HINTS FOR PROBLEM SET #8

 

Problem Set #8 – Problem 1

This problem involves a mass on a spring in a vertical position. 

The spring is compressed by the mass sitting on top of it

and then it is compressed an additional amount (150mm) before it is released.

 

You are asked to find a position (maximum height)

and a velocity (the maximum velocity) once the system is released from rest. 

Since this is about positions and velocities

and you are not asked for acceleration

you should realize that you need to use the Work-Energy Theorem.

 

First identify some important positions of the system:

     Let position 1 be the unstretched position of the spring.

     Let position 2 be the equilibrium position of the mass and spring.

Note: The spring is compressed in this position
             by the weight of the mass on it.
          The amount of compression
             will be determined by the spring constant. 

     Let position 3 be where the spring is compressed an additional 150mm.

     Let position 4 be where the collar reaches its maximum height.

 

If you take the origin to be the collar at its unstretched position, then you have:

             y1  =   0 mm

             y2  =  -11.332 mm  ( = F/ k)

             y3  =   y2 - 150mm  =  -161.32 mm

             y4  =   ymax

Note: By choosing the origin to be the unstretched position of the spring, 

the y coordinate is the spring extension, i.e. the "∆x" in Hooke's law. (Fe= k∆x)

 

Part a)

To find the maximum height the mass reaches after the system is released,

apply the W-E Theorem between position 3, the initial position

and position 4, the position where the block reaches the maximum height. 

             T3 + W3→4 = T4

 

The only forces acting on the mass between these positions is the gravity force and the spring force, both conservative forces.  (The spring force does not act over the entire displacement, but that does not matter.)

 

So you can use this form of the W-E Theorem:

             T3 + Vg3 + Ve3 = T4 + Vg4 + Ve4

 

             T3     =   1/2 mv32 = 0

             Vg3  =   mgy3

             Ve3 =    1/2 ky32

             T4    =    0

             Vg4  =    mgy4

             Ve4  =    0

Plugging all this into the equation will allow you to solve for y4 (= ymax)

 

Note that y4 is the maximum position above the unstretched position of the spring (which was defined to be the datum.)  But you are asked to find the maximum position above the equilibrium position.  So you need to add 11.32 mm to this answer.

 

Part b)

Now to find the maximum velocity.

Apply the W-E Theorem between position 3, the initial position

and position 2, the position where velocity is maximum.

 

Why is position 2 the position where the maximum velocity occurred?

(Refer to Problem Set #7 – Problem 3.)

Because from calculus you know that

the maximum (or minimum) of a function occurs when the slope is zero.

Since a = dv/dt, the slope of the velocity curve is equal to the acceleration.

So the velocity will be a maximum (or minimum) when the acceleration is zero.

Zero acceleration corresponds to when the net force is zero. 

In this case the net force is zero

when the weight is balanced by the spring force. 

This is position 2.

 

Since the only forces acting are gravity and the spring force

(both conservative forces)

the W-E Theorem between position 3 and position 2 takes this form:

             T3 + Vg3 + Ve3 = T2 + Vg2 + Ve2

 

             T3    =   1/2 mv32 = 0

             Vg3  =   mgy3

             Ve3  =   1/2 ky32

             T2    =   1/2 mv22

             Vg2  =   mgy2

             Ve2  =   1/2 ky22

Plug this into the equation and solve for v2  (= vmax)

 

 
 

Problem Set #8 – Problem 2

This problem involves a mass on a spring in a vertical position and the mass is held just on top of the unstretched spring and then released. 

You are asked to find the maximum deflection of the spring when the mass is

     (a) released slowly

     (b) released suddenly.

 

Part a)

To find the maximum deflection when the mass is slowly released,

just find the position when the weight of the mass balances the spring force.

 

Part b)

To find the maximum deflection when the mass is suddenly released,

use the Work-Energy Theorem. 

 

Let position 1 be the position when the mass is just touching the spring. 

Let position 2 be the position of the spring when it has reached its maximum deflection.

 

Since the only forces acting are gravity and the spring force,

 the W-E Theorem takes this form between position 1 and 2:

                  T1 + Vg1 + Ve1 = T2 + Vg2 + Ve2

 

                  T1    =  0

                  Vg1  =  0

                  Ve1  =  0

                  T2    =  0

                  Vg2  = mg y2

                  Ve2  = 1/2 ky22

Plug and chug to get y2

 
 

 

Problem Set #8 – Problem 3 

Part a)

Case 1 (with bottom):

For the package to reach point C in the loop with the bottom,

you only need to ensure that the package makes it to point B, the top of the loop.

To ensure that the package makes it to the top of the loop

you need to ensure that the package will have some finite velocity there.

Worst case would be zero.

 

So apply the Work-Energy Theorem

between the initial position (point A)

and the top (point B).

Set the velocity at point B to be zero.

This will correspond to the minimum initial velocity at A, vo.

You should get vo = sqrt (2gh)

 

Case 2 (no bottom):

For the package to reach point C in the loop without the bottom, it is not enough to ensure that it merely makes it to the top. You need to ensure that there is a normal force from the loop acting on the package between B and C. Then you are assured that the package will be touching the loop and will not just fall out.

 

To ensure that a normal force is acting,

find an expression for it and require that it be greater than zero.

 

Draw the FBD of the package somewhere between B and C. 

(Call this unknown position B+.)

The only forces acting will be

the normal force, N,

and the weight, mg.

Apply Fnet = ma in the normal direction

and then solve for the normal force, N. 

You should get this:

                        N = mvB+2/r - mg sinq

where q is the angle between the normal force and the horizontal.

 

N must be larger than zero:

                        N > 0

 Ž             mvB+2/r - mg sinq > 0

Solve this inequality for vB+ :

                        vB+2 >  grsinq

 

Now apply the W-E Theorem between point A and point B+

to get an expression for vB+ .

You should get this:

                        vB+2 =  vo2 - 2gh + 2gr(1- sinq)

Substitute this expression for vB+2

into the inequality above and you will get:

                        vo2 - 2gh + 2gr(1- sinq)  >  grsinq

Solve this inequality for vo:

                        vo2  >  2gh - 2gr + 3grsinq

 

Now take the worst case scenario.

Worse case scenario occurs when sinq = 1

since this makes the right hand side as large as it can possibly be.

(Note that when sinq  = 1, q = 90o

which corresponds to point B,

the top of the loop.)

 

So the minimum initial velocity is found from:

                        vo2 >  2gh + gr

So the minimum initial velocity is:

                  vo = sqrt(2gh + gr)

 

Part b)

To find the normal force at point C that corresponds to the minimum velocities found in part a draw the FBD/IVD.

Apply Fnet = ma in the normal direction and solve for the normal force, N:

                        N = m vc2 /r

Apply the W-E Theorem between points A and C:

                  vc2  = vo2 - 5g

Hence:

                        N = m (vo2 - 5g) /r

 

Case 1 (with bottom)

Plug vo=  sqrt (2gh) into the equation.

 

Case 2 (no bottom)

Plug vo = sqrt(2gh + gr) into this equation.