HINTS FOR PROBLEM SET #17

 

Problem Set #17 – Problem 1

First of all note that the carriage is in pure translation.  So all the points on it have the same velocity and the same acceleration.  Call any point on the carriage, point C. 

Hence

              vC = 1.5 m/s to the left

              aC = 2.4 m/s^2 to the left

 

Let’s call the center of the caster point A.  Since the carriage is pin connected to the caster at point A, then point A must have the same acceleration and velocity as the carriage.  (Recall: Points on rigid bodies at pin connections have the same absolute velocity and accelerations.)

Hence

              vA = 1.5 m/s to the left

              aA = 2.4 m/s^2 to the left

 

If the caster does not slip, then the point on the ground where the caster touches is the instant center.  Hence the velocity of point A is given by:

              vA = (angular velocity of the caster)  x  (radius of the caster)

 

Since point A is in rectilinear motion, this equation for velocity is a scalar equation.  So you can differentiate it to give the acceleration of point A. 

Hence

              aA = (angular acceleration of the caster) x (radius of the caster) 

              (angular acceleration of the caster) =  aA / (radius of the caster) =  2.4/ 0.025

              (angular acceleration of the caster) = 96 rad/s^2 ccw

 

Now let’s consider how the cylinder moves.  Let’s call the center of the cylinder point B.  If both the caster and the cylinder move without slipping, then the motion of point B is related to the motion of any point C on the carriage.  Both are in rectilinear motion.  The relationship is this: 

              vB = 0.5vC

 

How did I get this?  Call the point of contact between the cylinder and the carriage point C.  Now consider the instant center of the cylinder.  It is the point on the ground that is in contact with the cylinder.  The velocity of every point on the cylinder is then given by the angular velocity of the cylinder times the distance the point is from this instant center.  Since point B is half as far from the instant center as point C then

              vB = 0.5vC

 

Since the point B is in rectilinear motion, this equation for velocity is a scalar equation.  So you can differentiate it to get the acceleration of point B.  Hence

              aB = 0.5aC 

              aB = 1.2 m/s^2

 

If the cylinder does not slip, then the point on the ground where the cylinder touches is the instant center.  Hence the velocity of point B is given by:

              vB = (angular velocity of the cylinder)  x  (radius of the cylinder)

 

Since point B is in rectilinear motion, this equation for velocity is a scalar equation.  So you can differentiate it to give the acceleration of point B. 

Hence

              aB = (angular acceleration of the cylinder) x (radius of the cylinder)

              (angular acceleration of the cylinder) = aB/(radius of the cylinder) = 1.2/0.025

              (angular acceleration of the cylinder) = 48 rad/s^2 ccw

 

 

 
 

Problem Set #17 – Problem 2

We are looking for the absolute accelerations of points A and B, given the absolute velocity and acceleration of point E on the rope.

 

The strategy: 

Find the absolute acceleration of point G (the geometric center and also the mass center of the cylinder) and then use the equation that relates the accelerations of two points on a rigid body (the cylinder) to find the absolute acceleration of points A and B:

              aA = aG +  aA/G

              aB = aG +  aB/G

 

First we need to get the acceleration of point G, aG.  If there is no slipping of the cylinder with respect to the rope, then  

              vG = 0.5vE 

              aG = 0.5 aE

 

If you don't see how I got these, read this paragraph.  Consider that point A is the instant center of the cylinder. 

Hence

              vG = 0.5vC 

And since vC = vE we have

              vG = 0.5vE

Now note that point G is in rectilinear motion. 

So the acceleration of G can be found by differentiating this equation. 

              aG = 0.5 aE

              aG =  240 mm/s^2 upward

 

Now we need to get the aA/G and the aB/G terms.  These consider the acceleration of points A and B relative to G.  So it is the acceleration of points A and B as if G were fixed.  This is just rotation about G.  So this acceleration has a tangential and a normal component.  To calculate the tangential component, we need the angular acceleration of the cylinder.  (a tangential = alpha x r)  To calculate the normal component, we need the angular velocity of the cylinder.  (a normal = v^2/r = omega ^2 r)

 

First get the angular velocity, omega.  You can do it using the fact that you know the velocity of two points on the body. 

              vE = vA + vE/A

Since vA = 0  and  vE = vC 

              vE = 0 + vC/A

              vE = (omega x rC/A)

              omega = vE / 2r = 300/2*75 = 2 rad/sec

 

To get the angular acceleration of the cylinder, alpha, use aG. 

              aG = alpha x r

              alpha = aG / r  = 240/75

              alpha = 3.2rad/s^2

 

Now plug all this into the equation that relates the accelerations of two points on a rigid body and solve for the absolute acceleration of points A and B.
 

Problem Set #17 – Problem 3

Consider the rigid body BC.  We will use the equation that relates the accelerations of two points on a rigid body to find the absolute acceleration of point C:

              aC = aB +  aC/B

 

What are the knowns in this equation? 

 

First of all we know the direction of aC, (the absolute acceleration of point C). 

It must be along the rod. We don’t know it’s magnitude.

 

We also know something about the absolute acceleration of point B, aB, since it is a point on a rigid body in rotation about the fixed point A.  So we calculate its tangential and normal components.  The tangential component must be zero (since the disk has a constant angular velocity) and we can calculate the normal component from the angular velocity of the disk and its radius.

 

The aC/B term which is the acceleration of C with respect to B can be found by looking at its tangential and normal components. 

The tangential component is:

              (angular acceleration of BD) x rC/B

We don’t know the angular acceleration.  Hold on.

 

We can calculate the normal component from the angular velocity of the rod BC and the distance from B to C. But we don’t know the angular velocity of the rod! So we must find it! This is a separate little problem. Use the method of instant center to convince yourself that this angular velocity must be zero.

 

Now we can put all of the together into the equation that relates the accelerations of two points on a rigid body to find the unknowns.  (Our unknowns are the angular acceleration of BD the magnitude of aC.)

You can solve the equation graphically or analytically, whatever your personal preference.