What Determines the Rider's Fate in Different Speed Domains of a Vertical Loop?

[S_Flaherty]

3. Consider the motorcyclist riding on the inside of a vertical circle of radius b. His initial speed at the bottom of the circle is u.

a) Explain why the three domains (i) u^2 < 2gb, (ii) 5gb > u^2 > 2gb and (iii) u^2 > 5gb are qualitatively different.

b) Describe what happens in domain (ii). Give a sketch of the trajectory, indicating the height h where the rider is released from the hoop. Calculate h(u).So for (a) I said that the domains are the speed of the motorcyclist at different points along the circle, domain i being the top of the track since speed would be smallest, domain being the bottom where speed is greatest, and domain ii being everywhere else. Would I be correct in saying this?

For (b), if my explanation above is correct then I know what happens in domain ii, where gravity affects the speed of the motorcyclist in such a way that it decreases his speed as he travels from the bottom to the top and increases it as he goes back from the top to the bottom. I'm just confused on the wording of the rest of (b) and not sure how to approach solving this since I thought the rider is in a closed hoop so how can he be released from it. Am I not reading the problem correctly?

 

[Spinnor]

There are three types of motion depending on the initial velocity. For the first type of motion let the initial velocity be very very small,

(mv^2/2 << 2mgb).

What happens?

Now keep adding energy and imagine or draw what happens. Then do the math.

 

[bossman27]

So for (a) I said that the domains are the speed of the motorcyclist at different points along the circle, domain i being the top of the track since speed would be smallest, domain being the bottom where speed is greatest, and domain ii being everywhere else. Would I be correct in saying this?

Be careful not to oversimplify energy by only talking about speed(velocity). Energy (as you know) includes potential as well as kinetic energy. I'd try recasting the domains as explicit energy expressions.

 

[S_Flaherty]

There are three types of motion depending on the initial velocity. For the first type of motion let the initial velocity be very very small,

(mv^2/2 << 2mgb).

What happens?

Now keep adding energy and imagine or draw what happens. Then do the math.

The total energy is essentially the potential energy right? Since kinetic energy is negligible when it is much smaller than potential.

 

[S_Flaherty]

Be careful not to oversimplify energy by only talking about speed(velocity). Energy (as you know) includes potential as well as kinetic energy. I'd try recasting the domains as explicit energy expressions.

Ok so domain i is where energy is essentially potential energy, which is at the top of the hoop, domain iii is where energy is essentially just kinetic energy which is at the bottom and domain ii is everywhere else.

 

[bossman27]

The total energy is essentially the potential energy right? Since kinetic energy is negligible when it is much smaller than potential.

Keep in mind that the rider starts at the bottom of the circle.

 

[S_Flaherty]

Keep in mind that the rider starts at the bottom of the circle.

So if initial velocity is really small he won't have enough kinetic energy to counteract gravity, and just stay at the bottom?

 

[bossman27]

The questions aren't asking, "what point is he at, when x > y?" They are asking for you to explain what happens if the given relationships are true *when he enters the circle*.

 

[bossman27]

So if initial velocity is really small he won't have enough kinetic energy to counteract gravity, and just stay at the bottom?

Yup, now we're getting somewhere. He won't necessarily stay right at the bottom, but he will not reach the top.

 

[S_Flaherty]

The questions aren't asking, "what point is he at, when x > y?" They are asking for you to explain what happens if the given relationships are true *when he enters the circle*.

Oh ok, so I need to say that in domain i he won't be able to move up the side of the hoop, in domain iii he can move in a complete circle and in domain ii he will stop before he gets past the top of the hoop?

 

[bossman27]

Oh ok, so I need to say that in domain i he won't be able to move up the side of the hoop, in domain iii he can move in a complete circle and in domain ii he will stop before he gets past the top of the hoop?

Almost, except for domain (ii). Here we need to consider the velocity he must have at the top of the circle in order to maintain contact with the track. Hint: at this point, the centripetal force is in the opposite direction of the gravitational force. In order to maintain contact, he must have enough centripetal force to cancel out the gravitational force.

Formula for centripetal force is:
$F_{c} = \frac{mv^{2}}{r}$

Edit: rereading the question, I suppose you don't have to prove this as it only asks for a qualitative answer.

 

[S_Flaherty]

Almost, except for domain (ii). Here we need to consider the velocity he must have at the top of the circle in order to maintain contact with the track. Hint: at this point, the centripetal force is in the opposite direction of the gravitational force. In order to maintain contact, he must have enough centripetal force to cancel out the gravitational force.

Formula for centripetal force is:
$F_{c} = \frac{mv^{2}}{r}$

Edit: rereading the question, I suppose you don't have to prove this as it only asks for a qualitative answer.

Ok so for part (a) what I said is fine? Would I need to show this relation where centripetal force equals potential at the top to find h in part (b) of the question?

 

[bossman27]

Ok so for part (a) what I said is fine? Would I need to show this relation where centripetal force equals potential at the top to find h in part (b) of the question?

Ahh yes, except remember that it's not $F_{c} = U_{g} = mgh$

but is instead $F_{c} = F_{g} = mg$

 

[S_Flaherty]

Ahh yes, except remember that it's not $F_{c} = U_{g} = mgh$

but is instead $F_{c} = F_{g} = mg$

So m(v^2)/r = mg which means v = sqrt(gr) for the rider to make it to the top, so for domain ii v < sqrt(gr). In order to calculate h(u) like the problem says, do I merely need to find some relationship like a > h(u) > b where a is 2 times the radius and b is some value determined by conservation of energy using v = sqrt(gr)?

 

[S_Flaherty]

I thought about my last post some more and thought of something that confuses me. If the speed required to make it to the top of the loop is sqrt(gr) then wouldn't the rider make the loop in domain ii and iii since u^2 > 2gr in both cases? Also since domain i isn't u^2 < gr, doesn't that mean that the rider could make it past the top of the loop for this domain as well?

 

[haruspex]

Wrt (i), I think you need to be a little more precise about "won't make it up the loop".

Wrt (ii), how much KE will the rider still have at the top? What speed does that imply? What is the min speed there to remain in contact with the track?

 

[S_Flaherty]

Wrt (i), I think you need to be a little more precise about "won't make it up the loop".

Wrt (ii), how much KE will the rider still have at the top? What speed does that imply? What is the min speed there to remain in contact with the track?

I did some calculations and found that the min speed required is sqrt(4gr). So for (a) of the problem I need to say that in domain i the rider will not reach the top of the hoop since u^2 < 2gb which is less than 4gb (where b = r). Domain iii the rider will definitely make it to the top of the hoop and will continue since his speed is greater than the min required to stay on the track, and in domain ii it depends on the speed, if u^2 > 4gb then he will make it past the top of the hoop, if u^2 < 4gb then he will not.

 

[haruspex]

I did some calculations and found that the min speed required is sqrt(4gr). So for (a) of the problem I need to say that in domain i the rider will not reach the top of the hoop since u^2 < 2gb which is less than 4gb (where b = r). Domain iii the rider will definitely make it to the top of the hoop and will continue since his speed is greater than the min required to stay on the track, and in domain ii it depends on the speed, if u^2 > 4gb then he will make it past the top of the hoop, if u^2 < 4gb then he will not.

For domain (i), there's more to it than merely "won't make it to the top". (Otherwise it would be the same as domain (ii).) There are two ways in which the rider might not make it to the top, with a considerable difference in the health & safety consequences.
For domain (ii), you should get a 5 in there, not 4. Please show your working.

 

[bossman27]

So m(v^2)/r = mg which means v = sqrt(gr) for the rider to make it to the top, so for domain ii v < sqrt(gr). In order to calculate h(u) like the problem says, do I merely need to find some relationship like a > h(u) > b where a is 2 times the radius and b is some value determined by conservation of energy using v = sqrt(gr)?

Careful, I think you have missed a step for v. The v we're talking about for centripetal force is what he must *have left* at the top. This means we need $KE_{o} \geq KE_{t} + PE_{t}$

I'll leave the expansion to you, but we can simplify this down to find the relationship $v_{o} \geq \sqrt{5gr}$

I'm working on the h(u) function, I'll get back to you in a few.

 

[S_Flaherty]

For domain (i), there's more to it than merely "won't make it to the top". (Otherwise it would be the same as domain (ii).) There are two ways in which the rider might not make it to the top, with a considerable difference in the health & safety consequences.
For domain (ii), you should get a 5 in there, not 4. Please show your working.

For the min speed velocity at the top will be zero, and at the bottom of the track potential energy is zero. So (1/2)m(u^2) = 2mgb, the m's cancel, and I multiplied both sides by 2 so
u^2 = 4gb.

 

[bossman27]

For the min speed velocity at the top will be zero, and at the bottom of the track potential energy is zero. So (1/2)m(u^2) = 2mgb, the m's cancel, and I multiplied both sides by 2 so
u^2 = 4gb.

You forgot that the KE at the top isn't zero in this case, it's just the threshold KE where Fg = Fc at that point.

Take a look at the equation in my last post, that should help you see what you forgot.

 

[S_Flaherty]

You forgot that the KE at the top isn't zero in this case, it's just the threshold KE where Fg = Fc at that point.

Take a look at the equation in my last post, that should help you see what you forgot.

Oh ok I see what I did, so since the min speed is v^2 = gr, KEt is .5mgr so
.5u^2 > .5gr + 2gr which leads to u^2 > 5gr.

 

[bossman27]

Oh ok I see what I did, so since the min speed is v^2 = gr, KEt is .5mgr so
.5u^2 > .5gr + 2gr which leads to u^2 > 5gr.

Although it isn't really good form to start with the answer and work your way backwards, yes, that is a correct statement.

Here's how I worked out a formula for $h(v_{0})$ (v₀ = initial velocity)

The height at which the motorcycle falls off the track is when the centripetal force is equal to the gravitational force perpendicular to the track.

This is simply $F_{c} = F_{g}sin\theta$

This can be simplified further by instead using $a_{c} = a_{g}sin\theta$

Recall that $a_{g} = g$ and $a_{c} = \frac{v_{t}^2}{r}$, where $v_{t}$ is the velocity at any given moment.

It will also behoove you to draw a graph of the circle, and convince yourself that $rsin\theta = (h - r)$

You will also need an expression for $v_{t}^2$ in terms of $v_{0}^2$ and $gh$ -- you can find this expression by simplifying that kinetic energy equation I gave you earlier.

Use your $v_{t}^2$ expression to solve the simplified version of the $a_{c} = a_{g}sin\theta$ equation. Solving for $h$, of course.

This should give you an equation for $h$ in terms of $v_{0}^2$ that will give you a height of $r$ and $2r$ when you plug in the respective boundary values of $v_{0}^2$.

You can interpret the meaning of those results for yourself.

Hope that helps!

 

[S_Flaherty]

Although it isn't really good form to start with the answer and work your way backwards, yes, that is a correct statement.

Here's how I worked out a formula for $h(v_{0})$ (v₀ = initial velocity)

The height at which the motorcycle falls off the track is when the centripetal force is equal to the gravitational force perpendicular to the track.

This is simply $F_{c} = F_{g}sin\theta$

This can be simplified further by instead using $a_{c} = a_{g}sin\theta$

Recall that $a_{g} = g$ and $a_{c} = \frac{v_{t}^2}{r}$, where $v_{t}$ is the velocity at any given moment.

It will also behoove you to draw a graph of the circle, and convince yourself that $rsin\theta = (h - r)$

You will also need an expression for $v_{t}^2$ in terms of $v_{0}^2$ and $gh$ -- you can find this expression by simplifying that kinetic energy equation I gave you earlier.

Use your $v_{t}^2$ expression to solve the simplified version of the $a_{c} = a_{g}sin\theta$ equation. Solving for $h$, of course.

This should give you an equation for $h$ in terms of $v_{0}^2$ that will give you a height of $r$ and $2r$ when you plug in the respective boundary values of $v_{0}^2$.

You can interpret the meaning of those results for yourself.

Hope that helps!

This helps a lot, thank you so much for all the time you spent helping me with this problem.

 

[bossman27]

This helps a lot, thank you so much for all the time you spent helping me with this problem.

No problem at all, I wouldn't be on here trying to help out if I didn't get a bit of enjoyment out of it myself. And hopefully now you'll have a better understanding when you see problems like this in the future.