HINTS FOR PROBLEM SET #4

Problem Set #4 – Problem 1

You need to find the magnitude of the total acceleration of a particle (the pin) for two specific instants of time, t = 0 sec and t = 2sec.

If you have the acceleration vector as a function of time, the magnitude is easily found.

(Just plug in the values of t and then take the square root of the sum of the components squared.)

Since you know the path of motion of the pin, you should use tangential and normal components, a_t and a_n, to represent the acceleration vector.

The tangential component of the acceleration vector, a_t, is equal to dv/dt. 

You are told that the speed increases at a constant rate equal to 0.8 in/s².

Hence dv/dt = 0.8 for all t.

Hence a_t = 0.8.

The normal component of the acceleration vector, a_n, is v²/rho. 

Integrate dv/dt = 0.8  to get an expression for the speed, v, as a function of time, v(t). 

(You'll need to use the given initial condition to determine the constant of integration.)

Hence a_n = (0.8t)²/(3.5)

Use this expression for v(t) to calculate the normal component of the acceleration at any time, t.

Now that you have the components of the acceleration vector,

the magnitude is easily obtained.
 

Problem Set #4 – Problem 2

Part a)

You need to find the speed of the car in terms of  b, q and qdot.

(Note: qdot is just dq/dt.)

Speed is just the magnitude of the velocity vector.

First, you need to get the velocity vector itself.

Then you can obtain the magnitude by taking the square of the sum of the components.

Since the position of the car is defined by a distance and an angle from a point,

use radial and transverse components to represent the velocity vector of the car.

           v  =  dr/dt e_r + r dq/dt e_q

From geometry, you can see that  r = b/cosq.

Differentiating this once will give dr/dt as a function of q and qdot.

Plug this into the velocity vector above.

Find the magnitude of this vector.

Part b)

You want to find the magnitude of the acceleration vector of the car

in terms of b, q, qdot and qdotdot.

(Note: qdotdot is just d²q/dt².)

First, you need to get the acceleration vector itself. 

Then you can obtain the magnitude by taking the square of the sum of the components.

Again, since the position of the car is defined by a distance and an angle from a point, use radial and transverse components to represent the acceleration vector of the car.

        a = { d²r/dt²  -  r (dq/dt)² } e_r   +  { r (d²q/dt²)  +  2 (dr/dt) (dq/dt) } e_q

Differentiating the expression you obtained for dr/dt in Part a will give d²r/dt² as a function of q, dq/dt and d²q/dt². 

Plug this into the acceleration vector above.

Find the magnitude of this vector.

Please note:

These are not trivial calculations -  in fact they will take you several pages.

However they are doable. It's just math!

Part c)

You need to get the average speed of the car during a 0.5 second interval

in which the angle, q, varies from 60^o to 35^o.

Recall the definition of average speed: 

       average speed = ∆y/∆t

In this case ∆t = 0.5 s

and ∆y can be easily found by geometry.

∆y = b tan60^o - b tan35^o

Problem Set #4 – Problem 3

You need to find rdot in terms of h, fdot and q.

This is essentially a geometry problem.

Observe the triangle made up of the sides r ,d, h and the two angles f and q.

Use the Law of Cosines to obtain an equation with r, d, h, and f. 

Note that r and f are the variables, h and d are constants.

Differentiate this equation to get r dot as a function of r, f and fdot. 

To get rid of r in this equation, go back to the same triangle and

use the Law of Sines to get an expression for r in terms of d, f and q. 

Substitute this into the above equation and solve for rdot.