General motion with constant speed

[Benzoate]

Homework Statement

A particle moves along any path in 3d space with constant speed. Show that its velocity and accelerations vectors must always be perpendicular to each other.[hint. Differentiate the formula v dot v = v^2 with respect to t.

Homework Equations

possible equations: v=dr/dt=dr/ds*ds/dt, a=dv/dt=(dv/dt)t +(v^2/rho)n,
v=v*t where t is the unit tangent vector and n is the normal unit vector.

The Attempt at a Solution

If a particle is moving in 3d space with constant speed, then the type of motion could be uniform circular motion and the acceleration is therefore equal to zero.

Probably, v dot v= v^2(x) + v^2(y)+ v^2(z) , x , y and z making up the dimensions of the 3-D plane; since the particle moves along a 3D plane. If the particle is in uniform motion, the position vector r probably is in polar coordinates: r= b*cos(theta)i + b*sin(theta)j and theta is equal to ut/b where u is the speed of the particle , b is the arc length and t is the time its traveled. Would I differentiate r and then plugged in dr/dt into the v dot v equation? I expect I also have to show that x hat times y hat or x hat times z hat or y hat times z is equal to zero, since the problem asks me to show that velocity and acceleration vectors are always perpendicular to each other.

Maybe the simplest solution is to show that v cross a = 0 since the acceleration of the particle is zero and therefore I would be done with the problem.

 

[Doc Al]

Homework Statement

A particle moves along any path in 3d space with constant speed. Show that its velocity and accelerations vectors must always be perpendicular to each other.[hint. Differentiate the formula v dot v = v^2 with respect to t.

Just follow the hint! Take the derivative of both sides of:
\vec{v}\cdot\vec{v} = v^2

Hint: If two vectors are perpendicular, what must their dot product be?

 

[Kurdt]

In uniform circular motion the acceleration is not zero. The acceleration is however orthogonal to velocity in uniform circular motion.

Try doing what the hint says first. You're making this way more complicated.

 

[Benzoate]

Just follow the hint! Take the derivative of both sides of:
\vec{v}\cdot\vec{v} = v^2

Hint: If two vectors are perpendicular, what must their dot product be?

I'm having trouble determining what must be the derivative of v dot v. Would the derivative of v dot v be dv/dt*v+v*dv/dt=0 => 2*(dv/dt *v)=0?, since v is constant , therefore dv/dt=0 and thus 2*(dv/dt*v) must be zero, right?

 

[Benzoate]

In uniform circular motion the acceleration is not zero. The acceleration is however orthogonal to velocity in uniform circular motion.

Try doing what the hint says first. You're making this way more complicated.

The problem doesn't say anything about the particle being in uniform circular motion, problem only states that v is constant, and therefore if their is no change in velocity, the acceleration o the particle must be zero.

 

[Kurdt]

I was responding to the first paragraph in your attempt at the answer.

Anyway if the derivative of \mathbf{v} \cdot \mathbf{v} = v^2 is almost as you have it.

\mathbf{v} \cdot \frac{d\mathbf{v}}{dt} + \frac{d\mathbf{v}}{dt} \cdot \mathbf{v} = 2v \frac{dv}{dt}

As you have said the derivative of the speed must be zero if it is constant and if that is true how will the left hand side be made zero? Use Doc Al's hint above.

EDIT: oops made the same speed/velocity mistake

 

[Doc Al]

I'm having trouble determining what must be the derivative of v dot v. Would the derivative of v dot v be dv/dt*v+v*dv/dt=0 => 2*(dv/dt *v)=0?, since v is constant , therefore dv/dt=0 and thus 2*(dv/dt*v) must be zero, right?

Yes, but it's essential to distinguish the velocity (a vector) from the speed (a scalar). In this problem you are told that the speed is constant, not the velocity. All the v's on the left side of your equation stand for velocity, not speed. (See Kurdt's post.)

 

[HallsofIvy]

The problem doesn't say anything about the particle being in uniform circular motion, problem only states that v is constant, and therefore if their is no change in velocity, the acceleration o the particle must be zero.

No, the problem said the speed was constant, not the velocity!

 

[Benzoate]

I was responding to the first paragraph in your attempt at the answer.

Anyway if the derivative of \mathbf{v} \cdot \mathbf{v} = v^2 is almost as you have it.

\mathbf{v} \cdot \frac{d\mathbf{v}}{dt} + \frac{d\mathbf{v}}{dt} \cdot \mathbf{v} = 2v \frac{dv}{dt}

As you have said the derivative of the speed must be zero if it is constant and if that is true how will the left hand side be made zero? Use Doc Al's hint above.

EDIT: oops made the same speed/velocity mistake

since the acceleration is zero, then v*dv/dt+dv/dt*v=2*v* dv/dt => v*0+0*v=2*v*0=> 0=0 which implies acceleration vector is perpendicular to velocity vector

 

[Doc Al]

since the acceleration is zero, then v*dv/dt+dv/dt*v=2*v* dv/dt => v*0+0*v=2*v*0=> 0=0 which implies acceleration vector is perpendicular to velocity vector

No. The acceleration is NOT zero.

\vec{a} = \frac{d\vec{v}}{dt} \ne 0

But:

\frac{dv}{dt} = 0

 

[Benzoate]

No. The acceleration is NOT zero.

\vec{a} = \frac{d\vec{v}}{dt} \ne 0

But:

\frac{dv}{dt} = 0

so if dv/dt ,without the vector, is not did acceleration, is dv/dt the magnitude of the acceleration?

 

[Kurdt]

Its the rate of change of speed. The question asks about the velocity and acceleration vector though. So how can the differentiated dot product side equal zero? Consider Doc Al's first hint.

 

[HallsofIvy]

Not exactly |d\vec{v}/dt| would be the magnitude of the acceleration. If the acceleration is not 0 then its magnitude also cannot be 0. Doc Al is using v to mean |\vec{v}| so dv/dt is the "derivative of the magnitude of the velocity" which is NOT the same as the "magnitude of the derivative of the velocity". The order is important!

 

[Doc Al]

so if dv/dt ,without the vector, is not did acceleration, is dv/dt the magnitude of the acceleration?

No. v, without the vector sign, stands for speed. dv/dt is the rate of change of the speed.