Question #119299
Consider a pendulum of mass “m” attached to a spring of mass “M” that is free to move in
single dimension along a frictionless horizontal surface. Take the gravity g = 10 m/s2
and the

gravitational potential energy is equal to zero at the level of block (y = 0).
a) Write the equations of constraints.
b) Determine the degree of freedom (S = ??)
c) Write the expression of r⃗⃗⃗M⃗ as a function of X and unit vector i
d) Write the expression of r′m

⃗⃗⃗⃗⃗⃗ as a function of unit vector i , e′r
⃗⃗⃗⃗⃗ and e′θ
⃗⃗⃗⃗⃗
e) Find the expression of kinetic energy of the system as a function of (M, m, Ẋ
, l, θ, θ̇)
f) Write the expression of potential energy PE of the system as a function of (m, l, θ)
g) Write the Lagrangian equation
h) Deduce the equations of motion from Euler-Lagrange equations
1
Expert's answer
2020-06-08T10:30:23-0400

Here,m is the mass of pendulum and M is the mass of spring , g is the acceleration due to gravity and at level block potential energy is equal to 0.

(a) F=-kx

(m+M)d2xdt2=kx(m+M) \frac{d^2x}{dt^2}=-kx

(b) for degree of freedom

s=n-m

s=2-1 =1

so,here degree of freedom will be 1

(c) F= -mgθ\theta

here, θ=sl\theta= \frac{s}{l}

For the position vector of mass

(d)

r= R +l(sinθi^+cosθj^)sin\theta \hat{i} +cos \theta \hat {j})

and the velocity is represented as

v= R˙+lθ˙(sinθi^+cosθj^)\dot{R} +l \dot\theta(sin\theta \hat{i} +cos \theta \hat {j})

(e)

KE=12mv2=12m(R˙2+l2θ˙2+2lθ˙(X˙cosθY˙sinθ)KE=\frac{1}{2} mv^2= \frac{1}{2} m (\mid{\dot{R}}\mid^2+ l^2\dot{\theta}^2+2l\dot{\theta}(\dot{X}cos \theta-\dot{Y}sin\theta)

(f) and for potential energy

PE= - (m+M)gcosθ(m+M)gY(m+M)g cos \theta - (m+M)g Y

(g)

L= KE-PE

L= 12m(R˙2+l2θ˙2+2lθ˙(X˙cosθY˙sinθ)+(m+M)gcosθ+(m+M)gY\frac{1}{2} m (\mid{\dot{R}}\mid^2+ l^2\dot{\theta}^2+2l\dot{\theta}(\dot{X}cos \theta-\dot{Y}sin\theta)+(m+M)g cos \theta +(m+M)g Y +






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