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)dt2d2x=−kx
(b) for degree of freedom
s=n-m
s=2-1 =1
so,here degree of freedom will be 1
(c) F= -mgθ
here, θ=ls
For the position vector of mass
(d)
r= R +l(sinθi^+cosθj^)
and the velocity is represented as
v= R˙+lθ˙(sinθi^+cosθj^)
(e)
KE=21mv2=21m(∣R˙∣2+l2θ˙2+2lθ˙(X˙cosθ−Y˙sinθ)
(f) and for potential energy
PE= − (m+M)gcosθ−(m+M)gY
(g)
L= KE-PE
L= 21m(∣R˙∣2+l2θ˙2+2lθ˙(X˙cosθ−Y˙sinθ)+(m+M)gcosθ+(m+M)gY +
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