HINTS FOR PROBLEM SET #21

 

Problem Set #21 – Problem 1

Since we are given information about the angular velocity and a displacement, 

we will use the Work Energy Theorem for the system of two rigid bodies: 

Kinetic Energy of the system at position 1 + Work done on system of rigid bodies from position 1 to 2 = Kinetic Energy at position 2

Position 1 is when the system is at rest.

Position 2 is when the system has rotated by 2 revolutions.  (= 4 pi radians.)

The two rigid bodies are the 9 inch disk and the rod AB of length L.  

(Why am I neglecting the shaft CD?) 

The kinetic energy at position 1 is zero.  (System is at rest.)

The kinetic energy at position 2 is found by summing the kinetic energy of the disk and the rod:

T2  = (½ mdisk vdisk^2 + ½ Idisk wdisk^2) + (½ mrod vrod^2 + ½ Irod wrod^2)

Be careful of your units.  

Idisk = 0.0175 ft lbs s^2   

Irod = 0.007764 L^3 ft lbs s^2

wdisk = wrod = 300 rpm = 31.416 radians/second

vdisk = vrod = 0   (Both rigid bodies are in pure rotation; there is no translation.)

Now calculate the work term.  After drawing the FBD of both rigid bodies, you should conclude that the only forces that do work is the couple, M acting on the system.

The work done by this couple is equal to M x 2revolutions = (4 ft-lbs) x (2)(2pi) = 50.265 ft-lbs

Plug this all in the Work Energy Theorem and solve for L.

L = 2.2 ft

 

Problem Set #21 – Problem 2

Part a)

Again, we have a problem that involves information about a velocity and a displacement.  We will use the Work Energy Theorem.

 Kinetic Energy of the system at position 1 + Work done on system from position 1 to 2 = Kinetic Energy at position 2

The system is just the one rigid body of the roller.

Position 1 is when the roller is at rest.

Position 2 is when the roller has advanced by 1.5 meters.

At position 1 the kinetic energy is zero.

At position 2 the kinetic energy is = (½ mroller vroller^2 + ½ Iroller wroller^2)

       Iroller = 0.625 kg m^2

       wroller and vroller are related because there is no slip:  

       wroller = vroller/ rroller

Now calculate the work term:  After drawing the FBD you should see that the only force that does work is the 90 N force acting on the roller.

The work done by this force is equal to 90N x 1.5m = 135Nm

Plug this all in the Work Energy Theorem and solve for vroller to get 3 m/s.

Part b)

 of this problem asks you to find the frictional force that must act in order to ensure no slip. Draw the FBD and equate it to a diagram showing the effective forces acting at G (= mass x acceleration vector) and the effective moment (= I x alpha).  Note that for no slip, the angular acceleration, alpha, of the roller and the acceleration of G are related by this equation:

aroller = rroller x alpha

Now get the x and y components of the equation: F = ma

Now get the k component of the equation: (Sum of the moments about G) = (Sum of the effective moments about G)  

These three equations will allow you to solve for the unknown frictional force.

 

Problem Set #21 – Problem 3

Again, we have a problem that involves information about a velocity and a displacement.  We will use the Work Energy Theorem.

 

Kinetic Energy of the system at position 1 + Work done on system of rigid bodies from position 1 to 2 = Kinetic Energy at position 2

 

The system is the two rigid bodies consisting of the flywheel and the rod AB.

Position 1 is when point B is to the direct left of point C.

Position 2 is when point B is directly below point C.

 

The kinetic energy is found by summing the kinetic energy of the flywheel and the rod:

       T  = (½ m_fly v_fly^2 + ½ I_fly w_fly^2) + (½ m_rod v_rod^2 + ½ I_rod w_rod^2)

(And note that v_fly  and v_rod  are the velocities of the mass centers.)

 

       m_fly and m_rod are given.

It is relatively straightforward to calculate I_fly and I_rod:

       I_fly = 0.5184 kg m^2

       I_rod = 0.1728 kg m^2

For position 1 and 2 the velocity of the flywheel, v_fly, is zero since it is in pure rotation.

For position 1 the angular velocity of the flywheel is given.

We want to solve this equation for the angular velocity of the flywheel in position 2.

 

The tricky part of this problem is getting the velocity of the rod, v_rod, and its angular velocity, w_rod, in each position.   For position 1 we can get these values by a kinematic analysis.  For position 2, we can get these values in terms of w_fly by a kinematic analysis.

 

So we need to do a kinematic analysis of the points A, B and G (the mass center of the rod) for both positions.

(Why G?  Because the velocity of G is v_rod!)

Use the kinematic equation that relates the velocity of two points on a rigid body in general plane motion: 

       vA = vB + vA/B = vB + (w x rA/B)

Also, we need to use the fact that we can get the velocity of B in terms of its velocity as part of the flywheel:

       vB = w_fly x r_fly

 

In position 1 you should find that the rod is in pure translation. (Use the method of instant center to quicken the process.) Since the rod has no rotation, the angular velocity of the rod is zero in position 1.  

 

To get the velocity of the mass center, v_rod, we note that it is the same as the velocity of point B since all points on a rigid body in pure translation have the same velocity. And the velocity of point B is easily calculated by vB = w_fly x r_fly.  (You should get v_rod = 1.508 m/s)

 

In position 2 you should be able to conclude that the instant center of the rod is point A.

So now you can get the velocity of point G in terms of the angular velocity of the rod:

       vG = v_rod = w_rod  x L/2

You can also get the velocity of point B in terms of the angular velocity of the rod:

       vB = w_rod  x L

You can also get the velocity of point B in terms of the angular velocity of the flywheel:

       vB = w_fly  x r_fly

 

With these three equations, you can put w_rod  and v_rod in terms of w_fly.

 

Now calculate the work term:

Draw the FBD and convince yourself that the only force that does work is the weight of the rod!

So the work term is just V1-V2.  (V is the potential energy function for the weight of the rod.)  The geometry here can get tricky here so be careful.

 

At position 1 the potential energy function for the weight of the rod is mgh = (4)(9.81)(0.3415) J = 13.4015 J

At position 2 the potential energy function for the weight of the rod is mgh = (4)(9.81)(0.12) J = 4.709 J

 

Now plug this all into the Work Energy Theorem and solve for the angular velocity of the flywheel in position 2.