Jun 26, 2013How to Find the Speed of a Block in an L-Shaped Track

In summary, the block's speed around the L-shaped track is determined by the tangential acceleration, which is related to the speed through the equation dv/dt = -μv^2/r. This can be solved using separation of variables. The normal force is equal to the centripetal force, and the equation for the centripetal force is N=mv^2/r.
  • #1
fogvajarash
127
0

Homework Statement



A small block of mass m is sliding around the inside of an L-shaped track of radius r. The bottom of the track is frictionless; the coefficient of kinetic friction and the wall of the track is μk. The block's speed is v0 at t0=0. Find an expression for the block's speed at a time t.

Homework Equations


The Attempt at a Solution



I'm not sure if my procedure is right. I think that the forces acting on the block on the three modified coordinate system are as follows: For the z-axis (up and down) we have the normal force and the weight of the block, and this net force ends up being zero. For the r-axis, we have that a normal force is the one that causes the centripetal acceleration (however, is it the same in magnitude as the normal force upwards?). And finally, in the tangential direction, we would have kinetic friction opposite in the direction of the block.

As well, I'm not sure which is the velocity we need to find (tangential or centripetal or the combined velocity?). I think that the radial and tangential acceleration can give us the magnitude of the total acceleration of the block, and then find the velocity at a time t based on this "general" acceleration.

Thank you very much in advance.
 

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  • #2
fogvajarash said:

Homework Statement



A small block of mass m is sliding around the inside of an L-shaped track of radius r. The bottom of the track is frictionless; the coefficient of kinetic friction and the wall of the track is μk. The block's speed is v0 at t0=0. Find an expression for the block's speed at a time t.

Homework Equations


The Attempt at a Solution



I'm not sure if my procedure is right. I think that the forces acting on the block on the three modified coordinate system are as follows: For the z-axis (up and down) we have the normal force and the weight of the block, and this net force ends up being zero. For the r-axis, we have that a normal force is the one that causes the centripetal acceleration (however, is it the same in magnitude as the normal force upwards?).

No, it is equal to the centripetal force.

fogvajarash said:
And finally, in the tangential direction, we would have kinetic friction opposite in the direction of the block.

Correct.

fogvajarash said:
As well, I'm not sure which is the velocity we need to find (tangential or centripetal or the combined velocity?). I think that the radial and tangential acceleration can give us the magnitude of the total acceleration of the block, and then find the velocity at a time t based on this "general" acceleration.

Thank you very much in advance.

The velocity along the circular path is tangential.It is the acceleration that has both tangential and radial (centripetal) components. The normal force does not change the speed. The speed would change because of the friction. The friction is proportional to the normal force from the wall. What is the equation relating the tangential acceleration to speed?

ehild
 
  • #3
ehild said:
No, it is equal to the centripetal force.



Correct.



The velocity along the circular path is tangential.It is the acceleration that has both tangential and radial (centripetal) components. The normal force does not change the speed. The speed would change because of the friction. The friction is proportional to the normal force from the wall. What is the equation relating the tangential acceleration to speed?

ehild
So this "total" acceleration is the one that changes the velocity of the block? (so it would be different comparing it to centripetal velocity). Isn't the equation that relates tangential acceleration to speed:

[itex]v=v_{o}+at[/itex]?

So, wouldn't we just plug in the acceleration value which is the magnitude of the tangential acceleration and the centripetal acceleration? I think that we have (as the only force in the radial component is the normal force):

[itex]a_{t}=-μ_{k}mg[/itex]
[itex]a_{r}=g[/itex]

The final answer at the end the book gives is:

[itex]v=\displaystyle\frac{rv_{0}}{r+v_{0}u_{k}t}[/itex]
 
  • #4
The bottom of the track is frictionless. mg is a vertical force, the block slides along a horizontal track. All forces are horizontal. mg does not come in.
The velocity is tangential, there is no "centripetal velocity".

The tangential acceleration is related to the speed as you wrote, v=vo+att. The speed decreases because of the force of friction with the wall. at=-μN, and N is not mg, but the force the wall pushes the block inward and forces the circular motion: that is, the normal force is equal to the centripetal force.
How is the centripetal force related to the speed?

ehild
 
  • #5
ehild said:
The bottom of the track is frictionless. mg is a vertical force, the block slides along a horizontal track. All forces are horizontal. mg does not come in.
The velocity is tangential, there is no "centripetal velocity".

The tangential acceleration is related to the speed as you wrote, v=vo+att. The speed decreases because of the force of friction with the wall. at=-μN, and N is not mg, but the force the wall pushes the block inward and forces the circular motion: that is, the normal force is equal to the centripetal force.
How is the centripetal force related to the speed?

ehild
The centripetal force would be:

[itex]N=F_{c}=\displaystyle\frac{mv^{2}}{r}[/itex]

And then v would be the equation we found, [itex]v=v_{0}-μNt[/itex]. So we would replace it in this equation? However, I find a second degree equation (with the mass I can't get rid of). What is the way to solve it?

Thank you ehild for your patience.
 
  • #6
fogvajarash said:
The centripetal force would be:

[itex]N=F_{c}=\displaystyle\frac{mv^{2}}{r}[/itex]

And then v would be the equation we found, [itex]v=v_{0}-μNt[/itex]. So we would replace it in this equation? However, I find a second degree equation (with the mass I can't get rid of). What is the way to solve it?

Your equation for the speed v(t) is not correct. The equation for the tangential acceleration is ma=mdv/dt=- μN. But N=mv2/r, plug it into the previous equation. You get mdv/dt=- μmv2/r. Simplify with m: dv/dt = -μv2/r. This is a separable differential equation for v. Can you solve it?

ehild
 

Related to Jun 26, 2013How to Find the Speed of a Block in an L-Shaped Track

1. What is a 2D dynamics problem?

A 2D dynamics problem is a mathematical problem that involves analyzing the motion of objects in a two-dimensional space. This can include calculating the position, velocity, acceleration, and forces acting on an object in both the x and y directions.

2. How is a 2D dynamics problem different from a 1D dynamics problem?

A 2D dynamics problem involves analyzing the motion of objects in two directions, while a 1D dynamics problem only involves one direction. This means that in a 2D problem, there are multiple forces and components to consider, making the problem more complex.

3. What are some common examples of 2D dynamics problems?

Some common examples of 2D dynamics problems include the motion of a projectile, such as a ball being thrown at an angle, the motion of a pendulum, and the motion of objects in circular motion, like a race car on a track.

4. How do you solve a 2D dynamics problem?

To solve a 2D dynamics problem, you first need to identify all the forces acting on the object in both the x and y directions. Then, you can use Newton's laws of motion and other equations, such as the equations of motion and the equations for calculating forces, to solve for the unknown variables, such as position, velocity, and acceleration.

5. What are some tips for solving 2D dynamics problems?

Some tips for solving 2D dynamics problems include drawing a clear diagram to visualize the problem, breaking the problem down into smaller steps, and labeling all the known and unknown variables. It is also important to carefully consider the direction and magnitude of forces and components in both the x and y directions.

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