the wheel in the simplified engine has a radius o A=0.250m and rotates with angular frequency omega=12.0 rad/s. At t=0, the piston is located at x=A. Calculate

Question

the wheel in the simplified engine has a radius o A=0.250m and rotates with angular frequency omega=12.0 rad/s. At t=0, the piston is located at x=A. Calculate the piston's a) position, b)velocity and c) acceleration at t=1.15 s

 

The answers in the back of my text book are:

a) 8.27x 10^-2m

b)-2.83 m/s

c)-11.9 m/s^2

 

I dont understand how they arrived at these answers. Please clarify the steps

Answer 1

Hi Nechama!

 

This questions looks like it is defining position as the "x" component of the piston's position as the wheel spins it around.  That x position is described by a simple harmonic motion expression.  This is demonstrated by the gif at the top of this page:

https://kaiserscience.wordpress.com/physics/waves/simple-harmonic-motion/shm-circular-motion-to-linear-motion-gif/

 

As such, we can attempt to describe the position function with either of the standard equations of motion for simple harmonic oscillation in one dimension:

 

x(t) = Asin(ωt + φ) or Acos(ωt + φ)

 

where

 

A = amplitude of the motion (maximum displacement from equilibrium)

ω = angular frequency of the motion (in radians per second (rad/s))

φ = phase shift (this is generally not needed if, at t = 0, the oscillating object is at either the equilibrium position or one of the amplitudes)

 

If the wheel has radius A, then A is the amplitude, the farthest the object can get from equilibrium in the x direction (as shown in the gif).  Since the object is at x = A at t = 0, we will not need the phase shift term φ.

 

What we do need is the form of the standard equation which gives x = A at t = 0.  The sin version cannot do this, since t = 0 would give us sin(0), which is 0.  But cos can, since at t = 0, we get cos(0) = 1.  Thus, if:

 

x(t) = Acos(ωt), then x(0) = A, as we require.

 

So we will take x(t) = Acos(ωt)

 

From the problem:

 

A = 0.250 m

ω = 12 rad/s

 

(a) To answer this part, we just evaluate the above expression at t = 1.15 s.

 

x(1.15 s) = (0.250 m)cos(12 rad/s * 1.15 s)

 

[note: since the argument of the cos function is in radians, you need to have your calculator in "radians" mode to get the correct result]

 

(b) and (c) There is an associated sequence of linear velocity and acceleration functions which follow from x(t) = Acos(ωt).  They are:

 

v(t) = -ωAsin(ωt)

a(t) = -ω²Acos(ωt)

 

(in calculus terms, these are just the first and second time derivatives of the position function)

 

To answer (b) and (c), just evaluate these expressions at t = 1.15 s.  Remember to keep that calculator in radians! :)

I hope this helps.  If you have any questions, please do not hesitate to ask.  Best of luck!