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Question: A wagon of mass $M$ can move without friction along horizontal rails. A simple pendulum consisting o...

A wagon of mass MM can move without friction along horizontal rails. A simple pendulum consisting of a sphere of mass mm suspended from the ceiling of the wagon by a string of length ll. At the initial moment the wagon and the pendulum are at rest and the string is deflected through an angle α\alpha from the vertical. Find the velocity of the wagon when the pendulum passes through its mean position.

A

vm=msin⁡α22glM(M+m)v_m = m \sin \frac{\alpha}{2} \sqrt{\frac{2gl}{M(M+m)}}

B

vm=2msin⁡α2glM(M+m)v_m = 2m \sin \frac{\alpha}{2} \sqrt{\frac{gl}{M(M+m)}}

C

vm=2msin⁡α22glM(M+m)v_m = 2m \sin \frac{\alpha}{2} \sqrt{\frac{2gl}{M(M+m)}}

D

vm=msin⁡αglM(M−m)v_m = m \sin \alpha \sqrt{\frac{gl}{M(M-m)}}

Answer

B

Explanation

Solution

Concepts: Conservation of horizontal momentum and conservation of mechanical energy.

Solution: Let MM be the mass of the wagon, mm be the mass of the pendulum bob, and ll be the length of the string.
Let vMv_M be the velocity of the wagon when the pendulum passes through its mean position.
Let vrelv_{rel} be the velocity of the pendulum bob relative to the wagon at its mean position. The bob moves horizontally relative to the wagon.
The absolute velocity of the bob relative to the ground will be vb=vM−vrelv_b = v_M - v_{rel} (assuming the bob swings opposite to the wagon's motion, which is the case for the bob to gain maximum speed relative to the wagon).

  1. Conservation of horizontal momentum:
    Initially, the system (wagon + bob) is at rest, so initial momentum is 0.
    Finally, the total horizontal momentum is MvM+m(vM−vrel)Mv_M + m(v_M - v_{rel}).
    0=MvM+m(vM−vrel)0 = Mv_M + m(v_M - v_{rel})
    (M+m)vM=mvrel(M+m)v_M = mv_{rel}
    vrel=(M+m)mvMv_{rel} = \frac{(M+m)}{m}v_M (Equation 1)

  2. Conservation of mechanical energy:
    Initial potential energy of the bob (taking the mean position as zero potential energy) is mgl(1−cos⁡α)mgl(1-\cos\alpha).
    Initial kinetic energy of the system is 0.
    Final potential energy of the bob is 0.
    Final kinetic energy of the system is 12MvM2+12m(vM−vrel)2\frac{1}{2}Mv_M^2 + \frac{1}{2}m(v_M - v_{rel})^2.
    mgl(1−cos⁡α)=12MvM2+12m(vM−vrel)2mgl(1-\cos\alpha) = \frac{1}{2}Mv_M^2 + \frac{1}{2}m(v_M - v_{rel})^2
    Substitute vrelv_{rel} from Equation 1 into the energy equation:
    mgl(1−cos⁡α)=12MvM2+12m(vM−(M+m)mvM)2mgl(1-\cos\alpha) = \frac{1}{2}Mv_M^2 + \frac{1}{2}m\left(v_M - \frac{(M+m)}{m}v_M\right)^2
    mgl(1−cos⁡α)=12MvM2+12m(mvM−MvM−mvMm)2mgl(1-\cos\alpha) = \frac{1}{2}Mv_M^2 + \frac{1}{2}m\left(\frac{mv_M - Mv_M - mv_M}{m}\right)^2
    mgl(1−cos⁡α)=12MvM2+12m(−MvMm)2mgl(1-\cos\alpha) = \frac{1}{2}Mv_M^2 + \frac{1}{2}m\left(\frac{-Mv_M}{m}\right)^2
    mgl(1−cos⁡α)=12MvM2+12mM2vM2m2mgl(1-\cos\alpha) = \frac{1}{2}Mv_M^2 + \frac{1}{2}m\frac{M^2v_M^2}{m^2}
    mgl(1−cos⁡α)=12MvM2+12M2vM2mmgl(1-\cos\alpha) = \frac{1}{2}Mv_M^2 + \frac{1}{2}\frac{M^2v_M^2}{m}
    mgl(1−cos⁡α)=12vM2(M+M2m)mgl(1-\cos\alpha) = \frac{1}{2}v_M^2 \left(M + \frac{M^2}{m}\right)
    mgl(1−cos⁡α)=12vM2(Mm+M2m)mgl(1-\cos\alpha) = \frac{1}{2}v_M^2 \left(\frac{Mm + M^2}{m}\right)
    mgl(1−cos⁡α)=12vM2M(m+M)mmgl(1-\cos\alpha) = \frac{1}{2}v_M^2 \frac{M(m+M)}{m}
    vM2=2m2gl(1−cos⁡α)M(M+m)v_M^2 = \frac{2m^2gl(1-\cos\alpha)}{M(M+m)}
    Using the trigonometric identity 1−cos⁡α=2sin⁡2(α/2)1-\cos\alpha = 2\sin^2(\alpha/2):
    vM2=2m2gl(2sin⁡2(α/2))M(M+m)v_M^2 = \frac{2m^2gl(2\sin^2(\alpha/2))}{M(M+m)}
    vM2=4m2glsin⁡2(α/2)M(M+m)v_M^2 = \frac{4m^2gl\sin^2(\alpha/2)}{M(M+m)}
    vM=4m2glsin⁡2(α/2)M(M+m)v_M = \sqrt{\frac{4m^2gl\sin^2(\alpha/2)}{M(M+m)}}
    vM=2msin⁡(α/2)glM(M+m)v_M = 2m \sin(\alpha/2) \sqrt{\frac{gl}{M(M+m)}}