A projectile of mass m is fired from the origin at an angle of phi to the x-axis

Question

A projectile of mass m is fired from the origin at an angle of phi to the x-axis with a speed v0. A spring is attached from the origin to the projectile, and it has a spring constant k and zero relaxed length. Neglect air friction and the mass of the spring. By determining the forces acting on the mass, derive equations of motion for the position of the mass, x(t), y(t). Substitute the trial functions x(t) = A cos (omega t) + B sin (omega t) y(t) = C cos (omega t) + D sin (omega t) + y0 into the equations of motion and determine omega and y0. Use the initial conditions to determine the constants A, B, C, and D. Show that in the limit of small k/m the trajectory reduces to normal projectile motion for short t.

Explanation / Answer

1.)

Net force by the spring = - kx i - ky j

Fx = -kx

ax = -kx/m

Fy = -ky - mg

ay = -ky/m - g

Hence, equations of motion are :

x'' = -kx/m   ....[Equ 1]

y'' = -ky/m - g ....[Equ 2]

2.)

x'' = -kx/m

( A cos(wt) + B sin(wt) )'' = -k( Acos(wt) + B sin(wt) ) / m

( -Aw2*cos(wt) - Bw2*sin(wt) ) = -k( Acos(wt) + B sin(wt) ) / m

w2 = k / m

w = sqrt(k/m)

y'' = -ky/m - g

( C cos(wt) + D sin(wt) + y0 )'' = -k( C cos(wt) + D sin(wt) + y0 )/m - g

( -Cw2*cos(wt) - Dw2*sin(wt) ) = -k( C cos(wt) + D sin(wt) ) / m - (ky0/m + g)

w2( Ccos(wt) + Dsin(wt) ) = k( C cos(wt) + D sin(wt) ) / m + (ky0/m + g)

w2 = sqrt(k/m)

Hence

k( C cos(wt) + D sin(wt) ) / m = k( C cos(wt) + D sin(wt) ) / m + (ky0/m + g)

0 = (ky0/m + g)

y0= -mg/k

3.)

x(0) = 0 = A cos(0) + B sin(0) = A

Hence A = 0

x'(0) = v0cos(phi) = -Aw*sin(0) + Bw*cos(0) = Bw

B = v0cos(phi) / w = v0cos(phi)*sqrt(m/k)

y(0) = 0 = C cos(0) + D sin(0) + y0 = C + y0

C = -y0 = mg/ k

y'(0) = v0sin(phi) = -Cw*sin(0) + Dw*cos(0) = Dw

D = v0*sin(phi) / w = v0*sin(phi)*sqrt(m/k)

4.)

For x(t) :

x(t) in projectile = v0cos(phi)*t

x(t) = v0cos(phi)*sqrt(m/k)*sin( sqrt(k/m)t ) = v0cos(phi)*sin( sqrt(k/m)t ) / sqrt(k/m)

For small sqrt(k/m) the sin( sqrt(k/m)t ) = sqrt(k/m)t

Hence,

x(t) = v0cos(phi)*sin( sqrt(k/m)t ) / sqrt(k/m) = v0cos(phi)*sqrt(k/m)t / sqrt(k/m) = v0cos(phi)*t

For y(t) :

y(t) in projectile = v0*sin(phi)*t - 0.5gt2

y(t) = gcos(sqrt(k/m)t)/ (k/m) + v0*sin(phi)*sqrt(m/k)sin(sqrt(k/m)t) - mg/ k

{ For small sqrt(k/m) the sin( sqrt(k/m)t ) = sqrt(k/m)t }

gcos(sqrt(k/m)t)/ (k/m) + v0*sin(phi)*t - mg/ k

= v0*sin(phi)*t - g( m/k - (m/k)cos(sqrt(k/m)t) )

= v0*sin(phi)*t - g(m/k)[ 1 - cos(sqrt(k/m)t) ]

= v0*sin(phi)*t - g(m/k)[ 2sin2( 0.5sqrt(k/m)t) ]

{ For small sqrt(k/m) the sin2( 0.5sqrt(k/m)t ) = 0.25(k/m)t2 }

= v0*sin(phi)*t - g(m/k)(2*0.25(k/m)t2 ) = v0*sin(phi)*t - g(m/k)(0.5(k/m)*t2 )

= v0*sin(phi)*t - 0.5gt2

Hence proved for small t

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