# Kinematics: Acceleration of a figure skater changing direction

• phantomvommand
phantomvommand
Homework Statement
A boy enters a patch of ice with a coefficient of friction µ with speed v.
By running on the ice, the boy turns his velocity vector by 90◦ in the minimum possible time, so
that his final speed is also v. What is the minimum possible time, and what kind of curve is the
trajectory? Assume the normal force with the ice is constant.
Relevant Equations
v = u + at
My guess was simply that as acceleration changes from the north to east direction, the total magnitude change of v is ##v \sqrt 2##.
Acceleration is ##\mu g##, so time would be ##\frac {v \sqrt 2} {\mu g}##. This agrees with the textbook solution.

What I do not understand is the trajectory taken. Supposedly, because acceleration is constant, the answer is parabola. How is this arrived at? For example, acceleration in circular motion (at constant speed v) is also constant, so I do not see why constant acceleration means that it must be a parabola (and not a circle, or anything else!).

Many thanks for any help.

Acceleration and Force are vectors with both magnitude and direction. In a circular path, the magnitude of the acceleration is constant, but the direction is always toward the center of the circle. So this is not constant acceleration.

scottdave said:
Acceleration and Force are vectors with both magnitude and direction. In a circular path, the magnitude of the acceleration is constant, but the direction is always toward the center of the circle. So this is not constant acceleration.
How did you know that acceleration's direction is constant in the above question?

And in that case, would it be correct to say that acceleration is constantly directed southeast (or 45 degrees south of east), despite the boy's direction changing? Why should that be so, if friction is supposed to be directly opposite the boy's velocity?

phantomvommand said:
Homework Statement: A boy enters a patch of ice with a coefficient of friction µ with speed v.
By running on the ice, the boy turns his velocity vector by 90◦ in the minimum possible time, so
that his final speed is also v. What is the minimum possible time, and what kind of curve is the
trajectory? Assume the normal force with the ice is constant.
Relevant Equations: v = u + at

My guess was simply that as acceleration changes from the north to east direction, the total magnitude change of v is ##v \sqrt 2##.
Acceleration is ##\mu g##, so time would be ##\frac {v \sqrt 2} {\mu g}##. This agrees with the textbook solution.

What I do not understand is the trajectory taken. Supposedly, because acceleration is constant, the answer is parabola. How is this arrived at? For example, acceleration in circular motion (at constant speed v) is also constant, so I do not see why constant acceleration means that it must be a parabola (and not a circle, or anything else!).

Many thanks for any help.
If the speed remains constant throughout, then the force must be perpendicular to the velocity, which implies a circular path. It's not clear, however, whether that minimises the time to complete the turn.

One extreme option is to decelerate to rest, then accelerate in a perpendicular direction. The other options are everything in between, involving both deceleration and turning, followed by acceleration and turning.

Have you been studying the calculus of variations?

phantomvommand said:
friction is supposed to be directly opposite the boy's velocity?
Friction opposes the relative motion of surfaces in contact.
If you are facing North and attempting to run straight ahead but you are currently sliding East, your feet are moving somewhere between South and East relative to the ice, so friction acts on your feet somewhere between North and West. The exact direction depends how fast you are pushing your feet backwards compared to how fast you are sliding East at a given instant.

Note what this means about your forward acceleration. Had you not been sliding East, it would not matter how fast you tried to run; your forward thrust, if slipping, is limited to ##N\mu_k##. But because you are sliding sideways, some of the thrust is going towards slowing that motion. The faster you try to run forwards the greater the share of the thrust is in that direction.

Last edited:
phantomvommand
PeroK said:
Have you been studying the calculus of variations?
Yes, it could be intended as an exercise in that, but it is overkill here.
We know what ##\int\vec F(t)\cdot dt## needs to be. If at any time ##\vec F## has a component normal to that, that component is wasted and has to be cancelled by an opposing component at some other time. Hence the direction is constant.

MatinSAR and jbriggs444
phantomvommand said:
How did you know that acceleration's direction is constant in the above question?
To change a velocity between any two given velocity values if acceleration is limited to a particular magnitude is the straight line in velocity space, i.e., when acceleration is constantly pointing from the initial to the final velocity.

phantomvommand said:
And in that case, would it be correct to say that acceleration is constantly directed southeast (or 45 degrees south of east), despite the boy's direction changing? Why should that be so, if friction is supposed to be directly opposite the boy's velocity?
Friction is not opposite to the boy's velocity. The boy is actively working to make it point in the correct direction.

PeroK said:
If the speed remains constant throughout, then the force must be perpendicular to the velocity, which implies a circular path. It's not clear, however, whether that minimises the time to complete the turn.
It does not. See #6.
PeroK said:
Have you been studying the calculus of variations?
I think that's generally severe overkill for realising that the shortest path between to point is a straight line.

As to the OP's question: Why is the path a parabola when there is constant acceleration?

For any constant acceleration ##\vec a##, initial velocity ##\vec v_0##, and initial position ##\vec x_0##, the path is given by
$$\vec x(t) = \vec x_0 + \vec v_0 t + \vec a \frac{t^2}{2},$$
which is a parabola.

MatinSAR and phantomvommand
phantomvommand said:
How did you know that acceleration's direction is constant in the above question?

And in that case, would it be correct to say that acceleration is constantly directed southeast (or 45 degrees south of east), despite the boy's direction changing? Why should that be so, if friction is supposed to be directly opposite the boy's velocity?
See this video for an interesting demonstration and working out for the acceleration in a circular path.

Note, for a circular path, the acceleration direction is changing - it points toward the center of the circle as the object moves.

Friction actz on the skate blade. So the boy can change the direction of his skate and change the direction of force relative to his motion.

scottdave said:
it points toward the center of the circle as the object moves.
Only if the speed is constant. If not the acceleration will also have a tangential component.

scottdave said:
Note, for a circular path, the acceleration direction is changing - it points toward the center of the circle as the object moves.

Friction actz on the skate blade. So the boy can change the direction of his skate and change the direction of force relative to his motion.
I'm not sure that was the OP's question. It wasn't immediately obvious why the direction of the acceleration was constant. But, looking at things in velocity space (or in the initial rest frame of the skater) makes that clear. I didn't see that immediately.

scottdave said:
Friction actz on the skate blade
The boy is not on skates. He is wearing boots or shoes. This is clear because a coefficient of friction is mentioned. Skates do not have a single relevant coefficient of friction. They are designed not to. In addition, the problem says "running" and "enters a patch of ice".

Last edited by a moderator:
jbriggs444 said:
the problem says "running"
Tonight Tre Kronor (Team Sweden) will be running on skates … all over Team Germany.

Edit: Yup. 6-1. Never even close.

Last edited:
Following on from my post #5, it is interesting to consider what the boy has to do.
Let his velocity be ##\vec v(t)##, the net velocity change required be ##\vec{\Delta v}##, and his acceleration be ##\vec a=\frac{\vec{\Delta v}}{|\vec{\Delta v}|}\mu g##.
If he "runs" with velocity ##\vec u##, by which I mean he slides the soles of his shoes on the ice at velocity ##-\vec u## relative to his body, then they slide at ##\vec v-\vec u## relative to the ice.
Since the magnitude of the acceleration does not depend on ##\vec u##, any component of that parallel to ##\vec{\Delta v}## is wasted effort. It follows that for least effort ##\vec u=\vec v-\frac{\vec v\cdot\vec{\Delta v}}{\vec{\Delta v}^2} \vec{\Delta v}## where ##\vec v=\vec v_0+\vec a t##.
Note that this is the component of ##\vec v## normal to ##\vec{\Delta v}##, which is unchanged by the acceleration. Hence
##\vec u=\vec v_0-\frac{\vec v_0\cdot\vec{\Delta v}}{\vec{\Delta v}^2} \vec{\Delta v}##.
Unsurprisingly, this breaks down in the special case of the initial and final velocities being parallel.

Last edited:
haruspex said:
Following on from my post #5, it is interesting to consider what the boy has to do.
Let his velocity be ##\vec v(t)##, the net velocity change required be ##\vec{\Delta v}##, and his acceleration be ##\vec a=\frac{\vec{\Delta v}}{|\vec{\Delta v}|}\mu g##.
If he "runs" with velocity ##\vec u##, by which I mean he slides the soles of his shoes on the ice at velocity ##-\vec u## relative to his body, then they slide at ##\vec v-\vec u## relative to the ice.
Since the magnitude of the acceleration does not depend on ##\vec u##, any component of that parallel to ##\vec{\Delta v}## is wasted effort. It follows that for least effort ##\vec u=\vec v-\frac{\vec v\cdot\vec{\Delta v}}{\vec{\Delta v}^2} \vec{\Delta v}## where ##\vec v=\vec v_0+\vec a t##.
Note that this is the component of ##\vec v## normal to ##\vec{\Delta v}##, which is unchanged by the acceleration. Hence
##\vec u=\vec v_0-\frac{\vec v_0\cdot\vec{\Delta v}}{\vec{\Delta v}^2} \vec{\Delta v}##.
This is interesting.

Consider the case where the velocity changes 90 degrees, starting East and ending North.

The diagram shows the initial and final velocities, as well as intermediate velocities at ##a, b##, and ##c##. The velocity at ##b## is perpendicular to the overall change in velocity ##\Delta \vec v##. I believe this velocity corresponds to your running velocity ##\vec u##.

The velocity that the foot slides relative to the ice is ##\vec v(t) - \vec u##. Thus, for the instant of time corresponding to ##a##, we have

The green arrow shows the velocity of the foot relative to the ice. This has the correct direction to cause the friction force on the body to be in the direction of ##\Delta \vec v##.

But if we consider the time corresponding to ##c##, we have

Now the foot is sliding in the wrong direction. Thus, the person would need to add a component of running velocity in the direction of ##\Delta \vec v## with a magnitude at least equal to that of the green vector.

Is this right?

phantomvommand and haruspex
TSny said:
But if we consider the time corresponding to ##c##, we have

View attachment 345232
Now the foot is sliding in the wrong direction. Thus, the person would need to add a component of running velocity in the direction of ##\Delta \vec v## with a magnitude at least equal to that of the green vector.

Is this right?
Well spotted. The flaw in my analysis is here:
haruspex said:
Since the magnitude of the acceleration does not depend on ##\vec u##, any component of that parallel to ##\vec{\Delta v}## is wasted effort.
But the direction does depend on ##\vec u##, and the boy has to ensure that is the right way along the line of ##\vec{\Delta v}##.
In fact, the problem starts at b, where, according to post #13, there would be no acceleration. The boy has to supply a minimal movement of the feet to maintain thrust.
From that point on, the velocity of the foot on the ice needs to be ##-\epsilon\frac{\vec{\Delta v}}{|\vec{\Delta v}|}## for some ##\epsilon>0##, so ##\vec u=\vec v+ \epsilon\frac{\vec{\Delta v}}{|\vec{\Delta v}|}##.
I think this also fixes the special case I mentioned.

TSny
I became interested in plotting the trajectory of the skater so this is what I did.

We have seen elsewhere(1) that when a body is displaced by ##\mathbf s## under constant acceleration ##\mathbf a## the expression $$\mathbf a\times \mathbf s=\mathbf v\times \mathbf u$$ relates the cross product of the two quantities to the cross product of the final velocity ##\mathbf v## and initial velocity ##\mathbf u.## We write the displacement as the sum of two vectors, perpendicular and parallel to the acceleration, ##\mathbf s=\mathbf s_{\perp}+\mathbf s_{\parallel}##(2). Then $$\mathbf a\times \mathbf s=\mathbf a\times \mathbf (\mathbf s_{\perp}+\mathbf s_{\parallel})=\mathbf a\times \mathbf s_{\perp}\implies s_{\perp}=\frac{|\mathbf v \times \mathbf u|}{a}.$$ Quantity ##s_{\perp}## is optimized (maximum) when the initial and final velocities are perpendicular. We write final speed as a fraction/multiple of the initial speed, ##v=\beta u## and find \begin{align}s_{\perp}^{\text{max}}=\frac{vu}{a}=\frac{\beta u^2}{a}.\end{align}
We will now find an expression for the trajectory of an object the changes direction from due North to due East with initial speed ##v_0=u## and final speed ##v_f=\beta u## under constant acceleration ##a.## The velocity diagram on the right will guide our thinking.

We identify ##\mathbf{\hat e}_{\perp}\rightarrow\mathbf {\hat x}##, ##\mathbf{\hat e}_{\parallel}\rightarrow\mathbf {\hat y}## and find the trajectory when the initial speed is ##u## and the projection angle is given by ##\tan\theta=\frac{u}{\beta u}=\frac{1}{\beta}.##

In order to sketch the trajectories, the idea first to generate standard projectile plot points ##(x,y)## in which the acceleration is along the vertical and then rotate the entire plot counterclockwise by ##\varphi=\frac{\pi}{2}-\theta## to put the initial velocity vector, which is at angle ##\theta## above the horizontal, in the vertical direction. Then plot rotated coordinate points ##(x', y')## according to $$\begin{pmatrix} x' \\ y' \end{pmatrix} =\begin{pmatrix} \cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi \\ \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}.$$The graph below shows three trajectory plots at exit speeds that are, (a) half (##\beta=0.5##), (b) equal to (##\beta=1.0##), and (c) twice (##\beta=2.0##) the entry speed ##u##. Each plot was terminated at the value of ##s_{\perp}^{\text{max}}## provided by equation (1).

(1) https://www.physicsforums.com/insights/how-to-solve-projectile-motion-problems-in-one-or-two-lines/
(2) In projectile motion ##s_{\perp}## is identified as the horizontal displacement ##\Delta x## and ##s_{\parallel}## as the vertical displacement ##\Delta y##.

Last edited:
kuruman said:
I became interested in plotting the trajectory of the skater so this is what I did.

We have seen elsewhere(1) that when a body is displaced by ##\mathbf s## under constant acceleration ##\mathbf a## the expression $$\mathbf a\times \mathbf s=\mathbf v\times \mathbf u$$ relates the cross product of the two quantities to the cross product of the final velocity ##\mathbf v## and initial velocity ##\mathbf u.## We write the displacement as the sum of two vectors, perpendicular and parallel to the acceleration, ##\mathbf s=\mathbf s_{\perp}+\mathbf s_{\parallel}##(2). Then $$\mathbf a\times \mathbf s=\mathbf a\times \mathbf (\mathbf s_{\perp}+\mathbf s_{\parallel})=\mathbf a\times \mathbf s_{\perp}\implies s_{\perp}=\frac{|\mathbf v \times \mathbf u|}{a}.$$ Quantity ##s_{\perp}## is optimized (maximum) when the initial and final velocities are perpendicular. We write final speed as a fraction/multiple of the initial speed, ##v=\beta u## and find \begin{align}s_{\perp}^{\text{max}}=\frac{vu}{a}=\frac{\beta u^2}{a}.\end{align}View attachment 345334We will now find an expression for the trajectory of an object the changes direction from due North to due East with initial speed ##v_0=u## and final speed ##v_f=\beta u## under constant acceleration ##a.## The velocity diagram on the right will guide our thinking.

We identify ##\mathbf{\hat e}_{\perp}\rightarrow\mathbf {\hat x}##, ##\mathbf{\hat e}_{\parallel}\rightarrow\mathbf {\hat y}## and find the trajectory when the initial speed is ##u## and the projection angle is given by ##\tan\theta=\frac{u}{\beta u}=\frac{1}{\beta}.##

In order to sketch the trajectories, the idea first to generate standard projectile plot points ##(x,y)## in which the acceleration is along the vertical and then rotate the entire plot counterclockwise by ##\varphi=\frac{\pi}{2}-\theta## to put the initial velocity vector, which is at angle ##\theta## above the horizontal, in the vertical direction. Then plot rotated coordinate points ##(x', y')## according to $$\begin{pmatrix} x' \\ y' \end{pmatrix} =\begin{pmatrix} \cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi \\ \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}.$$The graph below shows three trajectory plots at exit speeds that are, (a) half (##\beta=0.5##), (b) equal to (##\beta=1.0##), and (c) twice (##\beta=2.0##) the entry speed ##u##. Each plot was terminated at the value of ##s_{\perp}^{\text{max}}## provided by equation (1).
View attachment 345368

(1) https://www.physicsforums.com/insights/how-to-solve-projectile-motion-problems-in-one-or-two-lines/
(2) In projectile motion ##s_{\perp}## is identified as the horizontal displacement ##\Delta x## and ##s_{\parallel}## as the vertical displacement ##\Delta y##.
This seems a bit convoluted when the general solution is quite obviously ##\vec x(t) = \vec v t + \frac{\vec u - \vec v}{T} \frac{t^2}{2}## regardless of initial and final speed. Just insert values for ##t##.

kuruman

• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
55
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
957
• Introductory Physics Homework Help
Replies
3
Views
882
• Introductory Physics Homework Help
Replies
28
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
26
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
12
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
2K