free body diagram, but not using the expression attached. However, I get the gist of the expression attached as I see that omega squared is (d(theta)/ dt )^2, it's just actually using it to answer part a). And for the rest of the question, I feel I need to use the expression for part a) so am not confident on how to proceed.

I've done quite a few problems involving spherical coordinates and the like, but this one has really stumped me! Any help would be much appreciated

Hi Ciaran!

In polar coordinates the block has coordinates $(r(t),\theta(t))$. For short: $(r,\theta)$.

If the block is to remain at position R from the center, that implies that $r(t)=R=constant$.
Consequently the first and second derivatives with respect to time are: $\dot r = \ddot r = 0$.

Similarly, the angular velocity has to remain constant at angular velocity $\omega$.
That means that $\dot \theta = \omega = constant$ and thus $\ddot \theta = 0$.

What do you get if you fill all that in into the general expression for acceleration in polar coordinates that you have? (Wondering)
$\mathbf a = (\ddot r-r\dot\theta^2)\mathbf{\hat r} + (r\ddot \theta + 2\dot r \dot \theta)\boldsymbol{\hat \theta} \tag 1$

 

[Ciaran]

Hello! I do indeed get the expression needed ! So for b), do I need to use the two components of the acceleration from part a)? And for c), do I simply need to consider the component of weight down the slope, the centripetal force and the force of friction?

 

[I like Serena]

Hello! I do indeed get the expression needed ! So for b), do I need to use the two components of the acceleration from part a)? And for c), do I simply need to consider the component of weight down the slope, the centripetal force and the force of friction?

I suggest drawing a Free Body Diagram (FBD) showing the block on the incline and all forces working on the block.
They have to add up to the required centripetal acceleration that you've found in a).

Apparently you've already figured out which forces there are, except that you've missed one: the normal force.
So indeed, you need to express them in formulas and put them in 2 equations.

For c) you need to replace the force of friction by the maximum possible friction.

 

[Ciaran]

So for parallel to the slope, I have a= g sinθ (1-µ) as force up slope= force down slope as it isn't moving. For vertically, I feel like I'm missing something because isn't it just zero?

 

[I like Serena]

So for parallel to the slope, I have a= g sinθ (1-µ) as force up slope= force down slope as it isn't moving. For vertically, I feel like I'm missing something because isn't it just zero?

For b) you can't assume that the force of friction is equal to the maximum possible friction - it is just an unknown force.
Moreover, the normal force is not equal to the mg sinθ. For starters it should be a cosine instead of a sine. But even then, they are not equal.
The normal force will be bigger. That's because it has to bring in part of the centripetal force.

Furthermore, this is not the centripetal acceleration, but only its component that is parallel to the slope.
The real centripetal acceleration is horizontal instead of parallel to the slope.

For the perpendicular forces the same applies: they do not cancel, since they have to result in the perpendicular component of the centripetal force.

 

[Ciaran]

So, vertically, the component of centripetal force is the normal force minus mgcosθ. And, parallel to the slope, the component of centripetal force is mgsinθ minus the unknown frictional force?

 

[I like Serena]

So, vertically, the component of centripetal force is the normal force minus mgcosθ. And, parallel to the slope, the component of centripetal force is mgsinθ minus the unknown frictional force?

I suggest to avoid the word vertical here - it's confusing - I'd call it perpendicular (to the slope). (Nerd)

But yes, that's it. (Nod)

 

[Ciaran]

Ahh I see, I can see how it could lead to confusion! But then how am I meant to derive 2 expressions for acceleration if I have unknown forces? I don't know how to express the friction or the normal force

 

[I like Serena]

Ahh I see, I can see how it could lead to confusion! But then how am I meant to derive 2 expressions for acceleration if I have unknown forces? I don't know how to express the friction or the normal force

For b) you'll have to leave friction as an unknown force.
You can set up 2 equations though, including the centripetal acceleration of $\omega^2 R$.

For c) you can set the force of friction to its maximum of $\mu N$ and find $\omega$ as a function of $\theta$ and $R$.

 

[Ciaran]

Is it just a_ perp= (N/m)- gcosθ and a_ parallel= gsinθ- (F_f/m) ? And for c), I get ω= sqrt( (g/R)(μsinθ- tanθ))

 

[I like Serena]

Is it just a_ perp= (N/m)- gcosθ and a_ parallel= gsinθ- (F_f/m) ?

Yes.

And for c), I get ω= sqrt( (g/R)(μsinθ- tanθ))

I get a different trigonometric fraction, so one of us must have made a mistake.
But yes, it should be something like that.

 

[Ciaran]

Really? That seems quite messy and bitty! What did you get as your answer? I started off with mgsinθ +mRω^2 cosθ = μmg cosθsinθ and simplified for omega from there