Find Scalar Tangential/Normal Component of Acceleration Given Speed

Oct 2014
27
17
California
Hello! I have two different questions, one asking for the scalar tangential component of acceleration and the other asking for the scalar normal component of acceleration, both given the speed, ||v||. I could really use some help with how to start solving these specific types of problems as my teacher didn't go over them at all. Sorry in advance for the notation, I don't know how to make it look pretty (Doh)

1.) The speed ||v|| of a particle at an arbitrary time t is given, find the scalar tangential component of acceleration at the indicated time.
||v||=(t^2+e^(-3t))^(1/2); t=0

On this one what I tried to do was find v(t) from ||v||, but I got four variations of v(t)=ti + e^(-3t)j with a plus or minus sign in front of both the i and j components due to them being squared in the formula for ||v||.

My textbook gives the formula for the scalar tangential component of acceleration as:
aT is the dot product of v anda divided by the speed ||v||.

2.) The nuclear accelerator at Enrico Fermi Laboratory is circular with a radius of 1km. Find the scalar normal component of acceleration of a proton moving around the accelerator with a constant speed of 2.9 x 10^5 km/s.

On this one I didn't really know what to do, I thought the velocity should be some form of v(t)=costi + sintj but I didn't know where to go from there.

My book gives the formula for the scalar normal component of acceleration as:
aN=||v x a||/||v||

Thank you!
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Hello! I have two different questions, one asking for the scalar tangential component of acceleration and the other asking for the scalar normal component of acceleration, both given the speed, ||v||. I could really use some help with how to start solving these specific types of problems as my teacher didn't go over them at all. Sorry in advance for the notation, I don't know how to make it look pretty (Doh)

1.) The speed ||v|| of a particle at an arbitrary time t is given, find the scalar tangential component of acceleration at the indicated time.
If all you are given is the speed, then you can't find the separate components.

||v||=(t^2+e^(-3t))^(1/2); t=0

On this one what I tried to do was find v(t) from ||v||, but I got four variations of v(t)=ti + e^(-3t)j with a plus or minus sign in front of both the i and j components due to them being squared in the formula for ||v||.

My textbook gives the formula for the scalar tangential component of acceleration as:
aT is the dot product of v anda divided by the speed ||v||.
Yes, and that requires knowing the actual vectors v and a, not just the scalar values.

2.) The nuclear accelerator at Enrico Fermi Laboratory is circular with a radius of 1km. Find the scalar normal component of acceleration of a proton moving around the accelerator with a constant speed of 2.9 x 10^5 km/s.

On this one I didn't really know what to do, I thought the velocity should be some form of v(t)=costi + sintj but I didn't know where to go from there.
The position vector of an object going around a circle of radius R with constant angular velocity \(\displaystyle \omega\) is \(\displaystyle \vec{r(t)}= R cos(\omega t)\vec{i}+ R sin(\omega t)\vec{j}\) so that the velocity vector is \(\displaystyle \vec{v}(t)= -R\omega sin(\omega t)\vec{i}+ R\omega cos(\omega t)\vec{j}\)
The speed, then, is [itex]R\omega= 290000 km/s[/itex] and R= 1 km. So what is [itex]\omega[/itex]?

My book gives the formula for the scalar normal component of acceleration as:
aN=||v x a||/||v||

Thank you!
 
Last edited:
Dec 2012
1,145
502
Athens, OH, USA
I believe Hall's assertion that you need the specific components of $v$ is false. You are given $|v|$ and hence also $|v|^2$. So let the position vector be $r(t)=(x(t),y(t))$; you can write this in terms of i and j if you like. Then $v(t)=(x'(t),y'(t))$ and $a(t)=(x''(t),y''(t))$ Then $$|v(t)|^2=(x'(t))^2+(y'(t))^2$$
So the derivative of $|v(t)|^2$ is
$$2x'(t)x''(t)+2y'(t)y''(t)=2a\cdot v$$
I hope you can use your book's formula and finish from here. Aside, the same thing holds true in 3 dimensions.
 
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