Find Velocity and Position of Mass on Frictionless Table

[Cake]

Homework Statement

This Problem involves solving a simple Diff. Eq.
A block of mass m slides on a frictionless table. It is constrained to move inside a ring of radius L that is fixed to the table. At t=0 the block is moving along the inside of the ring (In the tangential direction) with velocity $v_0$. The coefficient of friction between the block and the ring is $\mu$

a) Find the velocity of the block at later times
b)Find the position at later times

The Attempt at a Solution

I don't have the ability to post a picture, but I'm using the radius inward from the block for the positive x direction, and direction of motion perpendicular to the radius as the positive y direction.

Sum of the forces:

SUM-X=$\frac{m(v_0)^2} {L}$
SUM-Y=$m\frac{dv} {dt}-m\mu\frac{dv} {dt}$

I'd like to know if I'm right so far. My problem is I'm not sure where to go from here. I think my sum of the forces in the x direction are faulty because dividing through, solving for dv/dt, and then working the diff. eq. doesn't give the right answer.

Thanks for any help :D

 

[Brian T]

How did you get that for sum-Y? Where should friction force be directed, where is normal force directed? Gravitational?

Also, M(dv/dt) is not a force but the net force

 

[Cake]

The coordinate system is looking down on the table at the circle, not the side. I directed the frictional force in opposition to the tangential force since the friction is against the ring, not the table. the gravity and normal force do not apply to this problem since the forces involved in its motion are only related to the circle it's traveling in.

 

[Chestermiller]

Hi Cake. Have you drawn a free body diagram of the block. What are the forces acting on the block in the radial (normal to the ring) direction? What are the forces acting on the block in the tangential direction? Kinematically, what is the radial acceleration? Kinematically, what is the tangential acceleration? What is your force balance in the radial direction (as an equation, not just a bunch of terms)? What is your force balance in the tangential direction (as an equation, not just a bunch of terms)?

Chet

 

[Cake]

Hi Chet :D

Here's a bit more detail

$a_t = \frac{dv} {dt}$
$a_r = \frac{v^2} {L}$

I figured the frictional force equals $m\mu\frac{dv} {dt}$ and opposes $ma_t$
so

$\sum F_t=m\frac{dv} {dt}-m\mu\frac{dv} {dt}$

I couldn't think of what to name as the opposing force to $ma_r$

 

[Cake]

Sorry, added the equations.

 

[Chestermiller]

Hi Chet :D

Here's a bit more detail

$a_t = \frac{dv} {dt}$
$a_r = \frac{v^2} {L}$

I figured the frictional force equals $m\mu\frac{dv} {dt}$ and opposes $ma_t$
so

$\sum F_t=m\frac{dv} {dt}-m\mu\frac{dv} {dt}$

I couldn't think of what to name as the opposing force to $ma_r$

The radial force that the ring exerts on the block is called the normal force N.

The frictional force is not equal to $m\mu\frac{dv} {dt}$. It is equal to $\mu N$.

If you had drawn a free body diagram, you would have seen this more easily.

Now, let's see your two force balance equations (in the radial and tangential directions).

Chet

 

[Cake]

Aha, so:

$\sum F_x=N=ma_c-N=\frac{mv^2} {L}-N=0$
$\sum F_y=m\frac{dv} {dt}-\mu N$Am I right that N opposes the centripetal force?
If so:
$N=\frac{mv^2} {L}$
and:
$\sum F_y=m\frac{dv} {dt}-\mu\frac{mv^2} {L}$

 

[Chestermiller]

Aha, so:

$\sum F_x=N=ma_c-N=\frac{mv^2} {L}-N=0$
$\sum F_y=m\frac{dv} {dt}-\mu N$

I don't see an equation for the tangential direction.

Am I right that N opposes the centripetal force?

No. N is the centripetal force.

If so:
$N=\frac{mv^2} {L}$
and:
$\sum F_y=m\frac{dv} {dt}-\mu\frac{mv^2} {L}$

Again, I don't see an equation for the tangential direction. The radial direction is correct.

Chet

 

[Cake]

I was using y for the tangential direction. $\frac{dv} {dt}$ is the tangential acceleration.

 

[Chestermiller]

I was using y for the tangential direction. $\frac{dv} {dt}$ is the tangential acceleration.

Where is the right hand side of the equation, and what is the friction force doing on the same side of the equation as mdv/dt?

Chet

 

[Cake]

When I drew my force diagram before I started the problem I thought since the tangential acceleration goes tangent to the circle it would cause a force in that direction, and that the frictional force would oppose it.

 

[Chestermiller]

When I drew my force diagram before I started the problem I thought since the tangential acceleration goes tangent to the circle it would cause a force in that direction, and that the frictional force would oppose it.

Newton's second law is ma = F, where F is the net force. Here's your version of the equation in this problem:

$$m\frac{dv}{dt}-m\mu \frac{v^2}{r}=0$$

Rewriting this, I get:

$$m\frac{dv}{dt}=m\mu \frac{v^2}{r}$$

According to this, the frictional force is causing the block to speed up. Is that what you intended? The equation should have read:
$$m\frac{dv}{dt}=-m\mu \frac{v^2}{r}$$

In the future, it would help if you put the net force on the opposite side of the equation from ma. That would have prevented making this error.

Do you know what to do next to get the velocity as a function of time?

Chet

 

[Cake]

Alright so isolating dv/dt and integrating:
$\int dv=\int -\mu\frac{v^2} {L} dt$
We get:
$\frac{dr} {dt}=-\mu\frac{v^2 t} {L}$
But the answer in the back is:
$\frac{v} {1+\frac{\mu v} {L}}$
What am I missing?

 

[Chestermiller]

Alright so isolating dv/dt and integrating:
$\int dv=\int -\mu\frac{v^2} {L} dt$
We get:
$\frac{dr} {dt}=-\mu\frac{v^2 t} {L}$
But the answer in the back is:
$\frac{v} {1+\frac{\mu v} {L}}$
What am I missing?

Where did the dr come from?
$$-\frac{dv}{v^2}=\frac{\mu}{L}dt$$
Can you integrate this, subject to the initial condition that v = v₀ at t = 0?

Chet

 

[Cake]

Chet, you're awesome. I handled the rest. Took a minute, but thank you so much.